Proof. Observe that \(A(u)\) and \(u-A(u)\) are nondecreasing functions in \(u\), since \(A'(\cdot) \in [0,1]\). Fix functions \(q, \tilde{q} \in L^{\infty}(\mathbb{Z}^d)\), and suppose \(k\in \mathbb{Z}^{d}\) is such that \(q(\ell)\leq q'(\ell)\) for all \(|\ell-k|_{1}\leq 1\). Then by the previous observation, \[\begin{align} \mathcal{S}q(k) &= q(k) + {1 \over 2d} \sum_{\ell\sim k} \left[A(q(\ell)) - A(q(k))\right] \\&= q(k) - A(q(k)) + {1 \over 2d} \sum_{\ell\sim k} A(q(\ell)) \\&\le q'(k) - A(q'(k))+ {1 \over 2d} \sum_{\ell\sim k} A(q'(\ell)). \\&=\mathcal{S}q'(k). \end{align}\] Since constants are solutions of the scheme \(\mathcal{S}\), it follows that \(0 \le \mathcal{S}q \le \|q\|_{L^\infty(\mathbb{Z}^{d})}\), which implies that \(\mathcal{S}q\in L^{\infty}(\mathbb{Z}^{d})\). Thus, \(\mathcal{S}\) is indeed locally monotone (and consequently monotone). ◻

Proof. To prove (i), notice that since \(A'(u) \in [0,1]\) for all \(u \in [0,\Lambda]\), it follows that \(0 \leq A(q) \leq q\), where \(q\in L^{1}(\mathbb{Z}^{d})\). Hence, we can rearrange the sum, \[\begin{align} \sum_{k\in \mathbb{Z}^{d}} (\mathcal{S}q(k) - q(k)) &= {1 \over 2d} \sum_{i=1}^d \sum_{\zeta \in \{-1, 1\}} \left[\sum_{k\in \mathbb{Z}^{d}} A(q(k+ \zeta e_i)) - \sum_{k\in \mathbb{Z}^{d}} A(q(k))\right] \\&= {1 \over 2d} \sum_{i=1}^d \sum_{\zeta \in \{-1, 1\}} \left[\sum_{\ell\in \mathbb{Z}^{d}} A(q(\ell)) - \sum_{k\in \mathbb{Z}^{d}} A(q(k))\right] \\&= 0, \end{align}\] which is precisely the statement of (i). The inequality in (ii) is a consequence of [16]. ◻

Proof. By Corollary 4, we simply need to check that \(A'(u)\in [0,1]\) whenever \(u\in [0, \frac{1}{2}]\). By direct computation, \(A'(u) = (m+1)u^m\), and since \(m\geq 1\), it follows that \(A'(0)\leq A'(u)\leq A'(\frac{1}{2})\). In the case when \(m=1\), it is clear that \(0\leq A'(u)\leq 1\), which implies the claim. Moreover, it can be seen that \(A'(\frac{1}{2})\) is decreasing in \(m\ge 1\), since \[{d \over dm} (m+1)\frac{1}{2^{m}} = (1 - (m+1)\log 2)\frac{1}{2^{m}},\] which is strictly negative when \(m>1\). The second two claims follow by Lemma 5. ◻

Proof. To show (i), we consider \(\tilde{A}(u):=\frac{\Delta_t}{(\Delta_x)^{2}}A(u)\). Then 15 can be rewritten as \[\mathcal{S}_{N}v(k\Delta_x)=v(k\Delta_x)+\frac{1}{2d}\sum_{k\sim \ell}\left[\tilde{A}(v(\ell \Delta_x))-\tilde{A}(v(k\Delta_x))\right].\] Using the identification with \(\mathcal{S}_{N}^{\Delta_x}: L^{\infty}(\mathbb{Z}^{d})\rightarrow L^{\infty}(\mathbb{Z}^{d})\), we now appeal to Lemma 3. For this, it is enough to check that \(\tilde{A}'(u)\in [0,1]\) for all \(u\in [0, \Lambda]\). Noting that \(\tilde{A}'(u)=\frac{\Delta_t}{(\Delta_x)^{2}}(m+1)u^{m}\), and recalling that \(\frac{\Delta_t}{(\Delta_x)^2} = \frac{N^{-1}}{N^{-2/(dm+2)}} = N^{-dm\beta}\), this implies that \[\tilde{A}'(u)=N^{-dm\beta}(m+1)u^{m}.\] For \(u\in [0, \Lambda]\), it follows by hypothesis ?? that \(\tilde{A}'(u)\in [0,1]\). Hence, an application of Lemma 3 yields (i).

To see (ii), note that the second conclusion of Lemma 3 gives that \(0 \leq \mathcal{S}_{N}w\leq \|w\|_{L^{\infty}(\mathbb{R}^{d})} \leq \Lambda\), meaning \(\mathcal{S}_{N}w \in \mathcal{B}^{+}_{\Lambda}(\mathbb{R}^{d})\cap L^{\infty}(\Delta_x\mathbb{Z}^{d})\). By iteration, we have (ii).

To prove property (iii), we remark that for a \(\Delta_x\mathbb{Z}^{d}\)-piecewise constant function \(w\), \(\int_{\mathbb{R}^{d}} |w(x)|\, dx=\sum_{k\in \mathbb{Z}^{d}} (\Delta_x)^{d}|w(k\Delta_x)|=(\Delta_x)^{d}|w(\cdot \Delta_x)|_{1}\). Therefore, the case \(n=1\) follows directly from Lemma 5. By setting \(w'_{N} \equiv 0\), we have \(\|\mathcal{S}_{N}w_{N}\|_{L^{1}(\mathbb{R}^{d})} \leq \|w_{N}\|_{L^{1}(\mathbb{R}^{d})}\). This fact, along with property (ii) says that \(\mathcal{S}_{N}w_{N} \in L^{1}(\mathbb{R}^{d}) \cap \mathcal{B}^{+}_{\Lambda}(\mathbb{R}^{d})\cap L^{\infty}(\Delta_x\mathbb{Z}^{d})\) whenever \(w_{N} \in L^{1}(\mathbb{R}^{d}) \cap \mathcal{B}^{+}_{\Lambda}(\mathbb{R}^{d})\cap L^{\infty}(\Delta_x\mathbb{Z}^{d})\). Hence, we can conclude property (iii) for all \(n\) by iteratively applying Lemma 2.4. ◻

Proof. Let \(q\) be a volcanic function. We now check that \(\mathcal{S}(q)\) is volcanic, under the hypotheses.

By the definition of \(\mathcal{S}\), if we compose \(q\) with some reflection or coordinate transposition, then \(\mathcal{S}q\) will undergo the same reflection or transposition, so \(\mathcal{S}q\) is symmetric.

Let \(k, \ell\) be neighbours with \(|k|_{1} > |\ell|_{1}\). The proof is complete if we can show \(\mathcal{S}q(k) \le \mathcal{S}q(\ell)\). By symmetry, we can assume without loss of generality that \(k\) and \(\ell\) differ in the first coordinate, and \(\ell_1 \ge 0\), so that the two points are \[\ell= (\ell_1, \ell_2, \ldots, \ell_d), \qquad k= (\ell_1 + 1, \ell_2, \ldots, \ell_d).\] We consider two cases.

Case 1. \(\ell_1 \ge 1\). We use the local monotonicity property of \(\mathcal{S}\). In this case, every coordinate \(\ell_i\) has the same sign as the corresponding coordinate \(k_i\). Let \(v\sim 0\) be a neighbour of the origin, and we will compare \(k+v\) and \(\ell+v\). If \(v = \pm e_i\) with \(i\neq 1\), then \(|\ell_{1}|<|k_{1}|\) and \(|\ell_{i}\pm1|=|k_{i}\pm1|\). If \(v=\pm e_{1}\), then since \(\ell_{1}\geq 1\), we still have \(|\ell_{1}\pm1|<|k_{1}\pm1|\), and \(|\ell_{i}|=|k_{i}|\).

Thus, \(|\ell+v|_{1}<|k+v|_{1}\), and they are clearly neighbors, so since \(q\) is volcanic, \(q(k+v)\leq q(\ell+v)=q(k+e_{1}+v)\). By the local monotonicity of \(\mathcal{S}\), this implies that \(\mathcal{S}q(k)\leq \mathcal{S}q(k+e_{1})=\mathcal{S}q(\ell)\), as desired.

Case 2. \(\ell_1 = 0\). In this case, the analysis is more delicate. Using the symmetry of \(q\), without loss of generality, we may write \(\ell= (0, \ldots, 0, \ell_{m+1}, \ldots, \ell_d)\), where \(m\geq 1\) denotes the number of zero coordinates of \(\ell\).

Since \(q\) is volcanic and \(k\) and \(\ell\) are neighbours with \(|k|_{1}> |\ell|_{1}\), \(q(k)\leq q(\ell)\). To prove the desired claim, we compute \[\begin{align} &\mathcal{S}q(\ell) - \mathcal{S}q(k)\\ &=q(\ell)-q(k)+{1\over2d}\sum_{i=1}^{m}[A(q(\ell\pm e_{i}))-A(q(\ell))]-{1\over2d}\sum_{i=1}^{m}[A(q(k\pm e_{i}))-A(q(k))]\\ &\quad +{1\over2d}\sum_{i=m+1}^{d}[A(q(\ell\pm e_{i}))-A(q(\ell))]-[A(q(k\pm e_{i}))-A(q(k))] \end{align}\]

We will analyze each of the above sums separately. By symmetry, we have that for all \(1\leq i\leq m\), \(q(\ell\pm e_{i}) = q(\ell+e_{1})=q(k)\), so \[\label{monotonezeropiecea} R_1 := {1\over2d}\sum_{i=1}^{m}[A(q(\ell\pm e_{i}))-A(q(\ell))] = {2m \over 2d} [A(q(k)) - A(q(\ell))].\tag{20}\]

For the next term, we note that for \(1< i\leq m\), \(|k\pm e_{i}|_{1}>|k|_{1}\) and \(|k+e_{1}|_{1}>|k|_{1}\). Since \(q\) is volcanic and \(A'\geq 0\), we have \(A(q(k+v))-A(q(k))\leq 0\) for all \(v\sim 0\) with \(v\neq -e_{1}\). This implies \[\begin{align} R_2 :=-{1\over2d}\sum_{i=1}^{m}[A(q(k\pm e_{i}))-A(q(k))]&\geq -\frac{1}{2d}[A(q(k-e_{1})-A(q(k))]\\ &=\frac{1}{2d}[A(q(k))-A(q(\ell))]. \end{align}\]

For the last term, we observe that for \(m+1\leq i\leq d\), \(|k\pm e_{i}|_{1}>|\ell\pm e_{i}|_{1}\), and thus by the same argument as in the last step, \(A(q(\ell\pm e_{i})) - A(q(k\pm e_{i})) \ge 0.\) This implies \[\begin{align} \label{monotonezeropiecec} R_3 &:={1\over2d}\sum_{i=m+1}^{d}[A(q(\ell\pm e_{i}))-A(q(\ell))]-[A(q(k\pm e_{i}))-A(q(k))]\notag\\ &\geq {1\over2d}\sum_{i=m+1}^{d}[A(q(k))-A(q(\ell))]={2d - 2m \over 2d} [A(q(k))-A(q(\ell))] \end{align}\tag{21}\] Adding up the three prior inequalities, we get the bound \[\begin{align} \mathcal{S}q(\ell)-\mathcal{S}q(k) &= q(\ell)-q(k) + R_{1}+R_{2}+R_{3} \\&\ge q(\ell)-q(k) + {2d+1 \over 2d} [A(q(k))-A(q(\ell))]. \end{align}\] Recall that \(q(k) \le q(\ell)\), and since \(0\leq A'(\cdot) \le 2/3\), we have \[0\leq A(q(\ell)) - A(q(k)) \le \frac{2}{3} (q(\ell) - q(k)).\] Therefore, since \(d>1\), we conclude \[\begin{align} \mathcal{S}q(\ell) - \mathcal{S}q(k) &\ge q(\ell)- q(k) - {(2d + 1) \over 3d}[q(\ell)-q(k)] \\ &\ge [q(\ell)- q(k)]\left[1 - {2d+1 \over 3d}\right] \\&\ge 0, \end{align}\] and this completes the proof. ◻

Proof of Lemma 11. To prove the unimodality claim, we recall that for fixed \(x\), \(\bar{u}(x,\cdot)\) could be initially 0, and once there exists \(T\) such that \(|x|<r(T)\), then \(\bar{u}(x,t)\) is strictly positive for all times \(t\geq T\) (so the time derivative must be positive at time \(T\)). We now compute \(\tfrac{\partial\overline{u}}{\partial t}(x,t)\) assuming that \(|x|\leq r(t)\) (since this is where the maximum of \(t\mapsto u(x,t)\) will occur). We introduce the shorthand \(\vartheta := \tfrac{|x|}{r(t)}\) and proceed by logarithmic differentiation. Since \[\log \bar{u}= \log(R^d \Gamma) - d \log r(t) + {1 \over m} \log(1 - \vartheta^2),\] using the fact that \(\tfrac{\partial \vartheta}{\partial t} = -\tfrac{|x|r'(t)}{r(t)^2} = -\tfrac{\vartheta r'(t)}{r(t)}=- \tfrac{\vartheta\beta}{(t + {T_0})}\), and \(md\beta+2\beta=1\), we obtain \[\begin{align} {1 \over \bar{u}} {\partial \bar{u} \over \partial t} &= {-d \beta \over t+T_0} - {1 \over m(1-\vartheta^2)} {\partial\vartheta^2 \over \partial t} \notag \\&={1 \over m(t+T_0)} \left(-md\beta + 2\beta {\vartheta^2 \over 1 - \vartheta^2}\right) \notag \\ &={1 \over m(t+T_0)} \left({1-md\beta \over 1 - \vartheta^2} - 1\right)\notag\\ &={1 \over m(t+T_0)} \left({2\beta \over 1 - \vartheta^2} - 1\right). \label{first46time46derivative} \end{align}\tag{24}\] We conclude that \(\tfrac{\partial\overline{u}}{\partial t}\) has the same sign as \(2\beta(1 - \vartheta^2)^{-1} - 1\), and this is decreasing in time for a fixed \(x\), because \(\vartheta = |x|/r(t)\) is decreasing in time (and less than 1). Therefore, for fixed \(x\), the time derivative of \(\bar{u}(x,t)\) is initially positive and then becomes negative for higher values of \(t\). This implies that \(\bar{u}(x,t)\) is unimodal in \(t\) for a fixed \(x\).

For the second claim, fix \(i\in [d]\), and we begin by rewriting \[{\partial^4 \over \partial x_i^4} \bar{u}^{m+1} = \left({R^d \Gamma \over r(t)^d}\right)^{m+1} {\partial^4 \over \partial x_i^4} \left(s-\left({x_i \over r(t)}\right)^2\right)^{(m+1)/m}_+,\] where \(s := 1 - (\sum_{j \ne i} x_j^2) / r(t)^2\) does not depend on the \(i\)-th coordinate, and thus \(s - x_i^2 / r(t)^2 = 1 - |x|^2/r(t)^2\) is greater than zero by the assumption that \(|x| < r(t)\).

For \(y\in \mathbb{R}\), we make the change of variables \(x_i = s^{1/2} r(t) y\). Then \[{\partial^4 \over \partial x_i^4} \bar{u}^{m+1} = C(R, \Gamma, t, x_1, \ldots, x_{i-1}, x_{i+1}, \ldots, x_d) {d^4 \over dy^4} (1-y^2)^{(m+1)/m}\] where \(C({\cdots})\) is a positive constant that depends on everything except \(x_i\). In other words, the sign of \(\partial_i^4 \bar{u}^{m+1}\) will be the same as the sign of \((d/dy)^4 (1-y^2)^{(m+1)/m}\). We also notice that \(s > x_i^2/r(t)^2\), so \(|y| < 1\).

We now calculate¹ the fourth derivative of \((1-y^2)^{(m+1)/m}\), given by \[{d^4 \over dy^4} (1-y^2)^{(m+1)/m} = {4(1+m) (1-y^2)^{1/m-3} \over m^4} (3m^2+(6m^2-12m)y^2+(4-m^2)y^4).\] Let \(\psi(y) := 3m^2 + (6m^2-12m)y^{2} + (4-m^2) y^4\) be the factor on the right. Then since \(m\geq 1\), it follows that \(\psi(y)\geq 0\) for all \(|y|\leq 1\).

The other factors are positive, so \((d/dy)^4 (1-y^2)^{(m+1)/m}\) is nonnegative, and \[{\partial^4 \over \partial x_i^4} \bar{u}^{m+1} = C {d^4 \over d^4y} (1-y^2)^{(m+1)/m} \ge 0.\] This inequality proves the claim about the fourth derivative. ◻

Proof. The first and third conditions are obviously satisfied. For the second condition, we must check that the probabilities \({\mathbf{P}}\left\{Z_n = k\right\}\) are volcanic when \(r \equiv b\).

By assumption, the initial condition \(h\) is volcanic. We observe that the sequence of functions \(q=(q_n(\cdot))_{n_{0}\leq n\leq n_{1}}\) given by \(q_{n}(k) := {\mathbf{P}}\left\{Z_n = k\right\}\) is a solution of 13 on \([n_{0}, n_{1}]\), with \(A(u) = bu\) and initial condition \(h\). Then since \(A'(\cdot) = b \in [0, 2/3]\), Lemma 9 tells us that volcanicity is preserved. Hence, \(q_n(k) = {\mathbf{P}}\left\{Z_n = k\right\}\) is volcanic for \(n_0 \le n < n_1\), and the second condition is also satisfied. ◻

Proof. Let \((Z_n)_{n_{0}\leq n\leq n_{1}}\) be a lazy random walk with jump probability \(r(k, n) = \tilde{p}(k, n)^m\) and initial distribution \(h\), as defined above. We first observe that \(\tilde{p}(k,n) = \sigma {\mathbf{P}}\left\{Z_n = k\right\}\), because the function \(f(k, n) := \sigma {\mathbf{P}}\left\{Z_n = k\right\}\) satisfies the relation (25 ) with initial condition \(f(k, n_{0}) = \sigma h(k) = \tilde{p}(k, n_{0})\). Since \(\tilde{p}\) also satisfies 25 with the same initial condition, we conclude that the two functions are equal: \(\sigma {\mathbf{P}}\left\{Z_n = k\right\} = \tilde{p}(k, n)\) for \(k\in \mathbb{Z}^d\) and \(n \in \{n_0, \ldots, n_1\}\).

We now argue that \(r(k, n) = \tilde{p}(k, n)^m\) satisfies 26 . Since \(\tilde{p}\) is volcanic for each \(n\in [n_{0}, n_{1}]\), the jump probability \(r(k, n) = \tilde{p}(k, n)^m\) and the distributions \({\mathbf{P}}\left\{Z_n = k\right\}\) are also volcanic for each \(n\in [n_{0}, n_{1}]\), so the first two conditions hold. We have assumed that \(\tilde{p}(k, n) \ge b^{1/m}\) when \(|k| \le \rho\), which implies that \(r(k, n) = \tilde{p}(k, n)^m \ge b\), so the third condition holds and \(r\) satisfies the three conditions 26 .

This tells us that \({\mathbf{P}}\left\{Z_{n_1} = 0\right\}\) is bounded above by \(V_{n_0 \to n_1}(h)\) by definition. Therefore \[\widetilde{p}(0, n_1) = \sigma {\mathbf{P}}\left\{Z_{n_1} = 0\right\} \le \sigma V_{n_0 \to n_1}(h).\] Our assumption that \(\tilde{p}(\cdot, n_{1})\) is volcanic implies in particular that \(\sup_k\tilde{p}(k, n_1) = \tilde{p}(0, n_1)\), so \(\tilde{p}(k,n_1) \le \tilde{p}(0,n_1) \le \sigma V_{n_0 \to n_1}(h)\) for all points \(k\in \mathbb{Z}^d\). ◻

Proof of (i). By symmetry and the definition of the discrete Laplacian, \[\begin{align} \sum_{k\in \mathbb{Z}^{d}} \Delta f(k) g(k) &= \sum_{k\in \mathbb{Z}^{d}} \sum_{\ell\sim k} f(\ell) g(k) - 2d\sum_{k\in \mathbb{Z}^{d}}f(k) g(k) \\&= \sum_{k\in \mathbb{Z}^{d}} \sum_{\ell\sim k} f(k) g(\ell) - 2d\sum_{k\in \mathbb{Z}^{d}} f(k) g(k) \\&=\sum_{k\in \mathbb{Z}^{d}} f(k) \Delta g(k) \end{align}\] The infinite sums are all well-defined by the assumption of bounded support. ◻

Proof of (ii). We check:\[\begin{align} \sum_e \nabla f(e) \nabla g(e) &=\sum_{k\in \mathbb{Z}^{d}} \sum_{\ell\sim k} (f(\ell) - f(k)) (g(\ell) - g(k)) \\&= 4d \sum_{k\in \mathbb{Z}^{d}} f(k) g(k) - 2\sum_{k\in \mathbb{Z}^{d}} \sum_{\ell\sim k} f(k) g(\ell) \\&= - 2\sum_{k\in \mathbb{Z}^{d}} f(k) \Delta g(k), \end{align}\] where in the second line we switch \(\ell\) and \(k\) in two of the sums. ◻

Proof of (iii). We expand both sides: \[\begin{align} \nabla(fg)(e) &= f(\ell)g(\ell) - f(k) g(k)\\ f(\ell) \nabla g(e) + g(k) \nabla f(e) &= f(\ell) g(\ell) - f(\ell) g(k) + g(k) f(\ell) - g(k) f(k) \end{align}\] and \(-f(\ell) g(k) + g(k) f(\ell) = 0\), so the two lines are the same. ◻

Proof of Theorem 16. For ease of notation, we write \(q_{n}(k) = q^{(b)}_n(k)\) and \(r_n(k) = r(k, n)\).

