Proof. For [itm:fY], \[\begin{align} f_Y(y) &= \expt{ f_{Z_\nc}(y - X) } = \frac{1}{2\nc} \expt{\one_{[-\nc,\nc)}(y - X)} \notag \\ &= \frac{1}{2\nc} \prob{ y-\nc < X \le y+\nc } = \frac{1}{2\nc} (F_X(y+\nc) - F_X(y-\nc)) . \notag \end{align}\]

For [itm:fY95sum], let us first use [itm:fY] to compute the sum truncated at \(\pm N\) for some \(N\in\bbZ_{>0}\): \[\begin{align} \sum_{k=-N}^N f_Y(s + 2k\nc) &= \frac{1}{2\nc} \sum_{k=-N}^N \paren{ F_X(s+2k\nc + \nc) - F_X(s+2k\nc - \nc) } \notag \\ &= \frac{1}{2\nc} \sum_{k=-N}^N \paren{ F_X(s+2(k+1)\nc - \nc) - F_X(s+2k\nc - \nc) } \notag \\ &= \frac{1}{2\nc} \paren{ F_X(s+2(N+1)\nc - \nc) - F_X(s-2N\nc - \nc) } , \notag \end{align}\] since it is a telescoping sum. Now sending \(N\to\infty\), we have \(F_X(s+2(N+1)\nc - \nc) \to 1\) and \(F_X(s-2N\nc - \nc) \to 0\), which establishes the result.

Finally, [itm:EY2] immediately follows from independence between \(X,Z_\nc\) and the fact \(\expt{Z_\nc^2} = \nc^2/3\). ◻

Proof. Denote \[\begin{align} && a_k(s) &\mathrel{\vcenter{:}}= s + 2k\nc , & p_{k,\lambda}(s) &\mathrel{\vcenter{:}}= \frac{1}{\Theta_{\lambda,\nc}(s)} \exp\paren{ -\lambda a_k(s)^2 } , & & \label{eqn:pk} \end{align}\tag{6}\] where we suppress the dependence on \(\nc\) in notation since \(\nc\) is fixed throughout. Then \(p_{\cdot,\lambda}(s)\) is the probability mass function of a probability distribution on \(\bbZ\). A straightforward calculation shows \[\begin{align} \frac{\partial}{\partial\lambda} p_{k,\lambda}(s) &= - \brack{ \frac{a_k(s)^2 e^{-\lambda a_k(s)^2}}{\Theta_{\lambda,\nc}(s)} - \frac{e^{-\lambda a_k(s)^2} \sum_{j\in\bbZ} a_j(s)^2 e^{-\lambda a_j(s)^2}}{\Theta_{\lambda,\nc}(s)^2} } \notag \\ &= - \brack{ a_k(s)^2 p_{k,\lambda}(s) - p_{k,\lambda}(s) \sum_{j\in\bbZ} a_j(s)^2 p_{j,\lambda}(s) } . \notag \end{align}\] Using this, we have \[\begin{align} \frac{\partial}{\partial\lambda} \sum_{k\in\bbZ} a_k(s)^2 p_{k,\lambda}(s) = - \brack{ \sum_{k\in\bbZ} a_k(s)^4 p_{k,\lambda}(s) - \paren{ \sum_{k\in\bbZ} a_k(s)^2 p_{k,\lambda}(s) }^2 } = - \var{ a_\bk(s)^2 } , \notag \end{align}\] where \(\bk \sim p_{\cdot,\lambda}(s)\). This quantity is strictly negative for every \(s\in[-\nc,\nc)\). Since \(Q_\nc\) can be written as \[\begin{align} Q_\nc(\lambda) &= \frac{1}{2\nc} \int_{-\nc}^\nc \sum_{k\in\bbZ} a_k(s)^2 p_{k,\lambda}(s) \diff s , \notag \end{align}\] we conclude the monotonicity of \(Q_\nc\).