We claim that for all \(n\in \left\{n_{0}, n_{0}+1, \ldots, n_{1}\right\}\), \[\label{optiminduction} V_{n \to n_1}(h) = \sum_{k\in \mathbb{Z}^{d}} h(k) q_{n_1-n}(k).\tag{28}\] We will prove this by induction downward in \(n\), starting from the base case \(n = n_1\) where \(V_{n_1 \to n_1}(h) = h(0)\) and the statement is obvious.

Suppose \(n\in [n_{0}, n_1)\). We know by the induction assumption that \[V_{(n+1) \to n_1}(h) = \sum_{k\in \mathbb{Z}^{d}} h(k) q_{n_1 - (n+1)}(k).\]

We will regard optimization of the jump probabilities \(r\) as an optimal control problem: instead of choosing \(r\) for all values of \(k\) and \(n\), we will choose \(r(k, n)\) at the current step \(n\) and then we will choose \(r(k, n')\) for \(n < n' \le n_1 - 1\). By the Bellman optimality principle, we can find \(V_{n \to n_1}(h)\) by choosing the jump probability \(r_n\) that gives us the best value at step \(n+1\).

If the lazy random walk \(Z_n\) has \({\mathbf{P}}\left\{Z_n = k\right\} = h(k)\), then the distribution of \(Z_{n+1}\) is \[\begin{align} {\mathbf{P}}\left\{Z_{n+1} = k\right\} = h(k) + {1 \over 2d} \sum_{\ell\sim k} \left[r(\ell, n) h(\ell) - r(k, n) h(k)\right]= h(k) + \frac{1}{2d} \Delta(r_nh)(k). \end{align}\] We can now ask about the value of the optimal control problems at time step \(n + 1\), starting from these new distributions. We consider the set of jump probabilities \(r_n: \mathbb{Z}^d \to [0, 1]\) that satisfy the following three conditions: \[\label{e46rcondstwo} \text{(i) r_n is volcanic; } \quad \text{(ii) h +\tfrac{1}{2d} \Delta(r_n h) is volcanic; } \quad \text{(iii) r_n(k) \ge b for |k| \le \rho}.\tag{29}\] We highlight that 29 is a condition for only \(r_{n}\), whereas 26 is a condition for \(r=(r_{n})_{n_{0}\leq n\leq n_{1}-1}\), as well as on \({\mathbf{P}}\left\{Z_{n}=\cdot\right\}\) for \(n\in \left\{n_{0}, \ldots, n_{1}-1\right\}\).

Let \[v := \sup_{r_n} V_{(n+1) \to n_1}\left(h + {1 \over 2d} \Delta(r_n h)\right) = \sup_{r_n} \sum_{k\in \mathbb{Z}^{d}} \left(h(k) + {1 \over 2d}\Delta(r_n h ) \right) q_{n_1 - (n+1)}(k),\] where we take the supremum over jump probabilities \(r_n\) that satisfy 29 , and the second equality holds by the induction hypothesis.

We claim that \(V_{n \to n_1}(h) = v\). To prove this, we show that \(V_{n \to n_1}(h) \le v\) and that \(v \le V_{n \to n_1}(h)\) separately. First, if \(r\) is a jump probability that satisfies the conditions 26 with initial time \(n\), then \(r\) satisfies 26 with initial time \(n+1\) and starting distribution \(h(k) = {\mathbf{P}}\left\{Z_{n+1} = k\right\} = h + \tfrac{1}{2d} \Delta(r_n h)\). Thus,\[{\mathbf{P}}\left\{Z_{n_1} = 0\right\} \le V_{(n+1) \to n_1}\left(h + {1 \over 2d} \Delta(r_n h)\right) \le v,\]because \(r_n\) satisfies the conditions 29 . Therefore \(V_{n \to n_1}(h) \le v\).

Conversely, if \((r_{n'})_{n+1\leq n'\leq n_{1}}\) satisfies 26 , and \(r_{n}\) satisfies 29 , then we claim that \((r_{n'})_{n\leq n'\leq n_{1}}\) satisfies 26 . Indeed, \({\mathbf{P}}\left\{Z_{n}=\cdot\right\}=h\) is volcanic, and \({\mathbf{P}}\left\{Z_{n+1}=\cdot\right\}=h+\frac{1}{2d}\Delta h\) is volcanic by property (ii) of 29 . Hence, we have that \[V_{(n+1) \to n_1}\left(h + {1 \over 2d} \Delta(r_n h)\right)\leq V_{n \to n_1}\left(h\right),\] and taking a supremum over \(r_{n}\) satisfying 29 , we obtain that \(v\leq V_{n \to n_1}\left(h\right)\).

We next prove that \(v = \sum_{k\in \mathbb{Z}^{d}} h(k) q_{n_1 -n}(k)\), which will give us the conclusion. By Lemma 13, constant jump probability \(r_n \equiv b\) satisfies all three conditions 29 , so \[\begin{align} \notag \sum_{k\in \mathbb{Z}^d} h(k) q_{n_1 - n}(k) &= \sum_{k\in \mathbb{Z}^{d}} h(k) \left( q_{n_1 - (n+1)}(k) + {b \over 2d} \Delta q_{n_1 - (n+1)}(k) \right) \\\label{e46achieved} &= \sum_{k\in \mathbb{Z}^{d}} \left(h(k) + {1 \over 2d} \Delta(bh)\right) q_{n_1 - (n+1)}(k), \end{align}\tag{30}\] where in the first line, we use the recurrence \(q_{n+1} = q_n + (b/2d) \Delta q_n\), and in the second line, we use part (i) of Lemma 15. Therefore, \(\sum_{k\in \mathbb{Z}^{d}} h(k) q_{n_1 - n}(k) \le v\).

We now prove that \(\sum_{k\in \mathbb{Z}^{d}} h(k) q_{n_1 - n}(k) \ge v\), which is more difficult and requires the hypothesis that \(n_1 - n_0 \le T(\rho, b)\). We will first show that it is enough to consider functions \(r_n\) with \(r_n \ge b\) pointwise everywhere, and then show that \(r_n \equiv b\) is an optimal choice.

Suppose \(r_{n}:\mathbb{Z}^d \to [0, 1]\) is a jump probability that satisfies conditions 29 (i) and 29 (iii). Our inductive hypothesis tells us that \[\label{e46valueinductivehypothesis}V_{(n+1) \to n_1}\left(h + {1 \over 2d}\Delta(r_nh)\right) = \sum_{k\in \mathbb{Z}^{d}} h(k) q_{n_1 - (n +1)}(k) + {1 \over 2d} \sum_{k\in \mathbb{Z}^{d}} \Delta(r_nh)(k) q_{n_1 - (n + 1)}(k).\tag{31}\] We will focus on analyzing the second term in the sum. We use the first part of Lemma 15(i) to move the discrete Laplacian to the other factor: \[\label{value46one}\frac{1}{2d}\sum_{k\in \mathbb{Z}^{d}} \Delta(r_nh)(k) q_{n_1 - (n + 1)}(k) = \frac{1}{2d}\sum_{k\in \mathbb{Z}^{d}} r_n(k) h(k) \Delta q_{n_1 - (n + 1)}(k).\tag{32}\]

Consider the modified jump probability \(r'(k) := r_n(k) \vee b\). Then we claim that for any \(k\in \mathbb{Z}^d\), \[r_n(k) h(k) \Delta q_{n_1 - (n + 1)}(k) \le r'(k) h(k) \Delta q_{n_1 - (n + 1)}(k).\] To see this, note that the two sides are equal for \(|k| \le \rho\), because \(r_n\) satisfies the third condition of 29 . On the other hand, if \(|k| > \rho\), then the discrete Laplacian \(\Delta q_{n_1 - n - 1}(k)\) is nonnegative by definition of \(T(\rho, b)\), and \(h \ge 0\) and \(r' \ge r_{n}\), so the relation also holds. Summing this up over all \(k\), we have \[\sum_{k\in \mathbb{Z}^{d}} r_n(k) h(k) \Delta q_{n_1 - (n + 1)}(k)\le \sum_{k\in \mathbb{Z}^{d}} r'(k) h(k) \Delta q_{n_1 - (n + 1)}(k).\] We now perform integration by parts on the term on the right, using Lemma 15(ii) and (iii), \[\begin{align} &\sum_{k\in \mathbb{Z}^{d}} r'(k) h(k) \Delta q_{n_1-(n+1)}(k) \\&\qquad\qquad = - \frac{1}{2}\sum_{e=(k,\ell)} \nabla(r'h)(e) \nabla q_{n_1-(n+1)}(e) \\&\qquad\qquad = - \frac{1}{2}\left(\sum_{e=(k,\ell)} r'(\ell) \nabla h(e) \nabla q_{n_1-(n+1)}(e) + \sum_{e=(k,\ell)} h(k) \nabla r'(e) \nabla q_{n_1-(n+1)}(e)\right), \end{align}\]

The functions \(r', h, q_{n_1-(n+1)}\) are volcanic, the first two by assumption and the last one by Lemma 13. This determines the sign of the discrete gradient across any edge. For example, \(r'(\ell) - r'(k)\) has the same sign as \(h(\ell) - h(k)\). Therefore, the products of two discrete gradients in the above sums are nonnegative. This means that the second sum is nonnegative, and \(r' \ge b\), so the whole expression in brackets is at least \(\sum_e b \nabla h(e) \nabla q_{n_1 - (n+1)}(e)\). Because there is a minus sign in front, we obtain the upper bound, \[\begin{align} \sum_{k\in \mathbb{Z}^d} r'(k) h(k) \Delta q_{n_1-(n+1)}(k) &\le -\frac{1}{2} \sum_{e=(k,\ell)} b \nabla h(e) \nabla q_{n_1-(n+1)}(e) \\&= b \sum_{k\in \mathbb{Z}^d} h(k) \Delta q_{n_1-(n+1)}(k), \end{align}\] where in the last step, we performed another integration by parts using Lemma 15(ii). We now put all this together. By 31 and the inequality we have just proved, \[\begin{align} V_{(n+1) \to n_1}\left(h + {1 \over 2d} \Delta(r_n h)\right) &\le \sum_{k\in \mathbb{Z}^{d}} h(k) q_{n_1-(n+1)}(k) + \frac{1}{2d}\sum_{k\in \mathbb{Z}^{d}} r'(k) h(k)\Delta q_{n_1-(n+1)}(k) \\&\le \sum_{k\in \mathbb{Z}^{d}} h(k) q_{{n_1}-(n+1)}(k) + \frac{1}{2d} \sum_{k\in \mathbb{Z}^{d}} b h(k) \Delta q_{{n_1}-(n+1)}(k) \\&=\sum_{k\in \mathbb{Z}^d} h(k) \left(q_{n_1 - (n+1)}(k) + {b \over 2d} \Delta q_{n_1 - (n+1)}(k)\right) \\&=\sum_{k\in\mathbb{Z}^d} h(k) q_{n_1 - n}(k). \end{align}\] We have proven the above bound for any jump probability \(r_n\) which satisfies 29 (i) and 29 (iii), so it certainly also holds for jump probabilities \(r_n\) which satisfy all three conditions of 29 . Taking the supremum over all such \(r_n\), we get \(v \le \sum_{k\in \mathbb{Z}^d} h(k) q_{n_1 - n}(k)\).

Finally, we conclude that \[V_{n \to n_1}(h) = v = \sum_kh(k) q_{n_1 - n}(k).\] This verifies the induction step, so the formula is true for all \(n \in \{n_0, \ldots, n_1\}\), and in particular it is true for \(n_0\), which gives us the result \(V_{n_0 \to n_1}(h) = \sum_{k\in \mathbb{Z}^{d}} h(k) q_{n_1 - n_0}(k)\). ◻

Proof. Let \(q_{n}^{(b)}\) be defined as in the text above 27 . The discrete Laplacian is \[\Delta q_{n-1}^{(b)}(k) = \sum_{\ell\sim k} q_{n-1}^{(b)}(\ell) - q_{n-1}^{(b)}(k) = \frac{2d}{b}\left(q^{(b)}_n(k) - q^{(b)}_{n-1}(k)\right)\] and by Theorem 17, this is nonnegative if \(|k| \ge k_2\) and \(n-1 \le \Lambda |k|^2/6b\). This implies the corollary with \(c = \Lambda/6 = 1/48d^4\) and \(\rho_0 = k_2\). ◻

Proof of Theorem 12. By the hypotheses, there are positive constants \(C, \beta\) and \(N\) so that for \(n \ge N\), \[\tilde{p}(k, n) \ge Cn^{-d\beta}\quad\text{for all |k| \le Cn^\beta},\] where \(\tilde{p}(\cdot, n)\) is volcanic for each \(n\).

Fix \(n_{1}\geq (2N)\vee (2C^{m})^{\tfrac{1}{dm\beta}}\) and let \(\rho := C(n_{1}/2)^\beta\). By the above, we have \[\tilde{p}(k,j)^m\geq \frac{C^{m}}{n_{1}^{dm\beta}}=:b\quad\text{for |k| \le \rho and n_{1}/2 \le j \le n_{1}}.\] Observe that by the choice of \(n_{1}\), \(b\leq \tfrac{1}{2}\).

Let \(n_0:=\tfrac{n_{1}}{2}\vee (n_{1} - T(\rho,b))\), where \(T(\rho,b)\) is as in (27 ), and let \(\sigma := \sum_{k\in \mathbb{Z}^{d}} \tilde{p}(k, n_0)\). Given that \(\tilde{p}(\cdot,n)\) is volcanic, all hypotheses of Lemma 14 are satisfied. Thus, \[\tilde{p}(k,n_{1})\leq \sigma V^{(\rho, b)}_{n_0 \to n_{1}}(\sigma^{-1}\tilde{p}(\cdot, n_{0})).\] By Theorem 16, this implies that for all \(k\in \mathbb{Z}^{d}\), \[\begin{align} \label{e46pqbd} \tilde{p}(k,n_{1})&\leq \sigma \sum_{k\in \mathbb{Z}^{d}} \sigma^{-1}\tilde{p}(k,n_{0})q^{(b)}_{n_{1} - n_0}(k) \notag \\&\le \sigma \sup_{k\in \mathbb{Z}^{d}} q_{n_{1} - n_0}^{(b)}(k)\notag \\&= \sigma q^{(b)}_{n_{1}-n_0}(0), \end{align}\tag{33}\] where in the last line, we used the fact that by Lemma 9 (with the function \(A(u) = bu\)), \(q_n^{(b)}(k)\) is volcanic, since \(b\leq \tfrac{1}{2}\).

In order to conclude, we simply need to estimate the probability that a \(b\)-lazy random walk \((Y_{n})_{n\geq 0}\) is at \(0\) after some number of steps. We seek an estimate which is uniform in \(b\), and consequently, we will consider the asymptotics of the characteristic function associated to the \(b\)-lazy random walk. After one step, the characteristic function \(\phi\) is given by \[\phi(\vartheta_1, \ldots, \vartheta_d) := {\mathbf{E}}\left[\exp(i Y_1 \cdot \vartheta)\right] = 1 - b + {b \over 2d} \sum_{j=1}^d \cos \vartheta_j.\] Due to the independent increments of the lazy random walk, \(\phi^n\) will be the characteristic function of \(Y_n\), and we can recover the probability that \(Y_n = 0\) by using Fourier series inversion and evaluating at 0, \[q^{(b)}_n(0)={\mathbf{P}}\left\{Y_n = 0\right\} = {1 \over (2\pi)^d} \int_{-\pi}^\pi \cdots \int_{-\pi}^{\pi} \phi^n(\vartheta_1, \ldots, \vartheta_d) \,d\vartheta_1 \cdots d\vartheta_d.\] Now we can use classic asymptotic analysis arguments. Since \(1-x \le e^{-x}\), \[\begin{align} q^{(b)}_n(0) &= {1 \over (2\pi)^d} \int_{-\pi}^\pi\cdots\int_{-\pi}^{\pi} \left(1-b+b\left(\frac{1}{2d}\sum_{i=1}^d \cos \vartheta_i\right)\right)^n \,d\vartheta_1 \cdots \,d\vartheta_d \\&\le {1 \over (2\pi)^d} \int_{[-\pi, \pi]^{d}} \exp\left(-bn\left(1 - \frac{1}{2d}\sum_{i=1}^d \cos \vartheta_i\right)\right) \,d\vartheta. \end{align}\]

Let \(f(\vartheta_1, \ldots, \vartheta_d) := 1 - {1 \over 2d} \sum_{i=1}^d \cos \vartheta_i\). The only minimum of \(f\) in the domain of integration is at \(\vartheta = 0\), where \(f = 0\). By Laplace’s method, we obtain the asymptotic estimate that for \(t\gg 1\), \[H(t) := \int_{[-\pi, \pi]^{d}} \exp(-tf(\vartheta)) \,d\vartheta \sim \frac{K}{t^{d/2}},\] where \(K=K(d)>0\). Since \(H(t)\) is bounded for all \(t\), we conclude that there exists \(K=K(d)\) such that \(H(t) \le K/t^{d/2}\) for all \(t \ge 0\). This constant depends only on \(d\), not on the probability \(b\), and hence \(q^{(b)}_{n}(0)\leq K(bt)^{-\tfrac{d}{2}}\). Continuing from 33 , this implies that \[\tilde{p}(k,n_{1})\leq \frac{K\sigma}{(b(n_{1}-n_{0}))^{d/2}}.\]

Finally, we estimate a lower bound on \(n_{1} - n_0\). By Corollary 18, we have that \(T(\rho, b) \ge c\rho^2/b\), where \(c=c(d,m)\). Therefore, by the choices of \(n_{1}\) and \(n_{0}\), we have \[b(n_{1}-n_0) \ge b\min\{n_{1}/2, T(\rho, b)\}\geq \min\{\tfrac{bn_{1}}{2}, c\rho^2\}=\min\{cn_{1}^{1-dm\beta}, cn_{1}^{2\beta}\}.\] But these are of the same order in \(n\), since \(1 - dm\beta = (dm+2)\beta - dm\beta = 2\beta\). Therefore, for all \(k\in \mathbb{Z}^{d}\), there exists \(C'=C'(C, d,m, \sigma)\) such that \[\tilde{p}(k,n_{1}) \le {K\sigma \over (b(n_{1}-n_0))^{d/2}} \le C'(n_{1}^{-2\beta})^{d/2} = {C' \over n_{1}^{d\beta}}.\] This works for all \(n_{1}\) sufficiently large, depending on \(C, d, m, N\). Finally, by enlarging the constant \(C'=C'(C, d,m, \sigma, N)\), we can obtain the desired estimate for all \(n_{1}\geq N\). ◻

Proof. Since \(d\geq 2\), \(p_1(k)=(2d)^{-1}\boldsymbol{1}_{\pm e_{i}}(k)\). Set \[\tilde{p}_0(k) = \begin{cases}\frac{1}{2d}&\text{ if k=0, \pm e_{i}}\\0&\text{ otherwise}.\end{cases}\] Then \(\tilde{p}_{0}\) is symmetric and nondecreasing towards the origin. Moreover, \(p_1 \le \tilde{p}_0\le1/3\). Let \(\tilde{p}=(\tilde{p}_{n})_{n\geq 0}\) be a solution of 14 with initial condition \(\tilde{p}_{0}\). Then by Corollary 6, \(p_{n+1} = \mathcal{S}^n p_1 \le \mathcal{S}^n \tilde{p}_0 = \tilde{p}_n\), and by Lemma 9, \(\tilde{p}_n\) is volcanic for every \(n\). This implies the result with \(N_{0} = 1\). Also by Corollary 6, for every \(n\), \[\sum_{k\in \mathbb{Z}^{d}} |\tilde{p}_n(k)| =\sum_{k\in \mathbb{Z}^{d}} |\tilde{p}_0(k)|= (2d)^{-1}(2d+1).\] ◻

Proof. We proceed by induction to show that for any \(n\in {\mathbb{N}_{0}}\), \[\label{eq:errorlemma}\mathrm{dist}[q_n, \tilde{p}_n] \le \mathrm{dist}[q_0, \tilde{p}_0] + \sum_{j=0}^{n-1} \mathrm{dist}[q_{j+1}, \mathcal{S}q_j].\tag{34}\] We consider the base case when \(n = 0\), where the sum is empty and this is the trivial statement that \(\mathrm{dist}[q_0, \tilde{p}_0] \le \mathrm{dist}[q_0, \tilde{p}_0]\).

Suppose (34 ) holds \(n\). Let \(r_n(k) := \min\{\tilde{p}_n(k), q_n(k)\}\). By monotonicity, \(\mathcal{S}r_n \le \mathcal{S}\tilde{p}_n\) and \(\mathcal{S}r_n \le \mathcal{S}q_n\).

Now using the monotonicity property of asymmetric distance, we have \[\mathrm{dist}[q_{n+1},\tilde{p}_{n+1}]=\mathrm{dist}[q_{n+1},\mathcal{S}\tilde{p}_n]\leq \mathrm{dist}[q_{n+1},\mathcal{S}r_n]\leq \mathrm{dist}[q_{n+1},\mathcal{S}q_n] + \mathrm{dist}[\mathcal{S}q_n,\mathcal{S}r_n].\] But \(\mathcal{S}r_n \le \mathcal{S}q_n\), so \(\mathrm{dist}[\mathcal{S}q_n,\mathcal{S}r_n] = |\mathcal{S}q_n - \mathcal{S}r_n |_1\leq |q_n - r_n|_1\) by Corollary 6(iii). Since \(r_n = \min\{\tilde{p}_n, q_n\}\), it follows that \(| q_n - r_n |_1 = \mathrm{dist}[q_n,\tilde{p}_n]\). Therefore, \[\begin{align} \mathrm{dist}[q_{n+1},\tilde{p}_{n+1}] &\le \mathrm{dist}[q_{n+1},\mathcal{S}q_n] + \mathrm{dist}[q_n,\tilde{p}_n] \\[-.9em]&\le \mathrm{dist}[q_{n+1},\mathcal{S}q_n] + \mathrm{dist}[q_0,\tilde{p}_0] + \sum_{j=0}^{n-1} \mathrm{dist}[q_{j+1},\mathcal{S}q_j]\\[-.9em] &=\mathrm{dist}[q_0,\tilde{p}_0]+\sum_{j=0}^{n} \mathrm{dist}[q_{j+1},\mathcal{S}q_j]. \end{align}\] Thus, 34 holds for all \(n\in {\mathbb{N}_{0}}\). Since \(q\) is an approximate subsolution to 14 on \([0, \infty)\) with initial condition \(q_{0}\), we conclude that \[\mathrm{dist}[q_n, \tilde{p}_n] \le \mathrm{dist}[q_0, \tilde{p}_0] + \frac{1}{2}\mathop{\rm smallest}(q). \hfill\qedhere\] ◻

Proof. By definition of \(\bar{u}\), we have \(\bar{u}_{n}(k)=0\) for all \(|k|\geq r(n)\). By the definition of \(\mathcal{S}\) in 36 , this implies that \(\mathcal{S}\bar{u}_{n}(k)=0\) for all \((k,n)\in \mathcal{O}\).

We now claim that \(\bar{u}_{n+1}(k)=0\) for all \((k,n)\in \mathcal{O}\). By the form of \(\bar{u}\), it suffices to show that \[\label{steponebound46parta} r(n+1)\leq r(n)+1,\tag{39}\] since in this case, \(|k|\geq r(n)+1\geq r(n+1)\) implies \(\bar{u}_{n+1}(k)=0\).

To verify 39 , we use 35 and the fact that \(\beta\in (0,1)\), which implies that \[r(n+1)\leq r(n)+\max_{t\geq 0}\left|\frac{dr}{dt}\right|= r(n)+\max_{t\geq 0}\frac{R\beta}{T_{0}}\bigg(1+\frac{t}{T_{0}}\bigg)^{\beta-1}\leq r(n)+\frac{R\beta}{T_{0}}\leq r(n)+\frac{\Gamma^{m}}{2d\gamma R}.\] Since \(\Gamma\in (0, \frac{1}{3})\) and \(R \geq (2d\gamma)^{-1}\), this implies 39 , and hence \(\bar{u}_{n+1}(k)=0\) for all \((k,n)\in \mathcal{O}\). Combining this with the first observation, we obtain the desired result. ◻

Proof. Fix \(k\) such that \((k,n)\in \mathcal{G}\). We first prove that there exists a constant \(c=c(d,m)\) such that \[\label{durationestimate} \mathop{\rm duration}(k) \le c{|k|^{dm+1} \over R^{dm} \Gamma^m}.\tag{42}\] In order to estimate \(\mathop{\rm duration}(k)\), we consider the function \(h: [0, \infty)\rightarrow [0, \infty)\) defined by \(h(s):=r^{-1}(s)=T_{0}((s/R)^{1/\beta}-1)={T_0}((s/R)^{dm+2} - 1)\).

The function \(h\) is convex, and there exists \(c=c(d,m)>0\) such that \[h'(s) = {{T_0}\over \beta R^{\frac{1}{\beta}}} s^{dm+1}= {2d\gamma R^{2-\frac{1}{\beta}} s^{dm+1} \over \beta \Gamma^m}= c{s^{dm+1} \over R^{dm} \Gamma^m}>0.\] Set \(t_{1}:=h(|k|-1)\) and \(t_{2}:=h(|k|+\sqrt{d})\), and note that \(\left\{n: (k,n)\in \mathcal{G}\right\}\subset \left\{n: t_1<n\leq t_{2}\right\}\). By convexity and the derivative estimate above, we estimate \[\begin{align} t_{2}-t_{1}\leq h'(|k|+\sqrt{d})\Big[(|k|+\sqrt{d})-(|k|-1)\Big]\leq {c \over R^{dm} \Gamma^m} (|k| + \sqrt{d})^{dm+1}. \end{align}\] This implies that \[\mathop{\rm duration}(k)\leq {c \over R^{dm} \Gamma^m} (|k| + \sqrt{d})^{dm+1}+1.\] We have assumed that \(R\geq 2\sqrt{d}\), so by 41 , \(|k|\geq \sqrt{d}\). Moreover, since \(\Gamma \le 1\), we have \(|k|^{dm+1} (R^{dm} \Gamma^m)^{-1}\ge 2^{-dm}|k|\Gamma^{-m}\geq \sqrt{d}2^{-dm}\), and hence, by increasing \(c\), we obtain 42 .

We now fix \(n\in V(k) := \{n : (k,n) \in \mathcal{G} \text{ or } (k,n-1) \in \mathcal{G}\}\). As we show in 43 below, \(V(k)\) can be used to control \(U(k)\).

We introduce some more notation to simplify (35 ). Set \(\vartheta=\vartheta(k) := |k| / r(n)\), so that \[\bar{u}_n(k) = {R^d \, \Gamma \over r(n)^d} \left(1 - \vartheta^2\right)_+^{1/m},\] and \(\mathcal{G}=\left\{(k,n): \vartheta \in \big[1- \tfrac{\sqrt{d}}{r(n)}, 1+\tfrac{1}{r(n)}\big)\right\}\). For \((k, n)\in \mathcal{G}\) such that \(|k|\leq r(n)\), \[1-\vartheta^2 = (1+\vartheta)(1-\vartheta) \le 2(1-\vartheta)\leq 2 \frac{\sqrt{d}}{r(n)}=:\frac{c'}{r(n)}.\]

If \((k,n-1)\in \mathcal{G}\), then by (39 ) and the fact that \(d> 1\), we have \[|k| \ge r(n-1) - \sqrt{d} \ge r(n) - 1 - \sqrt{d}\geq r(n)-2\sqrt{d}.\] In this case, we obtain the identical estimate above that if \(|k|\leq r(n)\), \[1-\vartheta^2\leq 2(1-\vartheta)\leq 2\frac{2\sqrt{d}}{r(n)}\leq \frac{2c'}{r(n)}.\]

Therefore, for \((k, n)\in \mathcal{G}\) or \((k, n-1)\in \mathcal{G}\) such that \(|k|\leq r(n)\), we have \[\bar{u}_n(k)={R^d\,\Gamma\over r(n)^d}\left(1-\vartheta^2\right)^{1/m} \le 2c'{R^d\,\Gamma \over r(n)^{d+1/m}} \le 2c'{R^d \, \Gamma \over |k|^{d+1/m}}.\] If \(|k|>r(n)\), then \(\bar{u}_n(k) = 0\) and the same inequality holds. Therefore, for \(k\) such that \((k, n)\in \mathcal{G}\), \[\label{estimateforu} U(k) = \max\{\bar{u}_{n}(k) \vee \bar{u}_{n+1}(k): (k, n) \in \mathcal{G}\}=\max_{n\in V(k)} \bar{u}_{n}(k)\le 2c'{R^d \,\Gamma \over |k|^{d+1/m}}.\tag{43}\] Combining this with (42 ), we obtain the existence of \(c_{u}=c_{u}(d,m)>0\) such that \[\label{durationanduestimate}\mathop{\rm duration}(k) \,U^m(k) \le c_{u}.\tag{44}\]  ◻

Proof of Proposition 24. Fix \(k\) such that \((k,n)\in \mathcal{G}\). Then from (37 ), dropping negative terms from the right hand side, and then taking the positive part, we obtain \[(\bar{u}_{n+1}(k) - \mathcal{S}\bar{u}_n(k))_{+} \le (\bar{u}_{n+1}(k) - \bar{u}_n(k))_{+} + \bar{u}_{n}(k)^{m+1}.\] Since \(r(t)\) is monotone, by the definition of \(\mathcal{G}\), \(\left\{n: (k,n)\in \mathcal{G}\right\}\) is a discrete interval. Moreover, since \(t\mapsto \bar{u}(k,t)\) is unimodal in \(t\) by Lemma 11(i), it follows that \(E:=\left\{n: \bar{u}_{n+1}(k)-\bar{u}_{n}(k)>0\right\}\) is also a discrete interval. Therefore, \[\begin{align} \sum_{n: (k,n) \in \mathcal{G}} (\bar{u}_{n+1}(k) - \bar{u}_{n}(k))_{+}\leq \sum_{n\in E: (k,n)\in \mathcal{G}} (\bar{u}_{n+1}(k) - \bar{u}_{n}(k)) \le \max_{n: (k,n) \in \mathcal{G}}\bar{u} (k, n+1)\le U(k), \end{align}\] where we used that the sum telescopes, and that \(\bar{u}\geq 0\) everywhere. By Lemma 25, this implies that there exists \(c=c(d,m)\) such that \[\begin{align} \sum_{n: (k,n) \in \mathcal{G}} (\bar{u}_{n+1}(k) - \mathcal{S}\bar{u}_n(k))_{+} &\le U(k)+\sum_{n: (k,n) \in \mathcal{G}} \bar{u}_{n}(k)^{m+1}\\ &\leq U(k)+\mathop{\rm duration}(k)U(k)^{m+1}\\ &\leq cU(k). \end{align}\] Combining this with Lemma 25 and 41 , we obtain that \[\label{gaperrorfinal} \begin{align} \sum_{(k,n)\in \mathcal{G}} (\bar{u}_{n+1}(k) - \mathcal{S}\bar{u}_n(k))_{+} &\leq \sum_{k\in \mathbb{Z}^d:|k| \ge R/2} cU(k)\\ &\le cR^d\,\Gamma \sum_{k\in \mathbb{Z}^d: |k| \ge R/2} \frac{1}{|k|^{d+1/m}} \\&\le cR^d \,\Gamma \int_{R/2}^{\infty} \frac{r^{d-1}}{r^{d+1/m}}\, dr=:c_{g}\Gamma R^{d-\frac{1}{m}}, \end{align}\tag{45}\] where \(c_{g}=c_{g}(d,m)\). ◻

Proof of Lemma 27. As a consequence of Taylor’s theorem, there exists \(s\in [n, n+1]\) such that \[\bar{u}(k, n+1) - \bar{u}(k, n) - \partial_t \bar{u}(k, n) = \frac{1}{2} \partial^2_t \bar{u}(k, s),\] and thus, we will be concerned with estimating \(\left(\partial^2_t \bar{u}(k, t)\right)_{+}\). Recall the parameterization \(\vartheta=\vartheta(x,t)=|x| / r(t)\); that in Lemma 11(ii), we computed \(\partial_{t}\bar{u}\) in 24 ; and that \(\tfrac{\partial \vartheta}{\partial t} =- \tfrac{\vartheta\beta}{(t + {T_0})}\). Thus, by a direct computation, we have that \[\label{eq:partial2t} \partial^{2}_{t}\bar{u}(x, t) = {R^d\Gamma(1 - \vartheta^2)^{\tfrac{1}{m}}\over r(t)^d m(t+T_0)^2} \left[-\left({2\beta \over 1 - \vartheta^2} - 1\right) + {1 \over m} \left({2\beta \over 1 - \vartheta^2} - 1\right)^2 - {4\beta^2 \vartheta^2 \over (1 - \vartheta^2)^2}\right],\tag{46}\] and combining terms, it follows that there exist positive constants \(K_{1}, K_{2}\), and \(K\), depending only on \(d,m\), such that \[\begin{align} \label{gbound} \partial^{2}_{t} \bar{u}(x, t)&\leq {R^d\Gamma\over r(t)^d m(t+T_0)^2}\left[K_{1} (1 - \vartheta^2)^{\tfrac{1}{m}} + K_2(1 - \vartheta^2)^{\tfrac{1}{m} - 1} + 4\beta^2 (\tfrac{1}{m} - \vartheta^2) (1 - \vartheta^2)^{\tfrac{1}{m}-2}\right]\notag\\ &\leq {KR^d\Gamma\over r(t)^d m(t+T_0)^2}(1 - \vartheta^2)^{\tfrac{1}{m} - 1}\notag\\&={KR^d\Gamma\over r(t)^d m(t+T_0)^2}\bigg(1 - \Big(\frac{|x|}{r(t)}\Big)^{2}\bigg)^{\tfrac{1}{m} - 1}=: g(x,t), \end{align}\tag{47}\] where for the last inequality, we used that \(m\geq 1\). We now simply need to obtain an upper bound on the function \(g\). First, since \(m\geq 1\), \(t\mapsto g(k,t)\) is a decreasing function, and hence by the first observation of the proof, \[\label{e46taylors} \bar{u}(k, n+1) - \bar{u}(k, n) - \partial_t \bar{u}(k, n)\leq \frac{1}{2}g(k,n).\tag{48}\] Furthermore, by definition of \(g\), it is clear that \(g\geq 0\) on \(\mathcal{C}\) and that for any \(t\) fixed, \(g(k,t)\) is monotone increasing with respect to \(|k|\) in \(\mathcal{C}\).

Fix \(n\) such that there exists \(k\) with \((k,n)\in \mathcal{C}\). We claim that \[\label{e46core1} \sum_{k: (k,n)\in \mathcal{C}} g(k,n) \le \int\limits_{B_{r(n)}} g(y, n) \,dy,\tag{49}\] where \(B_{r(n)}\) is the Euclidean ball in \(\mathbb{R}^{d}\) of radius \(r(n)\). In order to prove 49 , due to the symmetry of \(g\), it is enough to prove an analogous bound in the positive orthant \(E_{n}^{+}:=\mathbb{Z}^{d}_{+}\cap \left\{k: (k,n)\in \mathcal{C}\right\}\). More precisely, since \(k\in E_{n}^{+}\), we know \(|k| < r(n) - \sqrt{d}\), and hence \([k_{1}, k_{1}+1]\times \ldots [k_{d}, k_{d}+1]\subseteq \left\{|k|<r(n)\right\}=B_{r(n)}\cap \mathbb{Z}^{d}\). Owing to the monotonicity property of \(g\), we conclude \[\sum_{k\in E_{n}^{+}} g(k,n) \le \int_{\bigcup_{k\in E_{n}^{+}} [k_{1}, k_{1}+1]\times \ldots [k_{d}, k_{d+1}]}g(y,n) \,dy \le \int\limits_{\mathbb{R}^{d}_{+}\cap B_{r(n)}} g(y,n) \,dy,\] where \(\mathbb{R}^d_+ = [0, \infty)^d\). Summing over all orthants, we obtain 49 .

We now conclude by 49 that \[\begin{align} \sum_{k: (k,n)\in \mathcal{C}} g(k,n) &\leq {K R^d \Gamma \over r(n)^d(n+T_0)^2} \int_{B_{r(n)}} \Big(1 - \frac{|y|^{2}}{r(n)^{2}}\Big)^{\tfrac{1}{m}-1} \,dy \\&= {K R^d \Gamma \over (n+T_0)^2} \int_{B_{1}} (1 - z^2)^{\tfrac{1}{m}-1} \,dz \\&= {KR^d \,\Gamma \over (n+T_0)^2}, \end{align}\] by increasing the value of \(K=K(d,m)\) if necessary.

Combining this with 48 and repeatedly adjusting the constant \(K\) as needed, we obtain \[\sum_{(k,n)\in \mathcal{C}}\left(\bar{u}(k, n+1) - \bar{u}(k, n) - {\partial \bar{u} \over \partial t}(k,n)\right)_{+}\le \sum_{n=0}^\infty {KR^d \Gamma \over (n+T_0)^2} \le {KR^d \Gamma \over T_0}=KR^{d-2} \,\Gamma^{m+1},\] where we used the definition of \(T_{0}\) from 35 . The result follows. ◻

Proof of Lemma 28. For fixed \((k,n)\in \mathcal{C}\), we can re-express the space error as \[{1 \over 2d} \Delta(\bar{u}(k,n))^{m+1}- {1 \over 2d} \sum_{\ell\sim k} [\bar{u}(\ell, n)^{m+1} - \bar{u}(k,n)^{m+1}] = \sum_{i=1}^d \mathcal{E}_i(k, n)\] where \(\mathcal{E}_i\) is the error along a single coordinate: \[\mathcal{E}_i(k,n) := {1 \over 2d} \left[{\partial^2 \over \partial^2 x_i} \bar{u}(k, n)^{m+1} - \left(\bar{u}(k+e_i, n)^{m+1} - 2\bar{u}(k, n)^{m+1} + \bar{u}(k-e_i, n)^{m+1}\right)\right] .\]

To estimate the error, we observe the following: if \(f \in C^4(\mathbb{R})\), then \[\label{e46ibp} f''(x) - (f(x+1) - 2f(x) + f(x-1)) = -\int_{x-1}^{x+1} {(1 - |s-x|)^3 \over 6} f^{(4)}(s) \,ds.\tag{50}\] The proof of 50 relies on the integral version of Taylor’s theorem, and appears immediately after this argument.

Using 50 , we obtain the alternative formula \[\mathcal{E}_i(k,n) = -{1\over2d} \int_{k_i - 1}^{k_i + 1} {(1 - |s - k_i|)^3 \over 6} \partial^4_i(\bar{u}^{m+1}(k_1, \ldots, k_{i-1}, s, k_{i+1}, \ldots, k_d)) \,ds. \label{errori}\tag{51}\] By Lemma 11(ii), \(\bar{u}^{(4)}(k,n)\geq 0\) for all \((k,n)\in \mathcal{C}\). This completes the proof. ◻

Proof of 50 . By Taylor’s theorem in integral form, we have \[\begin{align} f(x+1)&=f(x)+f'(x)+\frac{1}{2}f''(x)+\frac{1}{6}f^{(3)}(x)+\int_{x}^{x+1}f^{(4)}(s)\frac{(1-|s-x|)^{3}}{6}\, ds,\\ f(x-1)&=f(x)-f'(x)+\frac{1}{2}f''(x)-\frac{1}{6}f^{(3)}(x)+\int_{x-1}^{x}f^{(4)}(s)\frac{(1-|s-x|)^{3}}{6}\, ds. \end{align}\] Adding the two expressions and rearranging, we arrive at 50 . ◻

Proof of Proposition 22. We now make explicit choices of \(R\) and \(\Gamma\). We allow arbitrarily small \(\Gamma\in (0,\frac{1}{3})\), while \(R\in [1, \infty)\) will be moderately large and fixed, given by \[\label{e46Rdef} R = \max\left\{2\sqrt{d}, (2d\gamma)^{-1}, \left( 4c_g\over c_0\right)^{m}, \sqrt{\frac{4c_c}{c_{0}}}\right\},\tag{52}\] where the constant \(c_0\) is from Lemma 29, \(c_g\) is from Proposition 24, and \(c_c\) is from Proposition 26. All these constants (and hence \(R\) itself) depend only on \(d\) and \(m\).

Choosing \(R\) as in 52 and \(\Gamma<1/3\), the hypotheses of Proposition 23, Proposition 24 and Proposition 26 hold simultaneously. It follows that \[\begin{align} \sum_{(k,n)\in \mathbb{Z}^{d}\times {\mathbb{N}}}& (\bar{u}_{n+1}(k) - \mathcal{S}\bar{u}_n(k))_+\\ &\leq \sum_{(k,n)\in \mathcal{O}}(\bar{u}_{n+1}(k) - \mathcal{S}\bar{u}_n(k))_+ +\sum_{(k,n)\in \mathcal{G}}(\bar{u}_{n+1}(k) - \mathcal{S}\bar{u}_n(k))_+\\ &\quad +\sum_{(k,n)\in \mathcal{C}}(\bar{u}_{n+1}(k) - \mathcal{S}\bar{u}_n(k))_+\\ &\leq c_{g}\Gamma R^{d-\frac{1}{m}}+c_{c}R^{d-2} \,\Gamma^{m+1}. \end{align}\] To bound each of these terms, we will use Lemma 29 which says \(\mathop{\rm smallest}(\bar{u})\geq c_0 R^d \Gamma\).