Next, we prove continuity of \(Q_\nc\). Recall \(\Theta_{\lambda,\nc}\) from 3 and define also \[\begin{align} \Omega_{\lambda,\nc}(s) &\mathrel{\vcenter{:}}= \sum_{k\in\bbZ} (s+2k\nc)^2 e^{-\lambda(s+2k\nc)^2} . \notag \end{align}\] Then \[\begin{align} Q_\nc(\lambda) &= \frac{1}{2\nc} \int_{-\nc}^\nc \frac{\Omega_{\lambda,\nc}(s)}{\Theta_{\lambda,\nc}(s)} \diff s . \label{eqn:QQ} \end{align}\tag{7}\] Note that for \(s\in[-\nc,\nc)\) and \(\abs{k}\ge1\), we have \(\abs{s + 2k\nc} \ge \nc (2\abs{k} - 1)\). Consider a compact interval \([a,b] \subset (0,\infty)\). Then for all \(\lambda\in[a,b]\) and \(\abs{k}\ge1\), \[\begin{align} e^{-\lambda(s+2k\nc)^2} &\le \exp\paren{ - a \nc^2 (2\abs{k} - 1)^2 } , \notag \end{align}\] the right-hand side of which is summable over \(k\in\bbZ\). Therefore, \(\Theta_{\lambda,\nc}(s)\) converges uniformly in \((\lambda,s) \in [a,b] \times [-\nc,\nc)\). A similar argument shows that \(\Omega_{\lambda,\nc}(s)\) also uniformly converges. Moreover, since each summand in \(\Theta_{\lambda,\nc}(s)\) and \(\Omega_{\lambda,\nc}(s)\) is continuous in \((\lambda,s)\), \(\Theta_{\lambda,\nc}(s)\) and \(\Omega_{\lambda,\nc}(s)\) are also continuous. The division by \(\Theta_{\lambda,\nc}(s)\) in 7 is legit since \(\Theta_{\lambda,\nc}(s)\) is bounded away from zero uniformly over all \((\lambda,s)\in[a,b]\times[-\nc,\nc)\). The latter can be seen by only looking at the term \(k=0\): \[\begin{align} \Theta_{\lambda,\nc}(s) &\ge e^{-\lambda s^2} \ge e^{-b\nc^2} > 0 . \notag \end{align}\] Therefore, \((\lambda,s) \mapsto \Omega_{\lambda,\nc}(s) / \Theta_{\lambda,\nc}(s)\) is continuous and bounded uniformly on \([a,b] \times [-\nc,\nc)\). Integrating this over \(s\in[-\nc,\nc)\) implies that \(Q_\nc\) is continuous on \((0,\infty)\).

As \(\lambda\to\infty\), the mass of \(p_{\cdot,\lambda}\) concentrates on points in the lattice \(\brace{a_k(s) : k\in\bbZ} = s + 2\nc \bbZ\) whose distance to \(0\) is minimized. For \(s\in(-\nc,\nc)\), the minimizer is unique and is given by \(a_k(s) = s\). If \(s=-\nc\), the minimizer is not unique, but this case can be ignored since it does not contribute to the integral in the definition 4 of \(Q_\nc\). Therefore, \[\begin{align} \lim_{\lambda\to\infty} Q_\nc(\lambda) &= \frac{1}{2\nc} \int_{-\nc}^\nc s^2 \diff s = \nc^2/3 . \notag \end{align}\]