For the first term, by the choice of \(R\), \(R \ge \left(\tfrac{4c_g}{c_{0}}\right)^{m}\), so \[c_{g}\Gamma R^{d-\frac{1}{m}}\leq c_{g}\Gamma R^{d}\frac{c_{0}}{4c_{g}}\leq \frac{1}{4}\mathop{\rm smallest}(\bar{u}).\]

Similarly, for the second term, by choice of \(R\), \(R^{2}\geq 4c_{c}c_{0}^{-1}\). Hence \[c_{c}R^{d-2} \,\Gamma^{m+1}\leq \frac{c_{0}}{4}R^{d}\Gamma^{m+1}\leq \frac{1}{4}\Gamma^{m}\mathop{\rm smallest}(\bar{u})\leq \frac{1}{4} \mathop{\rm smallest}(\bar{u}).\] This yields the desired claim. ◻

Proof. Let \(g(z) := R^d \Gamma (1 - |z|^2)^{1/m}_+\), and note that \(\int_{\mathbb{R}^{d}} g(z)\, dz>0\). We consider the set \(A(c) = \{z \in \mathbb{R}^d: |z_1|, \ldots, |z_d| \ge c\}\) which is simply the complement of the ball of radius \(c\) in the \(L^{\infty}\)-norm. As \(c \to 0+\), as a consequence of the monotone convergence theorem, we have \[\lim_{c \to 0+} \int_{A(c)} (g(z) - c)_+ \,dz = \int_{\mathbb{R}^d} g(z) \,dz.\] Consequently, we may fix a constant \(c=c(d,m)\in (0,1)\) such that \[\label{e46c3choice} \int_{A(c)} (g(z) - c)_+ \,dz \ge \frac{2}{3} \int_{\mathbb{R}^d} g(z) \,dz.\tag{55}\]

Let \(A_n(c) := \{k\in \mathbb{Z}^d: |k_1|, \ldots, |k_d| \ge cr(n)\}\). Recall that by Lemma 19, \(\tilde{p}_{n+n_{0}}\) is volcanic for every \(n\). It follows (by applying the volcanic property iteratively over neighbors) that for any \(k\in A_{n}(c)\) and \(\ell\in \mathbb{Z}^{d}\) with \(|\ell| \le cr(n)\), \(\tilde{p}_{n+n_{0}}(k) \le \tilde{p}_{n+n_{0}}(\ell)\). Therefore, by 54 , for any fixed \(|\ell| \le cr(n)\), \[\label{lowerboundonp46one} \sum_{k\in A_{n}(c)} \left(\bar{u}_{n}(k)-\tilde{p}_{n+n_{0}}(\ell)\right)_{+}\leq \sum_{k\in \mathbb{Z}^{d}} \left(\bar{u}_{n}(k)-\tilde{p}_{n+n_{0}}(k)\right)_{+}\leq \frac{1}{2}\mathop{\rm smallest}(\bar{u})\leq \frac{1}{2}|\bar{u}_{n}|_{1}.\tag{56}\]

By definition of \(\bar{u}_{n}\) in 35 \[\label{e46grel} \bar{u}_n(r(n)z) = {R^d \Gamma \over r(n)^d} \left(1 - |z|^2\right)^{1/m}_+ = \frac{g(z)}{r(n)^{d}}.\tag{57}\] Since \(r(n)\to \infty\) as \(n\to \infty\), we may apply a Riemann approximation to conclude that \[\lim_{n \to \infty} |\bar{u}_n |_1= \lim_{n \to \infty} \frac{1}{r(n)^{d}} \sum_{ (r(n)z)\in \mathbb{Z}^{d}} g(z) = \int_{\mathbb{R}^d} g(z) \,dz.\] As another consequence of 57 , we have \[\begin{align} \lim_{n \to \infty} \sum_{k\in A_n(c)} \left(\bar{u}_n(k) - \frac{c}{r(n)^d}\right)_+ = \lim_{n \to \infty} \frac{1}{r(n)^{d}} \sum_{r(n)z \in A_n(c)} (g(z) - c)_+ = \int_{A(c)} (g(z) - c)_+ \,dz. \end{align}\] By 55 and the prior display, we have \[\lim_{n \to \infty} \sum_{k\in A_n(c)} \left(\bar{u}_n(k) - \frac{c}{r(n)^d}\right)_+\geq \lim_{n \to \infty} \frac{2}{3}|\bar{u}_n |_1>0.\] Consequently, there is \(N>0\) so that for all \(n \ge N\), \[\sum_{k\in A_n(c)} \left(\bar{u}_n(k) - \frac{c}{r(n)^d}\right)_+\geq \frac{5}{9}|\bar{u}_n |_1>0.\] Now for the purposes of contradiction, fix \(n \ge N\), and suppose that there is a point \(\ell\) such that \(|\ell| \le cr(n)\) and \(\tilde{p}_{n+n_{0}}(\ell) < \tfrac{c}{r(n)^d}\). Then by the prior display, \[\sum_{k\in A_n(c)} (\bar{u}_n(k) - \tilde{p}_{n+n_{0}}(\ell))_+ \ge \sum_{k\in A_n(c)} \left(\bar{u}_n(k) - \frac{c}{r(n)^d}\right)_+ \ge \frac{5}{9} |\bar{u}_n |_1,\] which contradicts the inequality (56 ). This implies that \(\tilde{p}_{n+n_{0}}(k) \ge \tfrac{c}{r(n)^d}\) if \(|k| \le cr(n)\) and \(n \ge N\). ◻

Proof of Theorem 2(i). By Lemma 30, for \(n_{0}\) as in 53 , and by the definition of \(r\) in 35 , by adjusting \(c=c(d,m)\), we have that for all \(n\geq N\), \[\tilde{p}_{n+n_0}(k) \ge \frac{c}{r(n)^d}=\frac{c}{(T_{0}+n)^{d\beta}} \quad\text{for all |k|\leq cr(n).}\] Consequently, for all \(n\geq N\vee T_{0}\), by adjusting \(c=c(d,m)\) and since \(n_{0}>0\), we have \[\tilde{p}_{n+n_0}(k) \ge \frac{c}{(2n)^{d\beta}}\geq \frac{c}{(n+n_{0})^{d\beta}}.\] By Theorem 12, it follows that for all \(n\geq n_{0}+ (N\vee T_{0})\), \(\sup_{k} \tilde{p}(k,n)\leq C'n^{-d\beta}\). For \(N_{0}\) as in Lemma 19, this implies that there is \(C''=C''(d,m)\) such that for all \(n\geq n_{0}+ N\vee T_{0}\), we have \[p(k,n+N_{0})\leq \tilde{p}(k,n)\leq C'n^{-d\beta}\leq C''(n+N_{0})^{-d\beta}\quad\text{for all k\in \mathbb{Z}^{d}.}\] Since \(n_{0}, N_{0}, N, T_{0}\) all depend only on \(d,m\) and are all finite, this statement can then be upgraded to the statement that there exists \(C=C(d,m)>0\) such that \[p(k,n)\leq Cn^{-d\beta}\quad\text{for all n\in \mathbb{N} and all k\in \mathbb{Z}^{d}.} \qedhere\] ◻

Proof of Theorem 2(ii). If \(r\in (0,1]\), we obtain the desired bound for all \(n\in \mathbb{N}\), so long as \(C>1.\) Thus, let \(r > 1\), and let \((X_j)_{j\in \mathbb{N}}\) be \(\text{CM}(m)\)-distributed. Fix \(i \in [d]\), and let \(Y_j\) be the \(i\)-th coordinate of \(X_j\).

At any step of the \(\text{CM}(m)\), with probability \(p^m / d\), we move in the \(i\)-th coordinate, and when we do, we take a Bernoulli step that is conditionally independent of the past. Thus, \((Y_j)_{j\geq 0}\) is a martingale, and its conditional variance is given by \[V_{j}=\text{Var}[Y_j \mid \mathscr F_{j-1}] = \mathbb{E}[(Y_j - Y_{j-1})^2 \mid \mathscr F_{j-1}] = \frac{1}{d}p^m(X_{j-1}, j-1).\]

By Theorem 2(i), this has a deterministic upper bound, given by \[V_j = c p^m(X_{j-1}, j-1) \le {c \over j^{dm\beta}}.\] Since \(0 < dm\beta < 1\), we have \[\sum_{j=1}^{n}V_{j}=c\sum_{j=1}^n j^{-dm\beta} = cn^{1-dm\beta} + O(1),\] which implies \(\sum_{j=1}^n V_j \le cn^{1-dm\beta} = Cn^{2\beta}\), where \(C=C(d,m)\). Let \(b:=Cn^{2\beta}\), so that \(\sum_{j=1}^n V_j \le b\).

Let \(a:=rn^\beta/\sqrt{d}\). Then by properties of norm-equivalences on \(\mathbb{R}^{d}\), we have \[\{|X_n| \ge rn^\beta\} \subseteq \bigcup_{i=1}^d\{(X_n)_i \ge a\} \cup \bigcup_{i=1}^d \{-(X_n)_i \ge a\}.\] By a union bound, symmetry of the process, and the prior claim, this implies \[\mathbb{P}\left(|X_n| \ge rn^\beta\right) \le 2d\mathbb{P}\left((X_n)_i \ge a\right) = 2d \mathbb{P}\left(Y_n \ge a\right) = 2d \mathbb{P}\left(Y_n \ge a \text{ and }\sum_{j=1}^n V_j \le b\right).\] We now conclude by using Freedman’s inequality (Theorem 31). In particular, we have \[\mathbb{P}\left(|X_n| \ge rn^\beta\right) \le 2d\mathbb{P}\left(Y_n \ge a \text{ and } \sum_{j=1}^n V_j \le b\right) \le 2d\exp\left(-{r^2 n^{2\beta}/d \over 2(rn^{\beta}/\sqrt{d} + Cn^{2\beta})}\right).\] Using that \(r > 1\), we conclude with the lower bound \[{r^2 n^{2\beta}/d \over 2(rn^{\beta}/\sqrt{d} + Cn^{2\beta})} = \frac{r^2}{2d(rn^{-\beta}/\sqrt{d} + C)} \geq \frac{r^2}{2d(r/\sqrt{d} + C)} \geq \frac{r^2}{C(r+1)} \geq \frac{r}{C}.\] ◻

Proof. Recall that \(p=(p_{n})_{n\geq 0}\) solves the scheme 14 on \([0, \infty)\) with initial condition \(p_{0}=\mathbf{1}_{0}\), and by hypothesis, let us assume that \(0\leq p_{n}\leq 1/3\). Since \(A(u)=u^{m+1}\) in the scheme 14 , we have \(0\leq A'(\cdot)\leq \frac{2}{3}\) for \(0\leq u\leq \frac{1}{3}\).

By the same reasoning as in Lemma 9, if \(p_{n}\) is symmetric, then \(p_{n+1}\) is also symmetric. Therefore, it remains to verify that \(p_{n+1}\) satisfies the above two additional conditions necessary to be almost-volcanic.

We begin with the first claim, which is that \(p_{n+1}(k) \le p_{n+1}(\ell)\) for \(|k|_1 > |\ell|_1 \ge 1\). In the case of \(k, \ell\) with \(|\ell|_1 \ge 2\), this follows immediately from Case 1 in the proof of Lemma 9. Therefore, we simply need to check this inequality in the case when \(|k|_1 = 2, |\ell|_1 = 1\).

Suppose \(k\) is a point with \(|k|_1 = 2\). By symmetry, it is enough to consider the cases \(k= 2e_1\) or \(k= e_1 + e_2\).

Case 1. \(k= 2e_1\). If we look at all neighbors of \(e_1\), there are \(2d-2\) neighbours the form \(e_1 \pm e_j, j \ne 1\), one neighbor which is \(2e_{1}\) and one neighbor which is the origin. Thus, for \(\ell\sim e_{1}\), since \(p_{n}\) is almost-volcanic, the symmetry and monotonicity properties imply that \[p_{n}(\ell)\begin{cases} =p_n(e_1+e_2)&\text{if \ell=e_1 \pm e_j, j \ne 1},\\ =p_{n}(2e_{1})&\text{if \ell=2e_{1}},\\ \geq p_{n}(2e_{1})&\text{if \ell=0}. \end{cases}\] Therefore, since \(A\) is nondecreasing, we have \[\begin{align} p_{n+1}(e_1) &= p_n(e_1) + {1 \over 2d} \sum_{\ell\sim e_1} (A(p_n(\ell)) - A(p_n(e_1))) \\&\ge p_{n}(e_{1}) + {2d-2 \over 2d} (A(p_n(e_1+e_2)) - A(p_{n}(e_{1}))) + {2 \over 2d} (A(p_{n}(2e_{1})) - A(p_{n}(e_{1}))). \end{align}\]

On the other hand, \(2e_1\) has \(2d-2\) neighbours \(2e_1 \pm e_j, j \ne 1\), one neighbour that is \(e_1\), and one neighbour that is \(3e_1\). Using again the almost-volcanicity, we conclude \[p_{n}(\ell)\begin{cases} \leq p_n(e_1+e_2)&\text{if \ell=2e_1 \pm e_j, j \ne 1},\\ \leq p_{n}(2e_{1})&\text{if \ell=3e_{1}},\\ =p_{n}(e_{1})&\text{if \ell=e_{1}}. \end{cases}\]

Using the above and the scheme, we have that \[p_{n+1}(2e_1) \le p_{n}(2e_{1}) + {2d-2 \over 2d} (A(p_{n}(e_{1}+e_{2})) - A(p_{n}(2e_{1}))) + {1 \over 2d} (A(p_{n}(e_{1})) - A(p_{n}(2e_{1}))) + 0.\]

Subtracting these two equations, we get \[p_{n+1}(e_1) - p_{n+1}(2e_1) \ge p_{n}(e_{1})- p_{n}(2e_{1}) + {2d+1 \over 2d} (A(p_{n}(2e_{1})) - A(p_{n}(e_{1}))).\] Now, arguing as in the proof of Lemma 9, by the mean-value theorem, \[\begin{gather} (p_{n}(e_{1})- p_{n}(2e_{1}))\left[1 - {2d + 1 \over 2d} \left({A(p_{n}(2e_{1})) - A(p_{n}(e_{1})) \over (p_{n}(e_{1})- p_{n}(2e_{1}))}\right)\right]\\ = (p_{n}(e_{1})- p_{n}(2e_{1}))\left[1 - {2d+1 \over 2d} A'(t)\right], \end{gather}\] for some \(t \in [p_{n}(2e_{1}), p_{n}(e_{1})]\). Since \(A'(t) \le 2/3\) whenever \(0 \le t \le 1/3\) and \((2d+1)/2d \le 3/2\), the above expression is at least \((p_{n}(e_{1})- p_{n}(2e_{1})) [1 - (3/2) (2/3)] = 0\).

Case 2. \(k= e_1 + e_2\).

We perform a similar analysis as above. We have that for \(\ell\sim e_{1}+e_{2}\), by symmetry, \[p_{n}(\ell)\begin{cases} \leq p_n(2e_{1})&\text{if \ell=2e_1+e_{2}, e_{1}+2e_{2}},\\ =p_{n}(e_{1})&\text{if \ell=e_{1}, e_{2}},\\ \leq p_{n}(e_{1}+e_{2})&\text{otherwise}. \end{cases}\] This implies \[p_{n+1}(e_1 + e_2) \le p_{n}(e_{1}+e_{2}) + {2 \over 2d} (A(p_n(e_1)) - A(p_{n}(e_{1}+e_{2}))) + {2 \over 2d} (A(p_n(2e_1)) - A(p_{n}(e_{1}+e_{2}))).\] Subtracting this from the inequality for \(p_{n+1}(e_1)\) from the first case, we obtain the bound \[p_{n+1}(e_1) - p_{n+1}(e_1 + e_2) \ge p_{n}(e_{1}) - p_{n}(e_{1}+e_{2}) + {2d+2 \over 2d} (A( p_{n}(e_{1}+e_{2})) - A(p_{n}(e_{1}))).\] Since again, \(A'(t) \le 2/3\) whenever \(0\leq t \le 1/3\), and \((2d+2)/2d \le 3/2\) because \(d \ge 2\), this implies the right side is at least 0.

Finally, we must also show that \(p_{n+1}(k) \le p_{n+1}(0)\) if \(|k|_1 \ge 2\). In this case, because \(p_n\) is almost-volcanic, \(p_n(k) \le p_n(0)\) and all the neighbours \(\ell\sim k\) have \(p_n(\ell) \le p_n(e_1)\), so since \(A\) is increasing, \[\begin{align} p_{n+1}(k) &= p_n(k) + {1 \over 2d} \sum_{\ell\sim k} (A(p_n(\ell)) - A(p_n(k))) \\&\le p_n(k) + {1 \over 2d} \sum_{\ell\sim k} (A(p_n(e_1)) - A(p_n(k))). \end{align}\] If we remove all \(A(p_n(k))\) terms out of the sum, and using that \(t\mapsto t-A(t)\) is nondecreasing for \(0\leq t\leq 1/3\), we get \[\begin{align} p_{n+1}(k) &\le p_n(k) - A(p_n(k)) + {1 \over 2d} \sum_{\ell\sim k} A(p_n(e_1)) \\&\le p_n(0) - A(p_n(0)) + {1 \over 2d} \sum_{\ell\sim k} A(p_n(e_1)) \\&=p_{n+1}(0). \end{align}\] This proves the second condition, so \(p_{n+1}\) is indeed almost-volcanic. ◻

Proof of Theorem 2(iii). In order to estimate \([p_{n}]_{\mathop{\mathrm{TV}}(D)} := \tfrac{1}{2}\sum_{\{u,v\in D: u\sim v\}} |p_{n}(v)-p_{n}(u)|\) on the set \(D := B(rn^\beta) \cap \mathbb{Z}^d\), we observe that since both the set \(D\) and the almost-volcanic function \(p_{n}\) are invariant under reflection and coordinate transposition, we may reduce this expression to studying \[[ p_n ]_{\mathop{\mathrm{TV}}(D)} = d \sum_{k\in \mathcal{D}}\;|p_n(k+ e_1) - p_n(k)|,\] where \[\mathcal{D}:=\left\{k\in \mathbb{Z}^d: k, k+e_1 \in D\right\}.\]

Let \[\overline{p}_{n}(k) = \begin{cases} p_n(e_{1})&\text{if k=0},\\ p_{n}(k)&\text{otherwise.} \end{cases}\] This makes \(\overline{p}_{n}\) a volcanic function, and this implies that \[\overline{p}_{n}(k+ e_1) - \overline{p}_{n}(k)=\begin{cases}\geq 0&\text{if k_{1}<0},\\ \leq 0&\text{if k_{1}\geq 0.} \end{cases}\] For each \(k\in \mathcal{D}\), we rewrite \(k=(k_{1}, k')\) where \(k'\in \mathbb{Z}^{d-1}\), and for such \(k'\), we denote that \(k'\in \mathcal{D}'\). We now estimate the total variation of \(\overline{p}_{n}\), which by the prior observation and symmetry is given by, \[\begin{align} d\sum_{k\in \mathcal{D}}& |\overline{p}_{n}(k+ e_1) - \overline{p}_{n}(k)| \\ &= d \sum_{k'\in \mathcal{D}'} \Bigg[\sum_{k_{1}<0: (k_{1}, k')\in \mathcal{D}} \overline{p}_{n}(k+ e_1) - \overline{p}_{n}(k)+\sum_{k_{1}\geq 0: (k_{1}, k')\in \mathcal{D}} \overline{p}_{n}(k) - \overline{p}_{n}(k+ e_1)\Bigg]\\ &=d \sum_{k'\in \mathcal{D}'} [\overline{p}_{n}(0, k')-\overline{p}_{n}(-B+1, k')+\overline{p}_{n}(0, k')-\overline{p}_{n}(B+1, k')]\\ &\leq 2d \sum_{k'\in \mathcal{D}'} \overline{p}_{n}(0, k'), \end{align}\] where the point \(B\) is such that \((B, k')\in \mathcal{D}\) and \((B+1, k')\notin \mathcal{D}\).

Since \(\mathcal{D}\) (and hence \(\mathcal{D}'\)) has radius at most \(rn^\beta\), in order for \(k'\in \mathcal{D}'\), we must have \(|k_i| \le rn^\beta\) for \(i = 2, \ldots, d\). Therefore the sum has at most \((2rn^\beta)^{d-1}\) terms. By Theorem 2(i), each term is at most \(c/n^{d\beta}\), and thus \[[ \overline{p}_n ]_{\mathop{\mathrm{TV}}(\mathcal{D})}\leq cr^{d-1}n^{-\beta}.\]

We now compare the total variation of \(p_n\) to the total variation of \(\overline{p}_{n}\). Since only the origin of \(p_{n}\) differs from the origin of \(\overline{p}_{n}\), we have \[[ p_n ]_{\mathop{\mathrm{TV}}(D)}\leq [ \overline{p}_n ]_{\mathop{\mathrm{TV}}(D)}+2d|p_n(e_1) - p_n(0)|.\] Again by Theorem 2(i), the second term is bounded above by \(cn^{-d\beta}\), and hence \[[ p_n ]_{\mathop{\mathrm{TV}}(D)}\leq cr^{d-1}n^{-\beta}+cn^{-d\beta}.\]

If \(r < n^{-\beta}\), then \(D=\left\{0\right\}\), and by definition \([p_{n}]_{\mathop{\mathrm{TV}}(D)}=0\). Otherwise \(cn^{-d\beta} = c(n^{-\beta})^{d-1}n^{-\beta} \le cr^{d-1}n^{-\beta}\) and the total variation of \(p_{n}\) is at most \(cr^{d-1}n^{-\beta}\), which implies the claimed bound. ◻

Proof. We remark that \(u \in L^{\infty}(Q_{T})\) implies that \(u, u^{m+1} \in L^{1}_{\mathop{\mathrm{loc}}}(Q_{T})\). Therefore all that remains to be shown is the integration by parts formula 60 . Let \(\varphi\in C^{\infty}_{c}(Q_{T})\). First, we assume \(\varphi\geq 0\). By choosing \(c = 0\) in 59 , we have \[\int_{\mathbb{R}^{d}}\int_{0}^{T}\left(u\varphi_{t} +u^{m+1}\frac{1}{2d}\Delta \varphi\right)\,dt\,dx \geq 0.\] Next, by choosing \(c = \|u\|_{L^{\infty}(Q_{T})}\) in 59 , we have \[\int_{\mathbb{R}^{d}}\int_{0}^{T}\left((\|u\|_{L^{\infty}(Q_{T})}-u)\varphi_{t} +\left(\|u\|_{L^{\infty}(Q_{T})}^{m+1}-u^{m+1}\right)\frac{1}{2d}\Delta \varphi\right)\,dt\,dx \geq 0,\] which implies \[\begin{align} \int_{\mathbb{R}^{d}}\int_{0}^{T}\left(u\varphi_{t} +u^{m+1}\frac{1}{2d}\Delta \varphi\right)\,dt\,dx &\leq \int_{\mathbb{R}^{d}}\int_{0}^{T}\left(\|u\|_{L^{\infty}(Q_{T})}\varphi_{t} +\|u\|_{L^{\infty}(Q_{T})}^{m+1}\frac{1}{2d}\Delta \varphi\right)\,dt\,dx\\ & = 0, \end{align}\] where we used the fundamental theorem of calculus and the compact support of \(\varphi\) in the last step. Hence, we conclude that 60 holds for for all \(\varphi\in C^{\infty}_{c}(Q_{T})\) with \(\varphi\geq 0\). To account for general \(\varphi\in C^{\infty}_{c}(Q_{T})\), we decompose \(\varphi\) into \(\varphi= \varphi^{+} - \varphi^{-}\) for \(\varphi^{+}, \varphi^{-} \geq 0\) and \(\varphi^{+}\varphi^{-} \equiv 0\). We note that \(\varphi^{+}\) and \(\varphi^{-}\) may not be smooth at their \(0\)-level sets. However, by approximating each of \(\varphi^{+}\) and \(\varphi^{-}\) by smooth functions, we can still conclude 60 holds for both \(\varphi\) replaced by \(\varphi^{+}\) and for \(\varphi\) replaced by \(\varphi^{-}\). Finally, recalling \(\varphi= \varphi^{+} - \varphi^{-}\), we obtain the desired result for \(\varphi\). ◻

Proof. First, assume that \[\label{e46sliceloc} \|w_{N} - z_{N}\|_{L^{1}(D\times (a,b))} \xrightarrow{N \to \infty} 0.\tag{61}\] Fix \(t \in [a,b)\) and \(\varepsilon\in (0,b-t)\). We calculate \[\begin{align} &\|w_{N}(\cdot,t) - z_{N}(\cdot,t)\|_{L^{1}(D)} \\ &= \frac{1}{\varepsilon}\int_{t}^{t+\varepsilon}\|w_{N}(\cdot,t) - z_{N}(\cdot,t)\|_{L^{1}(D)}\,ds \\ &\leq \frac{1}{\varepsilon}\int_{t}^{t+\varepsilon}\|w_{N}(\cdot,t) - w_{N}(\cdot,s)\|_{L^{1}(D)}\,ds + \frac{1}{\varepsilon}\|w_{N} - z_{N}\|_{L^{1}(D\times(t,t+\varepsilon))} \\ &\quad\quad\quad\quad\quad\quad\quad+\frac{1}{\varepsilon}\int_{t}^{t+\varepsilon}\|z_{N}(\cdot,s) - z_{N}(\cdot,t)\|_{L^{1}(D)}\,ds. \end{align}\] We bound the final term in the upper bound by observing that \[\frac{1}{\varepsilon}\int_{t}^{t+\varepsilon}\|z_{N}(\cdot,s) - z_{N}(\cdot,t)\|_{L^{1}(D)}\,ds \leq \frac{1}{\varepsilon}\int_{t}^{t+\varepsilon}\omega(s-t+\tfrac{1}{N})\,ds \leq \omega(\varepsilon+\tfrac{1}{N}).\] A similar calculation holds for the first term in the upper bound, so we have \[\|w_{N}(\cdot,t) - z_{N}(\cdot,t)\|_{L^{1}(D)} \leq 2 \omega(\varepsilon+\tfrac{1}{N}) + \frac{1}{\varepsilon}\|w_{N} - z_{N}\|_{L^{1}(D\times(t,t+\varepsilon))}.\] Taking \(N \to \infty\) in the above, we apply 61 to obtain that for every \(t\in [a,b)\), \[\limsup_{N \to \infty}\|w_{N}(\cdot,t) - z_{N}(\cdot,t)\|_{L^{1}(D)} \leq 2 \omega(\varepsilon).\] Lastly, sending \(\varepsilon\to 0\) allows us to conclude that \(\|w_{N}(\cdot,t) - z_{N}(\cdot,t)\|_{L^{1}(D)} \to 0\). In the case \(t = b\), we can repeat a similar argument by instead integrating from \(t - \varepsilon\) to \(t\).

For the other implication, fix \(n \in {\mathbb{N}}\) and a partition \(\{t_{i}\}_{i = 0}^{n}\) of \([a,b]\) with \(t_{0} = a\) and \(t_{i}-t_{i-1} = \frac{b-a}{n}\). We then have \[\begin{align} &\|w_{N} - z_{N}\|_{L^{1}(D\times (t_{i-1},t_{i}))} \\ &\leq \int_{t_{i-1}}^{t_{i}}\|w_{N}(\cdot,s) - w_{N}(\cdot,t_{i})\|_{L^{1}(D)}\,ds + \int_{t_{i-1}}^{t_{i}}\|w_{N}(\cdot,t_{i})-z_{N}(\cdot,t_{i})\|_{L^{1}(D)}\,ds \\ & \quad\quad\quad\quad\quad\quad\quad+\int_{t_{i-1}}^{t_{i}}\|z_{N}(\cdot,t_{i}) - z_{N}(\cdot,s)\|_{L^{1}(D)}\,ds \\ &\leq 2\left(\tfrac{b-a}{n}\right)\cdot\omega\left(\tfrac{b-a}{n} + \tfrac{1}{N}\right) + \tfrac{b-a}{n}\|w_{N}(\cdot,t_{i})-z_{N}(\cdot,t_{i})\|_{L^{1}(D)} . \end{align}\] Therefore, \[\begin{align} \|w_{N} - z_{N}\|_{L^{1}(D\times (a,b))} &= \sum_{i = 1}^{n} \|w_{N} - z_{N}\|_{L^{1}(D\times (t_{i-1},t_{i}))} \\ &\leq 2\left(b-a\right)\cdot\omega\left(\tfrac{b-a}{n}+\tfrac{1}{N}\right) + \tfrac{b-a}{n}\sum_{i=1}^{n}\|w_{N}(\cdot,t_{i})-z_{N}(\cdot,t_{i})\|_{L^{1}(D)} \end{align}\] If for all \(t \in [a,b]\), \(\|w_{N}(\cdot,t) - z_{N}(\cdot,t)\|_{L^{1}(D)} \xrightarrow{N \to \infty} 0\), then it follows from the above that \[\limsup_{N \to \infty}\|w_{N} - z_{N}\|_{L^{1}(D\times (a,b))} \leq 2\left(b-a\right)\cdot\omega\left(\tfrac{b-a}{n}\right),\] and so we conclude by taking \(n \to \infty\). ◻

Proof. It suffices to prove the corollary assuming \(I\) is compact; so write \(I=[a,b]\) for \(a,b \in \mathbb{R}\). Fix a compact \(D \subset \mathbb{R}^{d}\). Because \(w, z \in C([a,b];L^{1}(\mathbb{R}^{d}))\), both \(w\) and \(z\) have temporal moduli of continuity on \(\mathbb{R}^{d}\times[a,b]\). By Remark 33, without loss of generality, we can assume these moduli to be the same. Thus, we apply Lemma 44 to the constant sequences \(w_{N} \equiv w\) and \(z_{N} \equiv z\) to obtain \[\|w - z\|_{L^{1}(D \times [a,b])} = 0 \text{ if and only if } \|w(\cdot,t)-z(\cdot,t)\|_{L^{1}(D)}=0 \text{ for all } t \in [a,b].\] Because \(D\) and \([a,b]\) are arbitrary, the result follows. ◻

Proof. To prove (i), fix \(t>0\), let \(n = \left\lfloor t/\Delta_t\right\rfloor\), \(\Lambda:= \|v_{0}\|_{L^{\infty}(\mathbb{R}^{d})}\), and define \[N_{0} := (4(m+1)\Lambda^{m})^{\frac{1}{dm\beta}}.\] For \(N \geq N_{0}\), \(N\) satisfies ?? , and so Lemma 8(i, ii) gives that \[0 \leq v_{N}(\cdot, t) = \mathcal{S}^{n}_{N}v_{0} \leq \bigg\|\frac{1}{\square_{N}}\int_{\square_{N}(\cdot)} v_{0}(x)\, dx\bigg\|_{L^{\infty}(\mathbb{Z}^{d}\Delta_x)}\leq \Lambda= \|v_{0}\|_{L^{\infty}(\mathbb{R}^{d})}.\] Furthermore, choosing \(w_{N} := v_{N}(\cdot,0)\) and \(w'_{N} \equiv 0\) in Lemma 8(iii) gives that \[\|v_{N}(\cdot,t)\|_{L^{1}(\mathbb{R}^{d})} = \|\mathcal{S}^n_{N}(v_{N}(\cdot,0))\|_{L^{1}(\mathbb{R}^{d})} \leq \|v_{N}(\cdot,0)\|_{L^{1}(\mathbb{R}^{d})} = \|v_{0}\|_{L^{1}(\mathbb{R}^{d})},\] and so (i) is proved.

To prove (ii), let \(N \in {\mathbb{N}}\). We remark that directly by definition 15 , \(\mathcal{S}_{N}\) is translation invariant, meaning that for \(y \in \mathbb{R}^{d}\), \[\big[\mathcal{S}_{N}v_{N}(\cdot + y)\big](x) = \big[\mathcal{S}_{N}v_{N}\big](x+y).\] We use the translation invariance of \(\mathcal{S}_{N}\), along with Lemma 8(iii), to obtain that for all \(t \geq 0\), \(y \in \mathbb{R}^{d}\), \[\|v_{N}(\cdot + y, t) - v_{N}(\cdot,t)\|_{L^{1}(\mathbb{R}^{d})} \leq \|v_{0}(\cdot+ y) - v_{0}(\cdot)\|_{L^{1}(\mathbb{R}^{d})} \leq \nu(|y|).\]

To show (iii), we will use Kružkov’s interpolation lemma (Lemma 46). Let \(\varphi\in C_{c}^{\infty}(\mathbb{R}^{d})\), and fix \(n_{1}, n_{2} \in {\mathbb{N}}_{0}\). We will show that there is a constant \(C>1\) such that \[\left| \int_{\mathbb{R}^{d}} (v_{N}(x,n_{2}\Delta_t)-v_{N}(x,n_{1}\Delta_t)) \varphi(x) \, dx\right| \leq C\cdot\left(\sum_{i = 1}^{d} \|\partial_{x_{i}}^{2}\varphi\|_{L^{\infty}(\mathbb{R}^{d})}\cdot\right) |n_{2}\Delta_t-n_{1}\Delta_t|.\] First, we use the fact that \(v_{N}(\cdot,t)\) is \(\mathbb{Z}^{d}\)-piecewise constant for all \(t \geq 0\) to calculate \[\begin{align} &\int_{\mathbb{R}^{d}} (v_{N}(x,n_{2}\Delta_t)-v_{N}(x,n_{1}\Delta_t)) \varphi(x) \, dx \\ &= \sum_{k \in \mathbb{Z}^{d}} \int_{\square_{N}(k\Delta_x)}(v_{N}(x,n_{2}\Delta_t)-v_{N}(x,n_{1}\Delta_t)) \varphi(x) \, dx \\ &= \sum_{k \in \mathbb{Z}^{d}} (v_{N}(k\Delta_x,n_{2}\Delta_t)-v_{N}(k\Delta_x,n_{1}\Delta_t)) \int_{\square_{N}(k\Delta_x)}\varphi(x) \, dx. \end{align}\] Define \(\overline{\varphi}_{N}: \mathbb{Z}^{d}\Delta_x\to \mathbb{R}\) by \(\overline{\varphi}_{N}(k\Delta_x) := \int_{\square_{N}(k\Delta_x)}\varphi(x)\,dx\). Introducing the shorthand \(V^{n}_{k} := v_{N}(k\Delta_x,n\Delta_t)\), we can rewrite the last display as \[\int_{\mathbb{R}^{d}} (v_{N}(x,n_{2}\Delta_t)-v_{N}(x,n_{1}\Delta_t)) \varphi(x) \, dx = \sum_{k \in \mathbb{Z}^{d}} (V^{n_{2}}_{k}-V^{n_{1}}_{k})\overline{\varphi}_{N}(k\Delta_x).\] Assume without loss of generality that \(n_{2} \geq n_{1}\). By a telescoping sum, we can write \[\label{e46KIL1} \int_{\mathbb{R}^{d}} (v_{N}(x,n_{2}\Delta_t)-v_{N}(x,n_{1}\Delta_t)) \varphi(x) \,dx = \sum_{n=n_{1}}^{n_{2}-1}\sum_{k \in \mathbb{Z}^{d}} (V^{n+1}_{k}-V^{n}_{k})\overline{\varphi}_{N}(k\Delta_x).\tag{63}\] For a fixed \(n \in {\mathbb{N}}_{0}\), we use the scheme satisfied by \(v_{N}\) to calculate \[\begin{align} \label{e46beforeFTC} &\sum_{k \in \mathbb{Z}^{d}} (V^{n+1}_{k}-V^{n}_{k})\overline{\varphi}_{N}(k\Delta_x) \notag\\ &= \frac{\Delta_t}{2d(\Delta_x)^{2}} \sum_{k \in \mathbb{Z}^{d}} \sum_{i=1}^{d} \left[(V^{n}_{k+e_{i}})^{m+1}-2(V^{n}_{k})^{m+1}+(V^{n}_{k-e_{i}})^{m+1}\right]\overline{\varphi}_{N}(k\Delta_x) \notag\\ &= \frac{\Delta_t}{2d(\Delta_x)^{2}} \sum_{k \in \mathbb{Z}^{d}} \sum_{i=1}^{d} (V^{n}_{k})^{m+1}\left[\overline{\varphi}_{N}((k+e_{i})\Delta_x) - 2\overline{\varphi}_{N}(k\Delta_x) + \overline{\varphi}_{N}((k-e_{i})\Delta_x)\right], \end{align}\tag{64}\] where in the last step we used summation by parts and the fact that \(\varphi\) has compact support. For a fixed \(i \in [d]\), we remark that \[\begin{align} &|\overline{\varphi}_{N}((k+e_{i})\Delta_x) - 2\overline{\varphi}_{N}(k\Delta_x) + \overline{\varphi}_{N}((k-e_{i})\Delta_x)| \\ & \leq \int_{\square_{N}(k\Delta_x)} \left|\left(\varphi(x+e_{i}\Delta_x)-\varphi(x)\right) - \left(\varphi(x) - \varphi(x-e_{i}\Delta_x)\right)\right|\,dx\\ &= \int_{\square_{N}(k\Delta_x)}\bigg| \int_{0}^{\Delta_x}\left[\partial_{x_{i}}\varphi(x + s_{1}e_{i}) - \partial_{x_{i}}\varphi(x + (s_{1}-\Delta_x)e_{i})\right]\,ds_{1}\bigg| dx \\ &= \int_{\square_{N}(k\Delta_x)}\bigg|\int_{0}^{\Delta_x}\int_{0}^{\Delta_x}\partial_{x_{i}}^{2}\varphi(x + (s_{1}+s_{2}-\Delta_x)e_{i})\,ds_{1}\,ds_{2}\bigg|\, dx. \end{align}\] Using now that \(|\square_{N}(k\Delta_x)|=(\Delta_x)^{d}\), we conclude that \[|\overline{\varphi}_{N}((k+e_{i})\Delta_x) - 2\overline{\varphi}_{N}(k\Delta_x) + \overline{\varphi}_{N}((k-e_{i})\Delta_x)| \leq \|\partial_{x_{i}}^{2}\varphi\|_{L^{\infty}(\mathbb{R}^{d})}(\Delta_x)^{d+2}.\] Returning to 64 , we deduce that \[\begin{align} \left|\sum_{k \in \mathbb{Z}^{d}} (V^{n+1}_{k}-V^{n}_{k})\overline{\varphi}_{N}(k\Delta_x)\right| &\leq \frac{\Delta_t}{2d(\Delta_x)^{2}} \sum_{k \in \mathbb{Z}^{d}} \sum_{i=1}^{d}(V^{n}_{k})^{m+1}\|\partial_{x_{i}}^{2}\varphi\|_{L^{\infty}(\mathbb{R}^{d})}(\Delta_x)^{d+2} \\ &= \frac{\Delta_t}{2d} \cdot\left((\Delta_x)^{d}\sum_{k \in \mathbb{Z}^{d}} (V^{n}_{k})^{m+1}\right) \cdot \left( \sum_{i=1}^{d}\|\partial_{x_{i}}^{2}\varphi\|_{L^{\infty}(\mathbb{R}^{d})}\right). \end{align}\] Because \(V^{n}_{k}:= v_{N}(k\Delta_x,n\Delta_t)\) and \(v_{N}(\cdot,n\Delta_t)\) is \(\mathbb{Z}^{d}\)-piecewise constant, \[\begin{align} (\Delta_x)^{d}\sum_{k \in \mathbb{Z}^{d}} (V^{n}_{k})^{m+1} &= \int_{\mathbb{R}^{d}}(v_{N}(x,n\Delta_t))^{m}v_{N}(x,n\Delta_t) \, dx \\ &\leq \|v_{N}(\cdot,n\Delta_t)\|_{L^{\infty}(\mathbb{R}^{d})}^{m} \|v_{N}(\cdot,n\Delta_t)\|_{L^{1}(\mathbb{R}^{d})}. \end{align}\] By property (i) since \(N\geq N_{0}\), we combine the previous two displays to obtain a constant \(C_{v_{0}}\), where \(C_{v_{0}} = C_{v_{0}}(d,m,\|v_{0}\|_{L^{\infty}(\mathbb{R}^{d})}, \|v_{0}\|_{L^{1}(\mathbb{R}^{d})}) > 0\), such that \[\left|\sum_{k \in \mathbb{Z}^{d}} (V^{n+1}_{k}-V^{n}_{k})\overline{\varphi}_{N}(k\Delta_x)\right| \leq C_{v_{0}} \Delta_t\left( \sum_{i=1}^{d}\|\partial_{x_{i}}^{2}\varphi\|_{L^{\infty}(\mathbb{R}^{d})}\right).\] Thus, returning to 63 , we have \[\begin{align} \left|\int_{\mathbb{R}^{d}} (v_{N}(x,n_{2}\Delta_t)-v_{N}(x,n_{1}\Delta_t)) \varphi(x) \,dx \right| &= \sum_{n=n_{1}}^{n_{2}-1} \left|\sum_{k \in \mathbb{Z}^{d}} (V^{n+1}_{k}-V^{n}_{k})\overline{\varphi}_{N}(k\Delta_x)\right| \\ &\leq C_{v_{0}}|n_{2}\Delta_t-n_{1}\Delta_t| \left( \sum_{i=1}^{d}\|\partial_{x_{i}}^{2}\varphi\|_{L^{\infty}(\mathbb{R}^{d})}\right). \end{align}\] Next, an application of Kružkov’s interpolation lemma (Lemma 46), with the choice \(\varepsilon= |n_{2}\Delta_t-n_{1}\Delta_t|^{1/3}\), yields, for all \(n_{1},n_{2} \in {\mathbb{N}}\), \[\begin{align} \int_{\mathbb{R}^{d}} |v_{N}(x,n_{2}\Delta_t)-v_{N}(x,n_{1}\Delta_t)| \varphi(x) \,dx&\leq 2C_{v_{0}}\big(|n_{2}\Delta_t-n_{1}\Delta_t|^{1/3} + \nu\big(|n_{2}\Delta_t-n_{1}\Delta_t|^{1/3}\big) \big) \\ &:= \omega(|n_{2}\Delta_t-n_{1}\Delta_t|), \end{align}\] where we remark that \(\omega= \omega(d,m,\|v_{0}\|_{L^{\infty}(\mathbb{R}^{d})}, \|v_{0}\|_{L^{1}(\mathbb{R}^{d})},\nu)\).

To conclude, we consider times which are not on the grid \({\mathbb{N}_{0}}\Delta_t\). Let \(t_{1},t_{2} \in [0,T]\). We have, by setting \(n_{1} = \left\lfloor t_{1}/\Delta_t\right\rfloor\) and \(n_{2} = \left\lfloor t_{2}/\Delta_t\right\rfloor\), that \[|n_{2}\Delta_t- n_{1}\Delta_t| \leq |t_{2}-t_{1}| + 2\Delta_t.\] Because \(v_{N}\) is piecewise constant in time, we conclude that, \[\begin{align} \int_{\mathbb{R}^{d}} |v_{N}(x,t_{2})-v_{N}(x,t_{1})| \,dx &= \int_{\mathbb{R}^{d}} |v_{N}(x,n_{2}\Delta_t)-v_{N}(x,n_{1}\Delta_t)| \,dx\\ &\leq \omega(|n_{2}\Delta_t-n_{1}\Delta_t|) \\ &\leq \omega(|t_{2}-t_{1}| + 2\Delta_t)\\ &\leq \omega(2(|t_{2}-t_{1}| +\tfrac{1}{N})), \end{align}\] where we used the fact that \(\Delta_t= N^{-1}\) in the final step. ◻

Proof. Taking \(\Lambda= \|v_{0}\|_{\infty}\), there is an a.e. representative for \(v_{0}\) which satisfies \(v_{0}\in \mathcal{B}_{\Lambda}^{+}(\mathbb{R}^{d})\). Let \(A(u) := u^{m+1}\). Because \(\max_{u\in[0,\Lambda]}A'(u) = A'(\Lambda) = (m+1)\Lambda^{m}\), we see that the condition ?? is satisfied with \(\varepsilon= \frac{1}{2}\), so long as \[N^{\frac{dm}{dm+2}} \geq 4(m+1)\Lambda^{m},\] which is also a sufficient condition to guarantee monotonicity of \(\mathcal{S}_{N}\) (as seen in Lemma 8). Therefore, the hypotheses of Theorem 48 are satisfied and thus \(v_{N} \to v\) in \(L^{1}_{\text{loc}}(Q_{T})\).