To study the \(\lambda\downarrow0\) limit of \(Q_\nc\), let us first estimate \(\Theta_{\lambda,\nc}\) for small \(\lambda>0\) using Fourier series (a.k.a.Poisson resummation upon periodizing \(\Theta_{\lambda,\nc}\) as below). Extending the domain of \(\Theta_{\lambda,\nc}\) from \([-\nc,\nc)\) to \(\bbR\), we obtain a \(2\nc\)-periodic function, i.e., \(\Theta_{\lambda,\nc}(s+2\nc) = \Theta_{\lambda,\nc}(s)\) for any \(s\in\bbR\). The Fourier coefficient of \(\Theta_{\lambda,\nc}\) at frequency \(n\in\bbZ\) can then be computed as \[\begin{align} \wh{\Theta}_{\lambda,\nc}(n) &= \frac{1}{2\nc} \int_0^{2\nc} \Theta_{\lambda,\nc}(s) \exp\paren{-\frac{2\pi\ii n s}{2\nc}} \diff s \notag \\ &= \frac{1}{2\nc} \int_0^{2\nc} \sum_{k\in\bbZ} \exp\paren{-\lambda(s+2k\nc)^2} \exp\paren{-\frac{2\pi\ii n s}{2\nc}} \diff s \notag \\ &= \frac{1}{2\nc} \sum_{k\in\bbZ} \int_{2k\nc}^{2(k+1)\nc} \exp\paren{-\lambda x^2} \exp\paren{-\frac{2\pi\ii n}{2\nc} (x - 2k\nc)} \diff x \tag{8} \\ &= \frac{1}{2\nc} \int_{-\infty}^\infty \exp\paren{-\lambda x^2} \exp\paren{ - \ii n \frac{\pi x}{\nc} } \diff x \tag{9} \\ &= \frac{1}{2\nc} \sqrt{\frac{\pi}{\lambda}} \exp\paren{-\frac{\pi^2 n^2}{4\lambda \nc^2}} , \tag{10} \end{align}\] where 8 is obtained from the change of variable \(s+2k\nc \mapsto x\); 9 follows since \(\exp\paren{2\pi\ii nk} = 1\); 10 is by the formula of the characteristic function of \(\cN(0,\sigma^2)\) evaluated at \(n\in\bbR\): \[\begin{align} \int_{-\infty}^\infty \frac{e^{-\frac{x^2}{2\sigma^2}}}{\sqrt{2\pi\sigma^2}} e^{-\ii n x} \diff x &= e^{-n^2 \sigma^2/2} . \notag \end{align}\] Summing up all phases modulated by these coefficients, we obtain the Fourier series of \(\Theta_{\lambda,\nc}\): \[\begin{align} \Theta_{\lambda,\nc}(s) &= \sum_{n\in\bbZ} \wh{\Theta}_{\lambda,\nc}(n) \exp\paren{ \frac{2\pi\ii n s}{2\nc} } \notag \\ &= \frac{1}{2\nc} \sqrt{\frac{\pi}{\lambda}} \sum_{n\in\bbZ} \exp\paren{-\frac{\pi^2 n^2}{4\lambda \nc^2}} \exp\paren{ \frac{2\pi\ii n s}{2\nc} } \notag \\ &= \frac{1}{2\nc} \sqrt{\frac{\pi}{\lambda}} \sum_{n\in\bbZ} \exp\paren{-\frac{\pi^2 n^2}{4\lambda \nc^2}} \cos\paren{ \frac{2\pi n s}{2\nc} } \tag{11} \\ &= \frac{1}{2\nc} \sqrt{\frac{\pi}{\lambda}} \brack{ 1 + 2 \sum_{n=1}^\infty \exp\paren{-\frac{\pi^2 n^2}{4\lambda \nc^2}} \cos\paren{ \frac{\pi n s}{\nc} } } , \tag{12} \end{align}\] where 11 follows since \(e^{\ii\theta} = \cos(\theta) + \ii \sin(\theta)\) and the imaginary part can be removed since \(\Theta_{\lambda,\nc}\) is real-valued. Denoting \[\begin{align} && q_\lambda &\mathrel{\vcenter{:}}= \exp\paren{-\frac{\pi^2}{4\lambda\nc^2}} , & R_\lambda(s) &\mathrel{\vcenter{:}}= 2 \sum_{n=1}^\infty q_\lambda^{n^2} \cos\paren{ \frac{\pi n s}{\nc} } , & & \notag \end{align}\] let us estimate \(R_\lambda(s)\) and its derivative with respect to \(\lambda\). Taking \(\lambda>0\) to be sufficiently small, we can make sure \(q_\lambda \in [0,1/8]\). Then we have the following estimate: \[\begin{align} \abs{ R_\lambda(s) } &= 2 \abs{ \sum_{n=1}^\infty q_\lambda^{n^2} \cos\paren{ \frac{\pi n s}{\nc} } } \le 2 \sum_{n=1}^\infty q_\lambda^{n^2} \le 2 \sum_{n=1}^\infty q_\lambda^n = 2 \frac{q_\lambda}{1 - q_\lambda} \le 4 q_\lambda \le 1/2 , \notag \end{align}\] and consequently, \[\begin{align} 1 + R_\lambda(s) &\in [1/2,3/2] . \label{eqn:R} \end{align}\tag{13}\] Similarly, \[\begin{align} \abs{\frac{\partial}{\partial\lambda} R_\lambda(s)} &= \frac{\pi^2}{2\nc^2\lambda^2} \abs{ \sum_{n=1}^\infty n^2 q_\lambda^{n^2} \cos\paren{\frac{\pi n s}{\nc}} } \le \frac{\pi^2}{2\nc^2\lambda^2} \abs{ \sum_{n=1}^\infty n^2 q_\lambda^n } = \frac{\pi^2}{2\nc^2\lambda^2} \frac{q_\lambda (1+q_\lambda)}{(1-q_\lambda)^3} \le \frac{6\pi^2}{\nc^2 \lambda^2} q_\lambda . \label{eqn:R39} \end{align}\tag{14}\] Since \(\Theta_{\lambda,\nc}(s) = (2\nc)^{-1} \sqrt{\pi/\lambda} \, \paren{ 1 + R_\lambda(s) }\), using [[eqn:R,eqn:R']](#eqn:R,eqn:R’){reference-type=“ref+Label” reference=“eqn:R,eqn:R’”}, we have \[\begin{align} \frac{\partial}{\partial\lambda} \log(\Theta_{\lambda,\nc}(s)) &= -\frac{1}{2\lambda} + \frac{\frac{\partial}{\partial\lambda} R_\lambda(s)}{1 + R_\lambda(s)} = -\frac{1}{2\lambda} + O\paren{\frac{q_\lambda}{\lambda^2}} = -\frac{1}{2\lambda} + O\paren{\frac{e^{-c/\lambda}}{\lambda^2}} , \notag \end{align}\] where \(c \mathrel{\vcenter{:}}= \pi^2/(4\nc^2)\). Recalling the definition 4 of \(Q_\nc\), we then have \(Q_\nc(\lambda) = (2\lambda)^{-1} (1+o(1))\). Therefore, \(Q_\nc(\lambda) \to \infty\) as \(\lambda\downarrow0\). ◻

Proof. Let \(f\) be a probability density function satisfying the constraints in ?? . By nonnegativity of KL divergence, \[\begin{align} 0 &\le D(f \Mid g_{\lambda,\nc}) = \int f(y) \log(f(y)) \diff y - \int f(y) \log(g_{\lambda,\nc}(y)) \diff y . \label{eqn:D} \end{align}\tag{15}\] This implies \[\begin{align} h(f) &\le - \int f(y) \log(g_{\lambda,\nc}(y)) \diff y \notag \\ &= \lambda \int y^2 f(y) \diff y + \log(2\nc) + \int f(y) \log(\Theta_{\lambda,\nc}([y]_\nc)) \diff y \tag{16} \\ &\le \lambda q + \log(2\nc) + \sum_{k\in\bbZ} \int_{-\nc}^\nc f(s+2k\nc) \log(\Theta_{\lambda,\nc}([s+2k\nc]_\nc)) \diff s \tag{17} \\ &= \lambda q + \log(2\nc) + \int_{-\nc}^\nc \paren{ \sum_{k\in\bbZ} f(s+2k\nc) } \log(\Theta_{\lambda,\nc}(s)) \diff s \tag{18} \\ &= \lambda q + \log(2\nc) + \frac{1}{2\nc} \int_{-\nc}^\nc \log(\Theta_{\lambda,\nc}(s)) \diff s . \tag{19} \end{align}\] Here, 16 follows since \(\log(g_{\lambda,\nc}(y)) = -\lambda y^2 - \log(2\nc) - \log(\Theta_{\lambda,\nc}([y]_\nc))\) by the definition ?? of \(g_{\lambda,\nc}\); in 17 , we use the first constraint in ?? on \(f\) and the assumption \(\lambda>0\) to upper bound the first integral from 16 by \(\lambda q\), we also split \(\bbR\) into the disjoint union of intervals \(\brace{[s+2k\nc-\nc,s+2k\nc+\nc) : k\in\bbZ}\) for the second integral from 16 ; 18 holds since \([s+2k\nc]_\nc = s\) for any \(s\in[-\nc,\nc)\) and \(k\in\bbZ\); 19 follows from the second constraint in ?? on \(f\).