Next, we show the final claim of \(L^{1}_{\mathop{\mathrm{loc}}}(\mathbb{R}^{d})\) convergence on time slices. By Lemma 47(iii), there is some \(N_{0} = N_{0}(m,d,\|v_{0}\|_{L^{\infty}(\mathbb{R}^{d})}) \in {\mathbb{N}}\) so that \(\{v_{N}\}_{N \geq N_{0}}\) admits an approximate temporal modulus of continuity \(\omega\) on \(\mathbb{R}^{d}\times[0,T]\). Because \(v\) is an \(L^{1}_{\text{loc}}(Q_{T})\)-limit point of \(\{v_{N}\}_{N \geq N_{0}}\), Lemma 42 gives that \(\omega\) is also a temporal modulus of continuity for \(v\). Thus, we apply Lemma 44 to the sequence \(w_{N} := v_{N}\) and the constant sequence \(z_{N} := v\) to conclude that we also have the convergence \(v_{N}(\cdot,t) \xrightarrow{L^{1}_{\text{loc}}(\mathbb{R}^{d})} v(\cdot,t)\), for all \(t \in [0,T]\). ◻

Proof.

(i) Let \(t \in [0,\infty)\). Because \(u_{N}(\cdot,t)\), is \(\mathbb{Z}^{d}\Delta_x\)-piecewise constant, \[\begin{align} \int_{\mathbb{R}^{d}}u_{N}(x,t)\,dx = (\Delta_x)^{d}\sum_{k \in \mathbb{Z}^{d}} u_{N}(k\Delta_x,t) = \sum_{k \in \mathbb{Z}^{d}} p(k, \left\lfloor t/\Delta_t\right\rfloor) = \sum_{k \in \mathbb{Z}^{d}}{\mathbf{P}}\left\{X^{\left\lfloor t/\Delta_t\right\rfloor} = k\right\} = 1. \end{align}\]

(ii) By Theorem 2[p46Linfty], there is a constant \(C = C(m,d) > 0\), such that \[0 \leq p(k,n) \leq Cn^{-d\beta} \text{ for all } k \in \mathbb{Z}^{d}, n \in {\mathbb{N}}.\] Fix \(t \in (0,\infty)\), \(x \in \mathbb{R}^{d}\), and integer \(N \geq 2/t\). Choose \(k \in \mathbb{Z}^{d}\) such that \(x \in \square_{N}(k\Delta_x)\) and let \(n = \left\lfloor t/\Delta_t\right\rfloor\). Since \(\Delta_x= (\Delta_t)^{\beta}\), we have \[0 \leq u_{N}(x,t) = (\Delta_t)^{-d\beta} p(k,n) \leq (\Delta_t)^{-d\beta} Cn^{-d\beta} = C (n\Delta_t)^{-d\beta}.\] We conclude using the fact that \(N \geq 2/t\), meaning \(\frac{1}{2} t \geq \Delta_t\), so \(n \Delta_t\geq t - \Delta_t\geq \frac{1}{2} t.\)

(iii) Since \(r > 2\sqrt{d}\Delta_x\), we have that \(r - \sqrt{d}\Delta_x> \frac{1}{2}r\), and thus, \[\int_{|x| >r} u_{N}(x,t)\,dx \leq (\Delta_x)^{d}\sum_{|k\Delta_x| > r - \sqrt{d}\Delta_x} u_{N}(k\Delta_x,t)\leq \sum_{|k\Delta_x| \geq \frac{1}{2} r}p(k,\left\lfloor t/\Delta_t\right\rfloor).\] Let \(n = \left\lfloor t/\Delta_t\right\rfloor\). Since \(\Delta_x= \Delta_t^{\beta}\), \[\sum_{|k\Delta_x| \geq \frac{1}{2} r}p(k,\left\lfloor t/\Delta_t\right\rfloor) = \sum_{|k| \geq \frac{1}{2} r\Delta_t^{-\beta}}p(k,n) = \sum_{|k| \geq \frac{1}{2} r(n\Delta_t)^{-\beta}n^{\beta}}p(k,n).\] Combining the previous two displays, we can apply Theorem 2[p46conc] to conclude that there is some constant \(C = C(m,d)>0\), such that: \[\int_{|x| >r} u_{N}(x,t)\,dx \leq \sum_{|k| \geq \frac{1}{2} r(n\Delta_t)^{-\beta}n^{\beta}}p(k,n) \leq C\exp(-r(n\Delta_t)^{-\beta}/C).\] We conclude by noting that \(\Delta_t^{-1}= N \geq 2/t\) means \(\frac{1}{2} t \geq \Delta_t\) and so \(n\Delta_t\geq t - \Delta_t\geq \frac{1}{2} t\).

(iv) By Theorem 2[p46BV], there is a constant \(C = C(m,d) > 0\) such that \[\label{e46pBValpha} [p(\cdot,n)]_{\mathop{\mathrm{TV}}(B_{\alpha}\cap\mathbb{Z}^{d})} \leq C\alpha^{d-1}n^{-d\beta} \text{ for all } \alpha > 0, n \in {\mathbb{N}}.\tag{65}\] Define the \(s\)-superlevel set of \(u_{N}\) as \(E^{s}:= \{x \in \mathbb{R}^{d} : u_{N}(x,t) > s\}\), and \(E^{s}_{r} := E^{s} \cap B_{r}\). Using one of the equivalent definitions for total variation, we have \[\label{e46TVdef2} [u_{N}(\cdot,t)]_{\mathop{\mathrm{TV}}(B_{r})} = \int_{-\infty}^{\infty} \mathcal{H}^{d-1}(\partial E^{s}_{r}) \, ds,\tag{66}\] where \(\mathcal{H}^{d-1}\) is the \((d-1)\)-dimensional Hausdorff measure (see for example [20]). Because \(u_{N}(\cdot,t)\) is \(\mathbb{Z}^{d}\Delta_x\)-piecewise constant, it follows that \[\partial E^{s}_{r} \subset \bigcup_{k \in \mathbb{Z}^{d}} \partial\square_{N}(k\Delta_x).\] For \(k \in \mathbb{Z}^{d}\), and \(\ell \sim k\), define \(\partial_{\ell}\square_{N}(k\Delta_x)\) to be the face of the \(d\)-hypercube \(\square_{N}(k\Delta_x)\) which has normal vector \(\ell-k\). We can re-write \[\bigcup_{k \in \mathbb{Z}^{d}} \partial\square_{N}(k\Delta_x) = \bigcup_{k \in \mathbb{Z}^{d}}\bigcup_{\ell \sim k} \partial_{\ell}\square_{N}(k\Delta_x).\] We remark that for every \(k \sim \ell\), \(\partial_{\ell}\square_{N}(k\Delta_x) = \partial_{k}\square_{N}(\ell\Delta_x)\). Hence, the right hand side union (up to a set of \(\mathcal{H}^{d-1}\) measure zero) exactly “double counts” each hypercube face. Therefore, we can re-write the integral 66 as \[[u_{N}(\cdot,t)]_{\mathop{\mathrm{TV}}(B_{r})} = \frac{1}{2}\sum_{k \in \mathbb{Z}^{d}} \sum_{\ell \sim k}\int_{-\infty}^{\infty} \mathcal{H}^{d-1}(\partial E^{s}_{r} \cap \partial_{\ell} \square_{N}(k\Delta_x)) \, ds.\] Since \(E^{s}_{r}:= E^{s} \cap B_{r} \subset B_{r}\), if \(|k\Delta_x| \geq 2r\) then \(|k\Delta_x|> r + 2\sqrt{d}\Delta_x\) and so for all \(\ell\sim k\) in this range, \(\partial E^{s}_{r} \cap \partial_{\ell} \square_{N}(k\Delta_x) = \emptyset.\) Thus, we can localize the sum over \(k\) in the above display as \[\begin{align} \label{e46TVupperbnd} [u_{N}(\cdot,t)]_{\mathop{\mathrm{TV}}(B_{r})} &= \frac{1}{2}\sum_{|k| < 2r(\Delta_x)^{-1}} \sum_{\ell\sim k}\int_{-\infty}^{\infty} \mathcal{H}^{d-1}(\partial E^{s}_{r} \cap \partial_{\ell} \square_{N}(k\Delta_x)) \, ds \notag\\ &\leq \frac{1}{2}\sum_{\substack{|k|,|\ell| < 4r(\Delta_x)^{-1} \\ \ell \sim k}} \int_{-\infty}^{\infty} \mathcal{H}^{d-1}(\partial E^{s} \cap \partial_{\ell} \square_{N}(k\Delta_x)) \, ds, \end{align}\tag{67}\] where we emphasize that we replaced \(E^{s}_{r}\) by \(E^{s}\) in the last step.

 Next, for a fixed $k \in \mathbb{Z}^{d}$ and $i \in [d]$, we
 analyze the term
 $\int_{-\infty}^{\infty} \mathcal{H}^{d-1}(\partial E^{s} \cap \partial_{k+e_{i}} \square_{N}(k\Delta_x)) \, ds$.
 Define
 $$\begin{align} s_{\min} &:= \min\{u_{N}(k\Delta_x,t), u_{N}((k+e_{i})\Delta_x,t)\}\\ s_{\max} &:= \max\{u_{N}(k\Delta_x,t), u_{N}((k+e_{i})\Delta_x,t)\},
 \end{align}$$ so $u_{N}(\cdot,t)$ has a jump discontinuity of size
 $s_{\max}-s_{\min}$ across the face
 $\partial_{k+e_{i}} \square_{N}(k\Delta_x)$. Because of this jump
 discontinuity, we have
 $$\partial E^{s} \cap \partial_{k+e_{i}} \square_{N}(k\Delta_x) = \begin{cases} \partial_{k+e_{i}} \square_{N}(k\Delta_x) &\text{ for } s \in (s_{\min},s_{\max}]\\ \emptyset &\text{ otherwise}. \end{cases}$$
 Hence, we calculate that
 $$\begin{align} \int_{-\infty}^{\infty} \mathcal{H}^{d-1}(\partial E^{s} \cap \partial_{k+e_{i}} \square_{N}(k\Delta_x))\,ds &= \int_{s_{\min}}^{s_{\max}} \mathcal{H}^{d-1}(\partial_{k+e_{i}} \square_{N}(k\Delta_x))\,ds \\ &= (s_{\max}-s_{\min})\mathcal{H}^{d-1}(\partial_{k+e_{i}} \square_{N}(k\Delta_x)) \\ &= |u_{N}((k+e_{i})\Delta_x,t) - u_{N}(k\Delta_x,t)| (\Delta_x)^{d-1} \\ &= (\Delta_x)^{-1} |p(k+e_{i},\left\lfloor t/\Delta_t\right\rfloor) - p(k,\left\lfloor t/\Delta_t\right\rfloor)|.
 \end{align}$$ Letting $n = \left\lfloor t/\Delta_t\right\rfloor$,
 we combine the above display with @eq:e46TVupperbnd , letting
 $\ell=k\pm e_{i}$, to obtain that
 $$[u_{N}(\cdot,t)]_{\mathop{\mathrm{TV}}(B_{r})} \leq (\Delta_x)^{-1} [p(\cdot,n)]_{\mathop{\mathrm{TV}}(B_{4r(\Delta_x)^{-1}}\cap\mathbb{Z}^{d})}.$$
 By applying @eq:e46pBValpha and $\Delta_x= (\Delta_t)^{\beta}$,
 this implies that
 $$[u_{N}(\cdot,t)]_{\mathop{\mathrm{TV}}(B_{r})} \leq (\Delta_x)^{-1} C (r (\Delta_x)^{-1})^{d-1}n^{-d\beta} = C r^{d-1}(n\Delta_t)^{-d\beta}.$$
 Finally, we conclude by again using that $n \Delta_t\ge t/2$.

 ◻

Proof. Let \(t \in [\eta,T]\). We first show that \(u_{N}(\cdot,t)\) admits a spatial modulus of continuity \(\nu_{\eta}\). Let \(|y| \leq 1\) and \(r \geq 4\). Then by partitioning \(\mathbb{R}^{d}\), \[\begin{gather} \label{e46uNsmod} \|u_{N}(\cdot + y,t) - u_{N}(\cdot,t)\|_{L^{1}(\mathbb{R}^{d})} = \|u_{N}(\cdot+y,t)-u_{N}(\cdot,t)\|_{L^{1}(B_{\frac{r}{2}})} \\+ \|u_{N}(\cdot+y,t)-u_{N}(\cdot,t)\|_{L^{1}(B_{\frac{r}{2}}^{c})}. \end{gather}\tag{68}\] We begin by bounding the second term of 68 . Note that for \(x \in \mathbb{R}^{d}\) with \(|x| > r/2\), we have \(|x+y| \geq |x|-|y| > r/2 - 1 \geq r/4\). Because \(u_{N}\) is nonnegative, Proposition 51[uN46conc] gives that there is a \(C = C(m,d) > 0\) such that \[\label{e46uNsmod1} \int_{|x| > r/2}|u_{N}(x+y,t)-u_{N}(x,t)|\,dx \leq 2\int_{|x| > r/4}u_{N}(x,t)\,dx \leq C \exp\left(-r\eta^{-\beta}/C\right),\tag{69}\] where we used \(t \geq \eta\) in the last step. (Note that Proposition 51[uN46conc] requires that \(N>2/t\), which is satisfied since \(N>2/\eta\).) Next, we bound the first term in the left hand side of 68 . Note that for \(x \in \mathbb{R}^{d}\) with \(|x| < r/2\) and since \(|y|<1\), we have \(|x+y| < r\). By Proposition 51[uN46BV], there exists a constant \(C = C(m,d) > 0\) such that \([u_{N}(\cdot,t)]_{\mathop{\mathrm{TV}}(B_{r})} \leq Ct^{-d\beta}r^{d-1} \leq C\eta^{-d\beta}r^{d-1}\), so by Proposition 51(i), we conclude that \(u_{N}(\cdot,t) \in \mathop{\mathrm{BV}}(B_{r})\). Therefore, we can approximate \(u_{N}(\cdot,t)\) by a sequence of smooth functions \(\{\varphi^{i}\}_{i \in {\mathbb{N}}} \subset C^{\infty}(B_{r})\) (see, for example, [20]) such that \[\lim_{i\to \infty}\|\varphi^{i}-u_{N}(\cdot,t)\|_{L^{1}(B_{r})} = 0 \text{ and } \lim_{i\to \infty}\|D\varphi^{i}\|_{L^{1}(B_{r})} = [u_{N}(\cdot,t)]_{\mathop{\mathrm{TV}}(B_{r})},\] where \(D\varphi\) is the gradient of \(\varphi\). Note that for all \(i \in {\mathbb{N}}\), \[\begin{align} \|\varphi^{i}(\cdot+y)-\varphi^{i}\|_{L^{1}(B_{r/2})} &= \int_{B_{r/2}}\left|\int_{0}^{1}D\varphi^{i}(x+sy)\cdot y\,ds\right|\,dx \\ &\leq |y| \int_{B_{r/2}}\int_{0}^{1}\left|D\varphi^{i}(x+sy)\right|\,ds\,dx \\ &= |y| \int_{0}^{1}\|D\varphi^{i}\|_{L^{1}(B_{r/2}(sy))}\,ds \\ &\leq |y| \cdot \|D\varphi^{i}\|_{L^{1}(B_{r})}, \end{align}\] where in the last step we used that \(B_{r/2}(sy) \subset B_{r}\) because \(|sy| \leq |y| \leq 1 \leq r/2\). Taking \(i \to \infty\) then yields \[\|u_{N}(\cdot+ y,t) - u_{N}(\cdot,t)\|_{L^{1}(B_{r/2})} \leq [u_{N}(\cdot,t)]_{\mathop{\mathrm{TV}}(B_{r})}\cdot |y| \leq C\eta^{-d\beta}r^{d-1}|y|.\] Combining this with 69 , we return to 68 to obtain that for all \(|y| \leq 1\) and \(r \geq 4\), \[\|u_{N}(\cdot + y,t) - u_{N}(\cdot,t)\|_{L^{1}(\mathbb{R}^{d})}\leq C\eta^{-d\beta}r^{d-1}|y| + C \exp\left(-r\eta^{-\beta}/C\right).\] Choosing \(r = 4|y|^{-1/d}\) in the above gives \[\|u_{N}(\cdot + y,t) - u_{N}(\cdot,t)\|_{L^{1}(\mathbb{R}^{d})}\leq C\eta^{-d\beta}|y|^{1/d} + C \exp\left(-|y|^{-1/d}\eta^{-\beta}/C\right).\] Because the above bound tends to \(0\) as \(|y| \to 0\), we may define the spatial modulus \(\nu_{\eta}\) as \[\nu_{\eta}(|y|) = \begin{cases} 0 & |y|=0\\ C\eta^{-d\beta}|y|^{1/d} + C \exp\left(-|y|^{-1/d}\eta^{-\beta}/C\right) & 0 < |y| \leq 1\\ \max\left\{C\eta^{-d\beta} + C \exp\left(-\eta^{-\beta}/C\right), 2\right\} &|y|> 1. \end{cases}\] To obtain an approximate temporal modulus of continuity \(\omega_{\eta}\) on \(\mathbb{R}^{d} \times[\eta,T]\), we appeal to Kružkov’s interpolation lemma (Lemma 46) in a manner similar to Lemma 47(iii) albeit only for times in the interval \([\eta,T]\). The only difference is that that proof requires an initial condition \(v_{0} \in L^{\infty}(\mathbb{R}^{d})\) to obtain bounds of the form \[\|v_{N}(\cdot,t)\|_{L^{\infty}(\mathbb{R}^{d})} \leq \|v_{0}\|_{L^{\infty}(\mathbb{R}^{d})} \text{ for } t > 0.\] We can avoid such an assumption for \((u_{N})_{N \geq 2/\eta}\) by using Proposition 51[uN46Linfty] to replace these bounds by bounds of the form \[\|u_{N}(\cdot,t)\|_{L^{\infty}(\mathbb{R}^{d})}\leq Ct^{-d\beta} \leq C\eta^{-d\beta} \text{ for }t \in [\eta,T].\qedhere\] ◻

Proof. We remark that \(u_{N_{i}} \geq 0\), and so \(w_{t} \geq 0\) almost everywhere. We first prove ?? . Applying the hypotheses and Proposition 51(i), we have \[\int_{\mathbb{R}^{d}} w_{t}(x) \, dx = \lim_{r\to \infty} \int_{|x| \leq r} w_{t}(x) \, dx = \lim_{r\to \infty} \lim_{i\to\infty} \int_{|x| \leq r} u_{N_{i}}(x,t) \, dx \leq \lim_{r\to \infty} \lim_{i\to\infty} 1 = 1.\] Next, we use Proposition 51 (statements (i) and (iii)) to obtain that for all \(r > 0\), \[\begin{align} \label{e46wbulk} \int_{|x| \leq r} w_{t}(x) \,dx &= \lim_{i\to\infty} \int_{|x| \leq r} u_{N_{i}}(x,t) \, dx \notag\\ &= \lim_{i \to \infty}\left(1 - \int_{|x|> r} u_{N_{i}}(x,t) \, dx \right) \notag\\ &\geq 1 - C\exp(-rt^{-\beta}/C), \end{align}\tag{70}\] where we remark that the hypotheses \(N _{i}> 2/t\) and \(r > 2\sqrt{d}\Delta_x\) are satisfied for all \(N_{i}\) large enough. Taking \(r \to \infty\) in the above gives that \(\int_{\mathbb{R}^{d}} w_{t}(x)\,dx \geq 1\) and so ?? is proved. We note that ?? follows directly from ?? and 70 .

For the convergence in \(L^{1}(\mathbb{R}^{d})\), choose \(r \geq 2\sqrt{d}(2/t)^{-\beta}\) so that \(r \geq 2\sqrt{d}\Delta_x\) for all \(N \geq 2/t\). We combine Proposition 51[uN46conc] and ?? to conclude that for \(N_{i} \geq 2/t\), \[\begin{align} \|u_{N_{i}}(\cdot,t) - w_{t}\|_{L^{1}(\mathbb{R}^{d})} &= \|u_{N_{i}}(\cdot,t) - w_{t}\|_{L^{1}(B_{r})} + \|u_{N_{i}}(\cdot,t) - w_{t}\|_{L^{1}(B^{c}_{r})} \\ &\leq \|u_{N_{i}}(\cdot,t) - u^{\eta}(\cdot,t)\|_{L^{1}(B_{r})} + 2C\exp\left(-rt^{-\beta}/C\right). \end{align}\] Taking \(N_{i}\to \infty\) followed by \(r \to \infty\) in the above gives \(u_{N_{i}}(\cdot,t) \to w_{t}\) in \(L^{1}(\mathbb{R}^{d})\). ◻

Proof. Let \(N_{\eta} \in {\mathbb{N}}\) be such that \(N_{\eta} \geq \frac{2}{\eta}\). Let \(\nu_{\eta}\) and \(\omega_{\eta}\) be respectively a spatial modulus of continuity and an approximate temporal modulus of continuity for \((u_{N})_{N \geq N_{\eta}}\) on \(\mathbb{R}^{d} \times [\eta,T]\), the existence of which is guaranteed by Lemma 52. The existence of these moduli of continuity, along with Proposition 51 (statements (i) and (ii)), entail that \((u_{N})_{N \geq N_{\eta}}\) satisfies the conditions of Lemma 42 on the domain \(\mathbb{R}^{d} \times [\eta,T]\). Therefore, there is a subsequence \((u_{N^{\eta}_{n}})_{n \in {\mathbb{N}}}\) such that \(u_{N^{\eta}_{n}} \to u^{\eta}\) in \(L^{1}_{\mathop{\mathrm{loc}}}(\mathbb{R}^{d} \times [\eta,T])\) for some \(u^{\eta} \in L^{1}(\mathbb{R}^{d}\times[\eta,T])\cap L^{\infty}( \mathbb{R}^{d}\times[\eta,T])\cap C([\eta,T];L^{1}(\mathbb{R}^{d}))\). To simplify the notation, we drop the subsequence \((u_{N^{\eta}_{n}})_{n\in{\mathbb{N}}}\) and assume without loss of generality that \(u^{\eta}\) is a limit point of the entire sequence \((u_{N})_{N \geq N_{\eta}}\). The conclusion of Lemma 42 also yields that \(u^{\eta}\) has \(\nu_{\eta}\) and \(\omega_{\eta}\) as a spatial and temporal modulus of continuity, respectively. Hence, we apply Lemma 44 to the sequence \((u_{N})_{N\geq N_{\eta}}\) and the constant sequence \((u^{\eta})_{N \in {\mathbb{N}}}\), deducing that \[\label{e46uNetaslice} \text{for all }t\in [\eta, T], u_{N}(\cdot,t) \xrightarrow{L^{1}_{\mathop{\mathrm{loc}}}(\mathbb{R}^{d})}u^{\eta}(\cdot,t).\tag{71}\]

Now we show that \(u^{\eta}\) is a nonnegative entropy solution. By Proposition 51[uN46Linfty], there is a constant \(C = C(m,d) > 0\) such that \[\label{e46uNbdd} \text{for all }t\in [\eta,T], \quad0 \leq u_{N}(\cdot,t) \leq Ct^{-d\beta}.\tag{72}\] The convergence 71 then implies that \[\label{e46uetabdd} \text{for all }t\in [\eta, T], \quad0 \leq u^{\eta}(\cdot,t) \leq Ct^{-d\beta} \text{ a.e.\@ in } \mathbb{R}^{d}.\tag{73}\] Therefore, \(u^{\eta}(\cdot,\eta)\) is nonnegative and belongs to \(L^{\infty}(\mathbb{R}^{d})\), so by Theorem 48 (in particular, the last part of Remark 49), there exists a unique nonnegative entropy solution \(v\) solving \[\label{e46veta} \begin{cases} v_{t} - \frac{1}{2d}\Delta(v^{m+1}) = 0 &\text{in } \mathbb{R}^{d} \times (0,T-\eta) \\ v(\cdot,0) = u^{\eta}(\cdot,\eta) &\text{in } \mathbb{R}^{d}. \end{cases}\tag{74}\] By Remark 41, if we can show that \(u^{\eta}(x,t) = v(x,t-\eta)\) for a.e. \((x,t) \in \mathbb{R}^{d}\times(\eta,T)\), we can conclude that \(u^{\eta}\) is itself a nonnegative entropy solution as desired.

Let \(v_{N}\) be generated by \(\mathcal{S}_{N}\) with initial condition \(u^{\eta}(\cdot,\eta)\). Because of 73 , we can apply the convergence result for diffuse initial conditions, Corollary 50, to obtain that \[\label{e46vNetaslice} v_{N}(\cdot,t) \xrightarrow{L^{1}_{\mathop{\mathrm{loc}}}(\mathbb{R}^{d})}v(\cdot,t)\text{ for }t \in [0,T-\eta].\tag{75}\] Since \(u^{\eta}(\cdot,\eta)\) has a spatial modulus of continuity, Lemma 47(iii) implies that \((v_{N})_{N \geq N_{\eta}}\) has an approximate temporal modulus of continuity, possibly after increasing \(N_{\eta}\) depending on \(\|u^{\eta}(\cdot,\eta)\|_{L^{\infty}(\mathbb{R}^{d})}\). Without loss of generality (by Remark 33), we take this approximate temporal modulus of continuity to be \(\omega_{\eta}\).

Next, let \(D \subset \mathbb{R}^{d}\) be compact. For \(t \in [\eta,T]\), we calculate \[\begin{align} &\|u^{\eta}(\cdot,t) - v(\cdot,t-\eta)\|_{L^{1}(D)} \\ &\leq \| (u^{\eta} - u_{N})(\cdot,t)\|_{L^{1}(D)} + \|u_{N}(\cdot,t) - v_{N}(\cdot,t-\eta) \|_{L^{1}(D)} + \|(v_{N} - v)(\cdot,t-\eta)\|_{L^{1}(D)}. \end{align}\] By 71 and 75 , we have \[\label{e46uetav} \|u^{\eta}(\cdot,t) - v(\cdot,t-\eta)\|_{L^{1}(D)} \leq \limsup_{N\to\infty}\|u_{N}(\cdot,t) - v_{N}(\cdot,t-\eta)\|_{L^{1}(D)}.\tag{76}\] To analyze this upper bound, fix \(N \geq N_{\eta}\) and let \(\ell_{N} = \left\lfloor(t-\eta)/\Delta_t\right\rfloor\). Because \(\|v_{0}\|_{L^{\infty}(\mathbb{R}^{d})} = \|u^{\eta}(\cdot,\eta)\|_{L^{\infty}(\mathbb{R}^{d})} \leq C\eta^{-d\beta}\), Lemma 47(i) gives that \(\|v_{N}(\cdot,t)\|_{L^{\infty}(\mathbb{R}^{d})} \leq C\eta^{-d\beta}\) for all \(t \in [0,T-\eta]\). Furthermore, we have that \(\|u_{N}(\cdot,t)\|_{L^{\infty}(\mathbb{R}^{d})} \leq C\eta^{-d\beta}\) by 72 . Hence, by possibly taking \(N_{\eta}\) larger so that \(N \ge N_{\eta}\) satisfies ?? with \(\Lambda := C\eta^{-d\beta}\), we apply Lemma 8(iii) to conclude that \[\begin{align} \tag{77} &\|u_{N}(\cdot,t) - v_{N}(\cdot,t-\eta)\|_{L^{1}(D)} \\ \notag &\leq \|u_{N}(\cdot,t) - v_{N}(\cdot,t-\eta)\|_{L^{1}(\mathbb{R}^{d})} \\\notag &= \|\mathcal{S}^{\ell_{N}}_{N}u_{N}(\cdot - \ell_{N}\Delta_t)- \mathcal{S}^{\ell_{N}}_{N}v_{N}(\cdot, 0)\|_{L^{1}(\mathbb{R}^{d})} \\\notag & \leq \|u_{N}(\cdot, t-\ell_{N}\Delta_t) - v_{N}(\cdot, 0)\|_{L^{1}(\mathbb{R}^{d})} \\\ & \leq \|u_{N}(\cdot, t-\ell_{N}\Delta_t) - u_{N}(\cdot,\eta)\|_{L^{1}(\mathbb{R}^{d})} + \|u_{N}(\cdot,\eta) - v_{N}(\cdot, 0)\|_{L^{1}(\mathbb{R}^{d})}.\tag{78} \end{align}\] For the second term in the upper bound, the fact that \(u_{N}\) and \(v_{N}\) are piecewise constant gives \[\|u_{N}(\cdot,\eta) - v_{N}(\cdot,0)\|_{L^{1}(\mathbb{R}^{d})} = (\Delta_x)^{d} \sum_{k \in \mathbb{Z}^{d}} |u_{N}(k\Delta_x,\eta) - v_{N}(k\Delta_x,0)|.\] Since \(u_{N}\) is piecewise constant and \(v_{N}\) has initial condition \(u^{\eta}(0,\eta)\), we have \[u_{N}(k\Delta_x,\eta) = \frac{1}{|\square_{N}|}\int_{\square_{N}}u_{N}(y+k\Delta_x,\eta)\,dy , \text{and } v_{N}(k\Delta_x,0)=\frac{1}{|\square_{N}|}\int_{\square_{N}}u^{\eta}(y+k\Delta_x,\eta)\,dy.\] Combining the two displays above, we obtain \[\begin{align} &\|u_{N}(\cdot,\eta) - v_{N}(\cdot,0)\|_{L^{1}(\mathbb{R}^d)} \\&= (\Delta_x)^{d} \sum_{k \in \mathbb{Z}^{d}} \left|\frac{1}{|\square_{N}|}\int_{\square_{N}}u_{N}(y+k\Delta_x,\eta)\,dy - \frac{1}{|\square_{N}|}\int_{\square_{N}}u^{\eta}(y+k\Delta_x,\eta)\right| \\&= \sum_{k \in \mathbb{Z}^{d}} \left|\int_{\square_{N}} \bigg(u_{N}(y+k\Delta_x,\eta) - u^{\eta}(y+k\Delta_x,\eta)\bigg)\,dy\right| \\ &\leq \sum_{k \in \mathbb{Z}^{d}} \int_{\square_{N}} \left|u_{N}(y+k\Delta_x,\eta) - u^{\eta}(y+k\Delta_x,\eta)\right| \,dy\\ &= \int_{\mathbb{R}^{d}} \left|u_{N}(y,\eta) - u^{\eta}(y,\eta)\right| \,dy \to 0, \end{align}\] where in the last step, we use the final statement of Lemma 53 to bootstrap the convergence \(u_{N}(\cdot,\eta) \to u^{\eta}(\cdot,\eta)\) in \(L^{1}_{\mathop{\mathrm{loc}}}(\mathbb{R}^{d})\) to convergence in \(L^{1}(\mathbb{R}^{d})\).

For the first term in 78 , we use \(\omega_{\eta}\), an approximate temporal modulus of continuity for \((u_{N})_{N\geq\eta}\) on \(\mathbb{R}^{d}\times[\eta,T]\), to obtain \[\|u_{N}(\cdot,t-\ell_{N}\Delta_t) - u_{N}(\cdot,\eta)\|_{L^{1}(\mathbb{R}^{d})} \leq \omega_{\eta}(|t-\ell_{N}\Delta_t- \eta| + \tfrac{1}{N}) \xrightarrow{N \to \infty} 0,\] where we used that \(\ell_{N}\Delta_t= \left\lfloor(t-\eta)/\Delta_t\right\rfloor\Delta_t\xrightarrow{N \to \infty} (t-\eta)\). Therefore, combining the previous two displays, we return to 76 to obtain \[\|u^{\eta}(\cdot,t) - v(\cdot,t-\eta)\|_{L^{1}(D)} = 0 \text{ for all }t \in [\eta,T].\] Since \(D \subset \mathbb{R}^{d}\) was arbitrary, \(u^{\eta}(x,t) = v(x,t-\eta)\) for a.e. \((x,t) \in \mathbb{R}^{d}\times(\eta,T)\). Thus we conclude that \(u^{\eta}\) is a nonnegative entropy solution of the porous medium equation with initial condition \(\delta\). ◻

Proof. In this proof, we frequently work with restrictions of the functions \(u_N\) to a subdomain of the form \(\mathbb{R}\times I\) for some interval \(I\). We abuse notation by continuing to refer to such restrictions as \(u_N\), rather than writing \(u_N|_{\mathbb{R}\times I}\). Now that this convention has been pointed out, it should cause no confusion.

Fix any increasing sequence of positive integers \((M_i)_{i \ge 1}\). We first use a diagonalization argument to show that there exists a subsequence \((N_i)_{i \in{\mathbb{N}}}\) of \((M_i)_{i \in {\mathbb{N}}}\) such that \[\label{e46utildeconv} u_{N_{i}} \xrightarrow[]{L^{1}_{\mathop{\mathrm{loc}}}(Q_{T})} \tilde{u} \text{ and, for all }t \in (0,T], u_{N_{i}}(\cdot,t) \xrightarrow[]{L^{1}_{\mathop{\mathrm{loc}}}(\mathbb{R}^{d})} \tilde{u}(\cdot,t),\tag{79}\] where \(\tilde{u} \in L^{1}_{\mathop{\mathrm{loc}}}(Q_{T})\) is a distributional solution to \(\tilde{u}_{t} - \frac{1}{2d}\Delta(\tilde{u}^{m+1}) = 0\). To this end, fix a decreasing sequence \((\eta_{i})_{i\in{\mathbb{N}}} \subset (0,T)\) with \(\lim_{i\to\infty}\eta_{i} = 0\). By applying Lemma 54 for each \(\eta_{i}\), we have that, for each \(i \in {\mathbb{N}}\), there exists an increasing sequence \((N^{\eta_{i}}_{j})_{j \in {\mathbb{N}}} \subset {\mathbb{N}}\) such that, as \(j \to \infty\), \[u_{N^{\eta_{i}}_{j}} \xrightarrow{L^{1}_{\mathop{\mathrm{loc}}}(\mathbb{R}^{d}\times(\eta_{i},T))} u^{\eta_{i}} \text{ and, for all }t \in [\eta_{i},T], u_{N^{\eta_{i}}_{j}}(\cdot,t) \xrightarrow{L^{1}_{\mathop{\mathrm{loc}}}(\mathbb{R}^{d})} u^{\eta_{i}}(\cdot,t),\] where \(u^{\eta_{i}}\) is an entropy solution to \(u^{\eta_{i}}_{t} - \frac{1}{2d}\Delta((u^{\eta_{i}})^{m+1}) = 0\) in \(\mathbb{R}^{d}\times(\eta_{i},T)\). Furthermore, these subsequences may be constructed iteratively so that \((N^{\eta_{i+1}}_{j})_{j \in {\mathbb{N}}}\) is a subsequence of \((N^{\eta_{i}}_{j})_{j \in {\mathbb{N}}}\) for all \(i \in {\mathbb{N}}\). It follows that for all \(i \in {\mathbb{N}}\), \[u_{N^{\eta_{i+1}}_{j}} \xrightarrow[j\to\infty]{L^{1}_{\mathop{\mathrm{loc}}}(\mathbb{R}^{d}\times(\eta_{i},T))}u^{\eta_{i}}\quad\text{ and }\quad u_{N^{\eta_{i+1}}_{j}} \xrightarrow[j\to\infty]{L^{1}_{\mathop{\mathrm{loc}}}(\mathbb{R}^{d}\times(\eta_{i+1},T))}u^{\eta_{i+1}} .\] This means that \[\label{e46domainex} u^{\eta_{i+1}} \overset{a.e.}{=} u^{\eta_{i}}\text{ in }\mathbb{R}^{d} \times (\eta_{i},T).\tag{80}\] In other words, \(u^{\eta_{i+1}}\) extends \(u^{\eta_{i}}\) to the larger domain \(\mathbb{R}^{d}\times(\eta_{i+1},T)\) (up to a set of measure zero). Next, we define \(\tilde{u}: \mathbb{R}^{d} \times (0,T]\to \mathbb{R}\) as the a.e. limit of these domain extensions, meaning \[\tilde{u}(x,t) := \begin{cases} u^{\eta_{0}}(x,t) &\text{ if } \eta_{0} \leq t \\ u^{\eta_{i}}(x,t) &\text{ for } t \in [\eta_{i+1},\eta_{i}), i \in {\mathbb{N}}. \end{cases}\]

We now show that \(\tilde{u} \in C((0,T];L^{1}(\mathbb{R}^{d}))\). Because \(u^{\eta_{i}} \in C([\eta_{i},T];L^{1}(\mathbb{R}^{d}))\) for every \(i \in {\mathbb{N}}\), we combine 80 with Corollary 45 to conclude that for all \(i \in {\mathbb{N}}\), \(t \geq \eta_{i}\) and \(j \geq i\), \(u^{\eta_{j}}(\cdot,t) \overset{a.e.}{=} u^{\eta_{i}}(\cdot,t)\). Thus, for all \(i \in {\mathbb{N}}\), and \(t \geq \eta_{i}\), \(\tilde{u}(\cdot,t) \overset{a.e.}{=} u^{\eta_{i}}(\cdot,t)\), and so \(\tilde{u} \in C([\eta_{i},T];L^{1}(\mathbb{R}^{d}))\). Therefore, because \(\eta_{i}\to 0\), \(\tilde{u} \in C((0,T];L^{1}(\mathbb{R}^{d}))\).

Next, note that for all \(i \in {\mathbb{N}}\), using 80 and the definition of \(\tilde{u}\), we have that \(\tilde{u} \overset{a.e.}= u^{\eta_{i}}\) in \(\mathbb{R}^{d} \times (\eta_{i},T)\). Therefore, for all \(i \in {\mathbb{N}}\), \(\tilde{u}\) is an entropy solution of the porous medium equation in \(\mathbb{R}^{d} \times (\eta_{i},T)\), which means, by Proposition 39 that for all \(i \in {\mathbb{N}}\), \(\tilde{u}\) is a distributional solution in \(\mathbb{R}^{d} \times (\eta_{i},T)\). Hence, because \(\eta_{i} \to 0\), \(\tilde{u}\) is a distributional solution in \(Q_{T}\). Finally, setting \(N_{i}:=N^{\eta_{i}}_{i}\), the sequence of functions \((u_{N_i})_{i \in {\mathbb{N}}}\) then satisfies 79 .

Next, we show that \(\tilde{u}\) is a.e. equal to the Barenblatt solution \(\bar{u}\). Let \(\varphi\in C_{c}(\mathbb{R}^{d})\). Because \(\varphi\) is compactly supported and continuous, it is uniformly continuous and thus admits a modulus of continuity, which we denote by \(\rho\). Fix \(t \in (0,T]\) and \(\varepsilon> 0\). By Lemma 53, \[\int_{\mathbb{R}^{d}} \tilde{u}(x,t) \, dx = 1 \quad \text{ and }\quad \int_{|x| > \varepsilon} \tilde{u}(x,t) \,dx \leq C\exp(-\varepsilon t^{-\beta}/C).\] This implies \[\begin{align} \left|\int_{\mathbb{R}^{d}}\tilde{u}(x,t)\varphi(x)\,dx - \varphi(0) \right|&= \left|\int_{\mathbb{R}^{d}}\tilde{u}(x,t)(\varphi(x)-\varphi(0))\,dx \right| \\ &\leq \int_{|x| \leq \varepsilon}\tilde{u}(x,t)|\varphi(x)-\varphi(0)|\,dx + \int_{|x| > \varepsilon}\tilde{u}(x,t)|\varphi(x)-\varphi(0)|\,dx \\ &\leq \rho(\varepsilon) \int_{|x| \leq \varepsilon}\tilde{u}(x,t)\,dx + 2 \|\varphi\|_{L^{\infty}(\mathbb{R}^{d})} \int_{|x| > \varepsilon}\tilde{u}(x,t)\,dx \\ &\leq \rho(\varepsilon) + 2 \|\varphi\|_{L^{\infty}(\mathbb{R}^{d})} C\exp(-\varepsilon t^{-\beta}/C). \end{align}\] Taking \(t \to 0^{+}\), followed by \(\varepsilon\to 0^{+}\) in the above display gives \[\lim_{t\to0^{+}}\int_{\mathbb{R}^{d}}\tilde{u}(x,t)\varphi(x)\,dx = \varphi(0),\] and hence \(\tilde{u}\) has initial trace \(\delta\). Therefore, by the uniqueness of distributional solutions (Theorem 37), \(\tilde{u} = \bar{u}\) a.e. in \(\mathbb{R}^{d} \times (0,T)\).

To conclude that the two solutions are a.e. equal on a fixed time slice, note that earlier in the proof, we showed \(\tilde{u} \in C((0,T];L^{1}(\mathbb{R}^{d}))\). It is straightforward to use the definition of \(\bar{u}\) in 23 to see that that \(\bar{u} \in C((0,T];L^{1}(\mathbb{R}^{d}))\), so by Corollary 45, it follows that for \(t \in (0,T]\), \(\tilde{u}(\cdot,t) = \bar{u}(\cdot,t)\) a.e. in \(\mathbb{R}^{d}\). Hence, we can re-write 79 as \[u_{N_{i}} \xrightarrow[]{L^{1}_{\mathop{\mathrm{loc}}}(Q_{T})} \bar{u} \text{ and, for all }t \in (0,T], u_{N_{i}}(\cdot,t) \xrightarrow[]{L^{1}_{\mathop{\mathrm{loc}}}(\mathbb{R}^{d})} \bar{u}(\cdot,t).\] Thus, for any increasing sequence \((M_i)_{i \in {\mathbb{N}}}\) there exists a further subsequence \((N_i)_{i \in {\mathbb{N}}}\) along which the above convergences both hold; the result follows. ◻

Proof of Theorem 1. For \(N \in {\mathbb{N}}\) let \(u_N\) be generated by \({\mathcal{S}}_N[A]\) with \(A(r)=r^{m+1}\) and initial condition \(\delta\). Choosing any \(T \geq 1\), by Theorem 55 we have that \(u_{N}(\cdot,1) \xrightarrow{L^{1}_{\text{loc}}(\mathbb{R}^{d})} \bar{u}(\cdot,1)\).