The inequality in 15 becomes an equality if and only if \(f = g_{\lambda,\nc}\) almost everywhere. To show that the upper bound 19 can be achieved, it remains to verify that \(g_{\lambda,\nc}\) is a probability density function satisfying both constraints in ?? . Indeed, \[\begin{align} \sum_{k\in\bbZ} g_{\lambda,\nc}(s+2k\nc) &= \frac{1}{2\nc} \sum_{k\in\bbZ} \frac{e^{-\lambda(s+2k\nc)^2}}{\Theta_{\lambda,\nc}([s+2k\nc]_\nc)} = \frac{1}{2\nc} \cdot \frac{1}{\Theta_{\lambda,\nc}(s)} \sum_{k\in\bbZ} e^{-\lambda(s+2k\nc)^2} = \frac{1}{2\nc} , \notag \end{align}\] which also implies \[\begin{align} \int g_{\lambda,\nc}(y) \diff y &= \int_{-\nc}^\nc \sum_{k\in\bbZ} g_{\lambda,\nc}(s+2k\nc) \diff s = 1 . \notag \end{align}\] Moreover, \[\begin{align} \int y^2 g_{\lambda,\nc}(y) \diff y &= \sum_{k\in\bbZ} \int_{-\nc}^\nc (s+2k\nc)^2 g_{\lambda,\nc}(s+2k\nc) \diff s = \sum_{k\in\bbZ} \int_{-\nc}^\nc (s+2k\nc)^2 \frac{e^{-\lambda(s+2k\nc)^2}}{2\nc \Theta_{\lambda,\nc}(s)} \diff s = Q_\nc(\lambda) = q , \notag \end{align}\] where the last equality is by hypothesis. This completes the proof. ◻

Proof. For any \(x\in\bbR\), there exists a unique pair \((s,K)\in[-\nc,\nc)\times\bbZ\) such that \(x - \nc = s + 2K\nc\). Recalling the notation \(a_k, p_{k,\lambda}\) from 6 and the definition ?? of \(g_{\lambda,\nc}\), we have \[\begin{align} F_\lambda(x) &= \sum_{j\ge0} \frac{e^{-\lambda(x - \nc - 2j\nc)^2}}{\Theta_{\lambda,\nc}([x - \nc - 2j\nc]_\nc)} = \sum_{j\ge0} \frac{e^{-\lambda(s + 2(K - j) \nc)^2}}{\Theta_{\lambda,\nc}([s + 2(K - j)\nc]_\nc)} = \sum_{k\le K} \frac{e^{-\lambda(s+2k\nc)^2}}{\Theta_{\lambda,\nc}(s)} = \sum_{k\le K} p_{k,\lambda}(s) . \label{eqn:Flam} \end{align}\tag{20}\] Since \(K\to\pm\infty\) as \(x\to\pm\infty\), it is easy to see that the right-hand side above converges to \(0\) as \(x\to-\infty\) and to \(1\) as \(x\to\infty\).