Now we show convergence in distribution. Let \(D \subset \mathbb{R}^{d}\) be a continuity set, meaning \(|\partial D|=0\), and take \(N \geq 2\). Since \(u_{N}(k\Delta_x,n\Delta_t) = |\square_{N}|^{-1}{\mathbf{P}}\left\{X^{n} = k\right\}\), \[\begin{align} {\mathbf{P}}\left\{\frac{1}{N^{1/(dm+2)}}X^{N} \in D\right\} &= {\mathbf{P}}\left\{\Delta_xX^{N} \in D\right\} \\ &=\sum_{k\Delta_x\in (D \cap \mathbb{Z}^{d}\Delta_x)}{\mathbf{P}}\left\{X^{N}= k\right\} \\ &=\sum_{k\Delta_x\in (D \cap \mathbb{Z}^{d}\Delta_x)} u_{N}(k\Delta_x,1) |\square_{N}|, \end{align}\] where in the last step we used that \(|\square_{N}| = (\Delta_x)^{d}\) and \(N \Delta_t= 1\). To write this last term as an integral of \(u_{N}\) over \(D\), we need to account for the discretization of the domain. Define \[D_{N} := \bigcup_{\square_{N}(k\Delta_x) \subset D}\square_{N}(k\Delta_x).\] Continuing from our previous calculation, we have \[{\mathbf{P}}\left\{\frac{1}{N^{1/(dm+2)}}X^{N} \in D\right\} =\sum_{k\Delta_x\in D \cap \mathbb{Z}^{d}\Delta_x} u_{N}(k\Delta_x,1) |\square_{N}| =\int_{D_{N}}u_{N}(x,1)\,dx.\] On the other hand, by the definition of the random variable \(B\), \[{\mathbf{P}}\left\{B \in D\right\} = \int_{D}\bar{u}(x,1)\,dx.\] Combining the above two displays, this implies \[\begin{align} &\left| {\mathbf{P}}\left\{\frac{1}{N^{1/(dm+2)}}X^{N} \in D\right\} - {\mathbf{P}}\left\{B \in D\right\}\right| \\ &= \left| \int_{D_{N}}u_{N}(x,1)\,dx - \int_{D}\bar{u}(x,1)\,dx\right| \\ &\leq \left| \int_{D_{N}}u_{N}(x,1)\,dx - \int_{D}u_{N}(x,1)\,dx\right| + \left| \int_{D}u_{N}(x,1)\,dx - \int_{D}\bar{u}(x,1)\,dx\right|. \end{align}\] For the latter term in the upper bound, since \(u_{N}(\cdot,1) \xrightarrow[N\to\infty]{L^{1}_{\text{loc}}(\mathbb{R}^{d})} \bar{u}(\cdot,1)\), \[\left| \int_{D}u_{N}(x,1)\,dx - \int_{D}\bar{u}(x,1)\,dx\right| \xrightarrow[N\to \infty]{} 0.\] For the other term, for \(N \geq 2\), by taking \(t = 1\) in Proposition 51[uN46Linfty], we have that \(\|u_{N}(\cdot,1)\|_{L^{\infty}(\mathbb{R}^{d})} \leq C\) and so \[\left| \int_{D_{N}}u_{N}(x,1)\,dx - \int_{D}u_{N}(x,1)\,dx\right| = \left|\int_{D\setminus D_{N}}u_{N}(x,1)\,dx\right| \leq C |D \setminus D_{N}|.\] Since \(\limsup_{N\to\infty}|D\setminus D_{N}| = 0\), we combine the previous three displays to conclude that \[{\mathbf{P}}\left\{\frac{1}{N^{1/(dm+2)}}X^{N} \in D\right\} \xrightarrow[N\to\infty]{} {\mathbf{P}}\left\{B \in D\right\}\, ,\] which implies that \(N^{-1/(dm+2)}X^{N} \rightarrow B\) in distribution. ◻

Proof of Theorem 56. Suppose we are given a path \((\bar{Y}_0, \ldots, \bar{Y}_n)\in (\mathbb{Z}^{d})^{n}\). Let \(M(n) \in \mathbb{Z}^d\) denote the move vector of the path up to time \(n\), whose \(i\)th-coordinate is given by \[M(n)_i = \#\{0 \le j < n: \bar{Y}_{j+1} - \bar{Y}_j = \pm e_i\}.\] Observe that for each \(i=1,\ldots, d\), \(M(n)_i\) is the number of times that the \(i\)-th coordinate has changed, up to time \(n\).

For \(k,m\in \mathbb{Z}^{d}\) and \(n\in {\mathbb{N}}\), let \(P_n(k; m)\) be the probability that \(\bar{Y}_n = k\) and \(M(n) = m\). These probabilities have an exact formula under certain hypotheses on \(m\): if \[\label{e46movec} \begin{cases} \sum_{i=1}^{d} m_i = n,\\ \text{m_i \ge |k_i|, for i=1, \ldots, d},\\ \text{m_i+k_i is even for i=1,\ldots,d}, \end{cases}\tag{81}\] then \[P_n(k; m) = \left(\begin{gather} n\\[-.2em] {\displaystyle {m_1 + k_1 \over 2}, {m_1 - k_1 \over 2}, \ldots, {m_d + k_d \over 2}, {m_d - k_d \over 2}} \end{gather}\right) (2d)^{-n},\] where \(\genfrac{(}{)}{0pt}{}{n}{a,\,b,\,c, \ldots}\) denotes the multinomial coefficient. If any of those three conditions in 81 fail, then the probability \(P_n(k;m)\) is zero.

Given \(k,m \in \mathbb{Z}^d\), if \(P_{n-2}(k;m)>0\), then for any \(i\in \left\{1, \ldots, d\right\}\), \(P_{n}(k;m+2e_i)>0\), and the above exact formula yields the identity \[\begin{align} \label{eq:pnidentity} P_{n-2}(k; m) &= {\displaystyle 4d^2 \left({m_i + 2 + k_i \over 2}\right) \left({m_i + 2 - k_i \over 2}\right) \over n(n-1)} P_n(k; m + 2e_i) \nonumber\\ &= {d^2 \over n(n-1)} ((m_i + 2)^2 - k_i^2) P_n(k; m + 2e_i).\nonumber \end{align}\tag{82}\] Rearranging the above expression and taking an average over \(i = 1, \ldots, d\), we obtain that for any valid vector \(m\) for a path of length \(n-2\), \[\left[{n(n-1) \over d^3}\sum_{i=1}^{d} {1 \over ((m_i+2)^2 - k_i^2)}\right] P_{n-2}(k; m) = {1 \over d} \sum_{i=1}^{d} P_n(k; m + 2e_i).\]

By Lemma 57, for \(\Lambda= (8d^4)^{-1}\), since \(2 \le n \le \Lambda|k|^2\) by hypothesis, we conclude \[\left(1 + {1 \over n}\right) P_{n-2}(k; m) \le {1 \over d} \sum_{i=1}^{d} P_n(k; m + 2e_i).\] Summing over \(m\in \mathbb{Z}^{d}\) such that \(P_{n-2}(k,m)>0\), we obtain \[\left(1 + {1 \over n}\right) \sum_{m:P_{n-2}(k,m)>0} P_{n-2}(k; m) \le {1 \over d} \sum_{m:P_{n-2}(k,m)>0} \sum_{i=1}^{d} P_n(k; m + 2e_i).\] The left side is \((1+1/n) {\mathbf{P}}\left\{\bar{Y}_{n-2} = k\right\}\), and each move vector \(m'\) with \(\sum_{i=1}^{d} m'_i = n\) can be written as a sum \(m + 2e_i\) in at most \(d\) ways, so the right side is no greater than \(\sum_{m'} P_n(k; m') = {\mathbf{P}}\left\{\bar{Y}_n = k\right\}\). This proves the theorem. ◻

Proof of Lemma 57. For ease of notation, let \(y_i = m_i+2\). We begin by factoring \[\label{e46factor}\sum_{i=1}^d {1 \over y_i^2 - k_i^2} = \sum_{i=1}^d {1 \over y_i^2} + \sum_{i=1}^d {k_i^2 \over (y_i^2-k_i^2)y_i^2}.\tag{83}\] We estimate each term in the right hand side from below. Let \(\mu := d^{-1} \sum_{i=1}^{d} y_i\). We apply Jensen’s inequality with the function \(f(x) = 1/x^2\): \[\sum_{i=1}^{d} {1 \over y_i^2} = d \left({1 \over d} \sum_{i=1}^{d} {1 \over y_i^2}\right) \ge {d \over \mu^2} = {d^3 \over (\sum_{i=1}^{d} y_i)^2}.\]

Since \(\sum_{i=1}^{d}m_{i}=n-2\), \(\sum_{i=1}^{d} y_i=n+2d-2\), so\[\begin{align} {n(n-1) \over d^3} \sum_{i=1}^{d} {1 \over y_i^2} &\ge {n(n-1) \over (\sum_{i=1}^{d} y_i)^2} \notag \\&= {n(n-1) \over (n+2d-2)^2} \notag \\&= \left(1-{1 \over n}\right)\left(1+{2d-2 \over n}\right)^{-2}. \notag \end{align}\] Then, since \((1+(2d-2)/n)^{-2} \ge (1-(2d-2)/n)^{2}\geq 1-(4d-4)/n\), we have \[{n(n-1) \over d^3} \sum_{i=1}^{d} {1 \over y_i^2} \ge \left(1-\frac{1}{n}\right)\left(1-\frac{4d-4}{n}\right) \ge 1 - {4d-3 \over n}.\]

We estimate the second term in 83 in the following way. Since \(\sum_{i=1}^{d} m_i=n - 2\), for each \(i=1, \ldots, d\), \(y_i = m_i + 2 \le n\). Moreover, \(k_i^2/(y_i^2 - k_i^2)y_i^2\) is at least \(k_i^2/y_i^4\). Therefore\[\begin{align} {n(n-1) \over d^3} \sum_{i=1}^{d} {k_i^2 \over (y_i^2 - k_i^2)y_i^2} &\ge {n(n-1) \over d^3} \sum_{i=1}^{d} {k_i^2 \over y_i^4}. \notag\\&\ge {n(n-1) \over d^3 n^4} \sum_{i=1}^{d} k_i^2 \notag\\&\ge{1\over 2d^3n^2} |k|^2\label{simple46estimb} \end{align}\tag{84}\] where in the final line, we used the estimate \(n(n-1) \ge 1/2n^2\) because \(n \ge 2\). Our assumption is that \(n \le \Lambda|k|^2\), so the estimate in (84 ) is bounded from below by \(1/2\Lambda d^3 n = 4d/n\). Combining this with the prior display, we obtain \[{n(n-1) \over d^3} \sum_{i=1}^{d} {1 \over y_i^2 - k_i^2} \ge 1 - {4d-3 \over n} + {4d \over n} = 1 + {3 \over n}.\]This proves the lemma. ◻

Proof. Let \(\varepsilon_n\) be the scaled maximum difference between the two functions \(q_n^{(1/2)}\) and \(\varphi_n\) over lattice points in \(\mathbb{Z}^d\), i.e. \[\varepsilon_n := n^{d/2} \sup_{k\in \mathbb{Z}^d} \left|q_n^{(1/2)}(k) - \varphi_n(k)\right|.\] The supremum here is bounded above by the supremum in (85 ), so \(\varepsilon_n\) vanishes as \(n \to \infty\).

Let \(a := (1/2) (d/\pi)^{d/2} e^{-2d\eta}\). Let \(n_0\) be the smallest positive integer so that \(\varepsilon_n \le a\) for \(n \ge n_0\). Such an integer exists because \(\varepsilon_n \to 0\). Let \(k_1: = \sqrt{2\eta n_0}\) so that \(k_{1}^{2}\eta^{-1}=2n_{0}\).

Suppose \(|k| \ge k_{1}\). We first observe that, for any \(\alpha\geq 1\), then \(\lfloor \alpha\rfloor \ge \alpha/2\). Thus for \(|k|\geq k_{1}\), \(|k|^2\eta^{-1} \ge k_{1}^2\eta^{-1} = 2n_0 \ge 1\), so by definition \(n = \lfloor |k|^2\eta^{-1} \rfloor \ge |k|^2 (2\eta)^{-1}\).

We conclude that for \(n= \lfloor |k|^2\eta^{-1} \rfloor\), \(n\geq |k|^2(2\eta)^{-1} \ge k_{1}^2 (2\eta)^{-1}= n_0\), and thus \(\varepsilon_n\le{a}\). Furthermore, since \(n \ge |k|^2(2\eta)^{-1}\), this implies \(d|k|^2/n \le 2d\eta\), and hence, \[\varphi_n(k) = \left(d \over \pi n\right)^{d/2} e^{-d|k|^2/n} \ge \left(d \over \pi n\right)^{d/2} e^{-2d\eta} = {2a \over n^{d/2}}\]

The definition of the error \(\varepsilon_n\) implies \(|q_n^{(1/2)}(k) - \varphi_n(k)| \le \varepsilon_n/n^{d/2}\), and thus \(q_n^{(1/2)}(k) \ge \varphi_n(k) - \varepsilon_{n}/n^{d/2}\). Combine this with the upper bound on \(\varepsilon_n\) and the lower bound \(\varphi_n(k)\), we obtain \[\label{weightlowerbound46b} q_n^{(1/2)}(k) \ge \varphi_n(k) - {\varepsilon_n \over n^{d/2}} \ge \varphi_n(k) - {a \over n^{d/2}} \ge {a \over n^{d/2}}.\tag{86}\]  ◻

Proof of Theorem 17. Let \(\nu_1, \nu_2, \ldots\) be independent \(\left\{0,1\right\}\)-valued random variables that are \(\text{Bernoulli}(2b)\)-distributed. Consider the following process on \(\mathbb{Z}^{d}\). Starting from the origin, at step \(n\), if \(\nu_n = 1\), then we take a \(1/2\)-lazy random walk step, i.e. with probability \(1/2\), we take a unit step in one of the \(2d\) directions, or with probability \(1/2\), we stay where we are. If \(\nu_n = 0\), we stay put.

On the one hand, this is just a \(b\)-lazy random walk, because we jump to a uniformly chosen neighbour with probability \(b\) at each step. On the other hand, at the \(n\)-th step, we have taken \(\nu_1 + \cdots + \nu_n\) steps in a \(1/2\)-lazy random walk. Consequently, \[q_n^{(b)}(k) = \mathbb{E}\left[q_{\nu_1 + \cdots + \nu_n}^{(1/2)}(k)\right].\] We may equivalently express this as \[q^{(b)}_n(k) = \sum_{m=0}^\infty q^{(1/2)}_m(k) \,\mathbb{P}\{\nu_1 + \cdots + \nu_n = m\}.\] Let \(\delta q_j^{(1/2)}(k):= q^{(1/2)}_{j+1}(k) - q^{(1/2)}_j(k)\). Then since \(|k|\geq k_{2}\) to be chosen, by a telescoping series, we have \[\begin{align} \sum_{m=0}^\infty q^{(1/2)}_m(k) \mathbb{P}\{\nu_1 + \cdots + \nu_n = m\}&=\sum_{m=0}^{\infty} \sum_{j =0}^{m-1} \delta q^{(1/2)}_j(k) \mathbb{P}\{\nu_1 + \cdots + \nu_n = m\} \\&=\sum_{j=0}^{\infty} \delta q^{(1/2)}_j(k)\sum_{m=j+1}^{\infty} \mathbb{P}\{\nu_1 + \cdots + \nu_n = m\} \\&=\sum_{j=0}^{\infty} \delta q_j ^{(1/2)}(k)\mathbb{P}\{\nu_1 + \cdots + \nu_n \ge j+1\}. \end{align}\] That means that the difference of successive probabilities is \[\begin{align} q^{(b)}_{n+1}(k) - q^{(b)}_n(k) &= \sum_{j=0}^{\infty} \delta q_j^{(1/2)}(k) \big(\mathbb{P}\{\nu_1 + \cdots + \nu_{n+1} \ge j+1\} - \mathbb{P}\{\nu_1 + \cdots + \nu_n \ge j+1\}\big) \\&= \sum_{j=0}^{\infty} \delta q_j^{(1/2)}(k)\mathbb{P}\{\nu_1 + \cdots + \nu_n = j, \nu_{n+1} = 1\} \\&= 2b \sum_{j=0}^{\infty} \delta q_j^{(1/2)}(k) \mathbb{P}\{\nu_1 + \cdots + \nu_n = j\}. \end{align}\]This is a weighted sum of the differences \(\delta q_j^{(1/2)}(k)\). By Lemma 59, if \(j + 1 \le \Lambda|k|^2\), then \[\delta q^{(1/2)}_j (k)\geq{1 \over 8(j+1)} q_j^{(1/2)}(k) \ge {1 \over 8\Lambda|k|^2} q_j^{(1/2)}(k).\]

Applying this lower bound to the prior display, for the indices \(j = 0, \ldots, \lfloor \Lambda|k|^2 \rfloor - 1\), we obtain \[\begin{align}[t]\label{lowb} q^{(b)}_{n+1}(k) - q^{(b)}_n(k) &\ge {b \over 4\Lambda|k|^2} \sum_{j=0}^{\lfloor \Lambda|k|^2 \rfloor - 1} q^{(1/2)}_j(k) \,\mathbb{P}\{\nu_1 + \cdots + \nu_n = j\}\\&\qquad+2b\sum_{j=\lfloor \Lambda|k|^2 \rfloor}^\infty \delta q_j^{(1/2)}(k) \,\mathbb{P}\{\nu_1 + \cdots + \nu_n = j\}\\ &=: S_{1}+S_{2}.\end{align}\tag{87}\] We want to show that the right hand side is nonnegative. The rough idea is that \(\mathbb{P}\{\nu_1 + \cdots + \nu_n = j\}\) should decrease very quickly as \(j\) increases when \(j\) is much larger than \(\mathbb{E}[\nu_1 + \cdots + \nu_n] = 2bn \le \Lambda|k|^2/3\).

It turns out that we get exponential decay if \(j \ge \Lambda|k|^2/2\), as we now show. The probabilities in (87 ) have a simple formula, \(\mathbb{P}\{\nu_1 + \cdots + \nu_n = j\} = (2b)^j (1-2b)^{n-j} \genfrac{(}{)}{0pt}{}{n}{j}\). In particular, the ratio of consecutive probabilities is \[{\mathbb{P}\{\nu_1 + \cdots + \nu_n = j+1\} \over \mathbb{P}\{\nu_1 + \cdots + \nu_n = j\}} = {\genfrac{(}{)}{0pt}{}{n}{j+1} (2b)^{j+1} (1-2b)^{n-j-1} \over \genfrac{(}{)}{0pt}{}{n}{j} (2b)^j (1-2b)^{n-j}} = {2b(n-j) \over (1-2b)(j+1)}.\] If \(j + 1 \ge \Lambda|k|^2 / 2\), then since by assumption \(n \le \Lambda|k|^2 / 6b\) and \(b \le 1/8\), then \[{2b(n-j) \over (1-2b)(j+1)} \le {2bn \over (3/4) (\Lambda|k|^2/2)} \le \frac{\Lambda |k|^{2}}{3}\frac{4}{3}\frac{2}{\Lambda |k|^{2}}= {8 \over 9},\] which implies that \[{\mathbb{P}\{\nu_1 + \cdots + \nu_n = j+1\} \over \mathbb{P}\{\nu_1 + \cdots + \nu_n = j\}}\leq \frac{8}{9}.\] We now iterate this estimate. Let \(\gamma:= \lfloor \Lambda|k|^2 / 2 \rfloor\). If \(j \ge \gamma\), then \[\label{inc46chain}\mathbb{P}\{\nu_1 + \cdots + \nu_n = j\} \le (8/9) \mathbb{P}\{\nu_1 + \cdots + \nu_{n} = j-1\} \le \cdots \le (8/9)^{j-\gamma} \rho\tag{88}\] where \(\rho := \mathbb{P}\{\nu_1 + \cdots + \nu_n = \gamma\}\).

We will use this to compare \(|S_{2}|\) with \(S_{1}\). Let \(a, k_1\) be the constants from Lemma 58 with \(\eta=\Lambda^{-1}\). Let \(k_2\geq \max\{1, 4/\Lambda, k_1\}\), and sufficiently large so that, for \(c\) to be chosen, \[\label{ak0bound} {a \over 4\Lambda^{1+d/2} |k|^{2+d}} \ge c\left(\frac{8}{9}\right)^{\tfrac{\Lambda|k|^2}{2}}\qquad \text{ for }|k| \ge k_2.\tag{89}\] This is possible because the right-hand side decreases exponentially, while the left-hand side only decreases polynomially.

The term with index \(\gamma\) is part of the first sum, because \(\gamma\le \Lambda|k|^2/2 \le \lfloor \Lambda|k|^2 \rfloor\).² Every term in the first sum is nonnegative, so \(S_1\) is at least equal to the \(\gamma\) term. Then by an application of Lemma 58, and the choice of constants, we have \[\begin{align} S_1&\ge {b \over 4\Lambda|k|^2} q_{\gamma}^{(1/2)}(k) \,\rho \ge {ab \over 4\Lambda|k|^2 \gamma^{d/2}} \,\rho\ge {ab \over 4\Lambda^{1+d/2} |k|^{2 +d}} \,\rho.\label{s1bound}{} \end{align}\tag{90}\]

On the other hand, for \(|S_{2}|\), we claim there exists an absolute constant \(c\) such that, by (88 ), \[\begin{align} \label{s2bound} |S_2| &\le 2b\sum_{j=\lfloor \Lambda|k|^2 \rfloor}^\infty \delta q_j^{(1/2)}(k)\mathbb{P}\{\nu_1 + \cdots + \nu_n = j\}\le 4b\sum_{j=\lfloor \Lambda|k|^2 \rfloor}^\infty (\tfrac{8}{9})^{j-\gamma}\rho\le cb(\tfrac{8}{9})^{\Lambda|k|^2/2}\rho. \end{align}\tag{91}\] The first term of the sum is \((8/9)^{\lfloor \Lambda|k|^2 \rfloor - \lfloor \Lambda|k|^2/2 \rfloor} \rho\), so \(c = 4\times 9 \times (9/8)\) would make the inequality true, for example.

By (89 ), the lower bound (90 ) is larger than the upper bound (91 ), so \[q^{(b)}_{n+1}(k) - q^{(b)}_n \ge S_1 + S_2 \ge S_{1}-|S_{2}|\geq 0.\] Therefore \(q^{(b)}_n(k) \le q^{(b)}_{n+1}(k)\) as long as \(n \le \Lambda|k|^2/6b\). ◻