To show continuity of \(F_\lambda\), we only need to consider inputs around the lattice points \(x = 2(K+1)\nc\) for some \(K\in\bbZ\). It suffices to show that \[\begin{align} \lim_{\eps\downarrow0} F_\lambda(x + \eps) - F_\lambda(x - \eps) &= 0 . \notag \end{align}\] We decompose \(x-\nc\pm\eps\) into the sum of a lattice point in \(2\nc\bbZ\) and a residual in \([-\nc,\nc)\): \[\begin{align} && x - \nc - \eps &= 2K\nc + (\nc-\eps) , & x - \nc + \eps &= 2(K+1)\nc - (\nc-\eps) . & & \label{eqn:decomp} \end{align}\tag{21}\] Then by the representation 20 of \(F_\lambda\), \[\begin{align} F_\lambda(x + \eps) - F_\lambda(x - \eps) &= \sum_{k\le K+1} p_{k,\lambda}(-(\nc - \eps)) - \sum_{k\le K} p_{k,\lambda}(\nc-\eps) \notag \\ &= \sum_{k\le K} ( p_{k+1,\lambda}(-(\nc-\eps)) - p_{k,\lambda}(\nc-\eps) ) . \label{eqn:F-F} \end{align}\tag{22}\] Each term converges to zero as \(\eps\downarrow0\): \[\begin{align} \lim_{\eps\downarrow0} \, (p_{k+1,\lambda}(-(\nc-\eps)) - p_{k,\lambda}(\nc-\eps)) &= \lim_{\eps\downarrow0} \paren{ \frac{e^{-\lambda(-(\nc-\eps) + 2(k+1)\nc)^2}}{\Theta_{\lambda,\nc}(-(\nc-\eps) + 2(k+1)\nc)} - \frac{e^{-\lambda(\nc-\eps + 2k\nc)^2}}{\Theta_{\lambda,\nc}(\nc-\eps+2k\nc)} } \notag \\ &= \lim_{\eps\downarrow0} \paren{ \frac{e^{-\lambda(\nc+\eps + 2k\nc)^2}}{\Theta_{\lambda,\nc}(\nc+\eps + 2k\nc)} - \frac{e^{-\lambda(\nc-\eps + 2k\nc)^2}}{\Theta_{\lambda,\nc}(\nc-\eps+2k\nc)} } = 0 , \label{eqn:p-p} \end{align}\tag{23}\] by continuity of \(\Theta_{\lambda,\nc}(\cdot)\). Therefore, \[\begin{align} \lim_{\eps\downarrow0} \, ( F_\lambda(x+\eps) - F_\lambda(x-\eps) ) &= 0 , \notag \end{align}\] which, in view of 22 , implies the continuity of \(F_\lambda\).

We also need to check monotonicity of \(F_\lambda\). For \(x \notin 2\nc\bbZ\), there is a unique pair \((s,K)\in(-\nc,\nc)\times\bbZ\) such that \(x-\nc = s + 2K\nc\). In this case, a straightforward calculation shows \[\begin{align} p_{k,\lambda}'(s) &= \frac{-2\lambda a_k(s) e^{-\lambda a_k(s)^2}}{\Theta_{\lambda,\nc}(s)} +2 \lambda \frac{e^{-\lambda a_k(s)^2}}{\Theta_{\lambda,\nc}(s)^2} \sum_{j\in\bbZ} e^{-\lambda a_j(s)^2} a_j(s) \notag \\ &= -2\lambda p_{k,\lambda}(s) \paren{ a_k(s) - \sum_{j\in\bbZ} a_j(s) p_{j,\lambda}(s) } . \notag \end{align}\] Then by the representation 20 , we have: \[\begin{align} F_\lambda'(x) &= \sum_{k\le K} p_{k,\lambda}'(s) = -2\lambda \paren{ \sum_{k\le K} a_k(s) p_{k,\lambda}(s) - \sum_{\substack{k\le K \\ j\in\bbZ}} a_j(s) p_{k,\lambda}(s) p_{j,\lambda}(s) } \notag \\ &= -2\lambda \paren{ \sum_{\substack{k\le K\\j\in\bbZ}} a_k(s) p_{k,\lambda}(s) p_{j,\lambda}(s) - \sum_{\substack{k\le K \\ j\in\bbZ}} a_j(s) p_{k,\lambda}(s) p_{j,\lambda}(s) } \tag{24} \\ &= -2\lambda \sum_{\substack{k\le K \\ j>K}} (a_k(s) - a_j(s)) p_{k,\lambda}(s) p_{j,\lambda}(s) \tag{25} \\ &= 4\lambda\nc \sum_{\substack{k\le K\\ j>K}} (j-k) p_{k,\lambda}(s) p_{j,\lambda}(s) , \tag{26} \end{align}\] where 24 holds since \(\sum_{j\in\bbZ} p_{j,\lambda}(s) = 1\); 25 follows since \[\begin{align} \sum_{\substack{k\le K\\j\le K}} (a_k(s) - a_j(s)) p_{k,\lambda}(s) p_{j,\lambda}(s) &= 0 . \notag \end{align}\] Note that the right-hand side of 26 above is obviously nonnegative.

Next, we show that the derivative at any lattice point \(x = 2(K+1)\nc\) is in fact continuous. Again consider \(x-\nc\pm\eps\) which admit the decompositions 21 . By the result 26 just established, the derivative at \(x-\eps\) is \[\begin{align} F_\lambda'(x-\eps) &= 4\lambda\nc \sum_{\substack{k\le K \\ j>K}} (j-k) p_{k,\lambda}(\nc-\eps) p_{j,\lambda}(\nc-\eps) , \notag \end{align}\] and the derivative at \(x+\eps\) is \[\begin{align} F_\lambda'(x+\eps) &= 4\lambda\nc \sum_{\substack{k\le K+1 \\ j>K+1}} (j-k) p_{k,\lambda}(-(\nc-\eps)) p_{j,\lambda}(-(\nc-\eps)) \notag \\ &= 4\lambda\nc \sum_{\substack{k\le K \\ j>K}} (j-k) p_{k+1,\lambda}(-(\nc-\eps)) p_{j+1,\lambda}(-(\nc-\eps)) . \notag \end{align}\] Therefore, by 23 , \[\begin{align} \lim_{\eps\downarrow0} \, (F_\lambda'(x-\eps) - F_\lambda'(x+\eps)) &= 0 . \notag \end{align}\] Combining this with 26 and comparing the result with ?? , we conclude that \(F_\lambda\) is continuously differentiable and its derivative is precisely \(f_X\) defined in ?? .

Finally, we compute \[\begin{align} \frac{1}{2\nc} (F_\lambda(y+\nc) - F_\lambda(y-\nc)) &= \sum_{j=0}^\infty g_{\lambda,\nc}(y - 2j\nc) - \sum_{j=0}^\infty g_{\lambda,\nc}(y - 2\nc - 2j\nc) \notag \\ &= \sum_{j=0}^\infty g_{\lambda,\nc}(y - 2j\nc) - \sum_{j=1}^\infty g_{\lambda,\nc}(y - 2j\nc) = g_{\lambda,\nc}(y) , \notag \end{align}\] which completes the proof. ◻

Proof of 1.. Let \(X\) be any real-valued random variable satisfying \(\expt{X^2} \le \ic\). Let \(Y = X + Z_\nc\) for \(Z_\nc \sim \unif([-\nc,\nc))\) independent of \(X\). By 1, the output density \(f_Y\) must satisfy the identity ?? and have second moment at most \(q \mathrel{\vcenter{:}}= \ic + \nc^2/3\). Let \(\lambda>0\) be the unique solution to \(Q_\nc(\lambda) = q\) (existence and uniqueness of solution is guaranteed by 2). Then 3 ensures \[\begin{align} h(Y) &\le \lambda q + \log(2\nc) + \frac{1}{2\nc} \int_{-\nc}^\nc \log(\Theta_{\lambda,\nc}(s)) \diff s . \notag \end{align}\] It is easy to verify that \(h(Z_\nc) = \log(2\nc)\). Therefore, \[\begin{align} I(X;Y) &= h(Y) - h(Z_\nc) \le \lambda q + \frac{1}{2\nc} \int_{-\nc}^\nc \log(\Theta_{\lambda,\nc}(s)) \diff s . \notag \end{align}\]

On the other hand, we claim that the above upper bound can be attained by the input density \(f_X\) defined in ?? . Let \(X_\star \sim f_X\). By 4, the cumulative distribution function of \(X_\star\) is \(F_\lambda\). Let \(Z_\nc \sim \unif([-\nc,\nc))\) be independent of \(X_\star\). Then by [itm:fY] of 1, the density of \(Y_\star \mathrel{\vcenter{:}}= X_\star + Z_\nc\) equals \[\begin{align} f_{Y_\star}(y) &= \frac{1}{2\nc} \paren{ F_\lambda(y+\nc) - F_\lambda(y-\nc) } = g_{\lambda,\nc}(y) , \label{eqn:fY61g} \end{align}\tag{27}\] where the last equality follows from 4. We also check that \(f_X\) satisfies the power constraint. To see this, note that 3 ensures \[\begin{align} \expt{Y_\star^2} &= \int y^2 g_{\lambda,\nc}(y) \diff y = q . \notag \end{align}\] Since \(\expt{Y_\star^2} = \expt{X_\star^2} + \expt{Z_\nc^2}\), this implies \(\expt{X_\star^2} = q-\nc^2/3 = \ic\). This proves [itm:thm3].

Combining [eqn:fY=g,eqn:hg], we have \[\begin{align} I(X_\star;Y_\star) &= h(Y_\star) - h(Z_\nc) = h(g_{\lambda,\nc}) - \log(2\nc) = \lambda q + \frac{1}{2\nc} \int_{-\nc}^\nc \log(\Theta_{\lambda,\nc}(s)) \diff s , \notag \end{align}\] which proves [itm:thm2].

Finally, in view of 27 , uniqueness of the capacity achieving output density \(f_{Y_\star}\) in [itm:thm1] follows from uniqueness of the maximizer \(g_{\lambda,\nc}\) for ?? . To show uniqueness of \(f_X\), take any capacity achieving input distribution \(F_X\). Then its pushforward under the channel action must be the unique output density \(g_{\lambda,\nc}\). By [itm:fY] of 1, \[\begin{align} F_X(x) - F_X(x - 2N\nc) &= 2\nc g_{\lambda,\nc}(x - \nc) . \notag \end{align}\] For any \(N\in\bbZ_{>0}\), consider the telescoping sum: \[\begin{align} F_X(x) - F_X(x - 2N\nc) &= \sum_{k = 0}^{N-1} F_X(x-2k\nc) - F_X(x-2k\nc-2\nc) = 2\nc \sum_{k=0}^{N-1} g_{\lambda,\nc}(x-2k\nc-\nc) . \notag \end{align}\] Sending \(N\to\infty\), we have that \(F_X(x - 2N\nc)\) on the left-hand side above vanishes and the right-hand side tends to \(F_\lambda(x)\) in ?? . Therefore \(F_X = F_\lambda\), which, by 4, implies that the density of \(F_X\) must be \(f_X\). This completes the proof. ◻