Proof. We only establish the first claim. Integrating \[\begin{align} 0\leq (f(x) - f(y))(g(x) - g(y))= f(x)g(x)+ f(y)g(y) -f(x)g(y)- f(y)g(x) \end{align}\] over \(X \times X\) yields desired estimate since \[\begin{align} 0&\leq 2\mu(X) \int_X f(x)g(x) \,\mathrm{d}\mu(x)- 2\int_Xg(y) \,\mathrm{d}\mu(y) \int_X f(x)\;d \mu(x). \end{align}\] ◻

Proof. Assume \((f,g)\) is weakly \((b,c)\)-asynchronous. For fixed \(x, y \in X\), there are two possible cases. If \(f(x) \leq b f(y)\) and \(cg(x) \geq g(y)\) that is \((g(x) - c^{-1} g(y))(f(x) - b f(y)) \leq 0,\) then we have \[\begin{align} g(x)f(x) + b c^{-1} g(y)f(y) \leq bg(x)f(y) + c^{-1} g(y)f(x). \end{align}\] If \(f(y) \leq b f(x)\) and \(cg(y) \geq g(x)\) then similarly we derive \[\begin{align} g(x)f(x) + b^{-1} c g(y)f(y) \leq b^{-1} g(x)f(y) + c g(y)f(x). \end{align}\] Since \(f\geq 0\) and \(g\geq 0\), each case implies the inequality \[\begin{align} g(x)f(x)+(b_{*}\,c_{*})^{-1}g(y)f(y)\leq b_{*} g(x)f(y)+c_{*}g(y)f(x). \end{align}\] Integrating both sides with respect to \(x\) and \(y\) over \(X \times X\) implies \[\begin{align} \int_X f(x) g(x)\,\,\mathrm{d}\mu(x) \leq \frac{b_{*}+c_{*}}{1+b^{-1}_{*}c^{-1}_{*}}\frac{1}{\mu(X)} \Big(\int_X g(x)\,\,\mathrm{d}\mu(x)\Big)\Big(\int_X f(x)\,\,\mathrm{d}\mu(x)\Big). \end{align}\] Analogously, if \((f,g)\) is weakly \((b,c)\)-synchronous then we have \[\begin{align} \int_X f(x) g(x)\,\,\mathrm{d}\mu(x) \geq \frac{b^{-1}_{*}+c^{-1}_{*}}{1+b_{*}c_{*}}\frac{1}{\mu(X)}\Big(\int_X g(x)\,\,\mathrm{d}\mu(x)\Big)\Big(\int_X f(x)\,\,\mathrm{d}\mu(x)\Big). \end{align}\] ◻

Proof. This follows from Theorem 8. Since if \(f\) and \(g\) have the same monotonicity then \((f, g)\) is \((b, c)\)-weakly synchronous. Otherwise, \((f, g)\) is \((b, c)\)-weakly asynchronous. ◻

Proof. It is sufficient to apply Theorem 10 on measure spaces \((X=(0,r),\,\,\mathrm{d}\mu(t)= t^{a-1}\,\mathrm{d}t)\) and \((X=(r,\infty),\,\,\mathrm{d}\mu(t)= t^{-a-1}\,\mathrm{d}t)\). ◻

Proof. Note that, if \(g\) is monotone then the sought inequalities follow immediately from Theorem 11. Due to scaling invariance, it is enough to consider the case \(r=1\). For a function \(w:(0,\infty)\to[0,\infty)\), we use the following \(c\)-adic decomposition \[\begin{align} \int_0^1 t^{a -1}w(t) \,\mathrm{d}t &= \sum_{n \geq 0} \int_{c^{-(n+1)}}^{c^{-n}} t^{a -1} w(t) \,\mathrm{d}t = \int_{\tfrac{1}{c}}^{1} \tau^{a -1} \sum_{n \geq 0} \frac{1}{c^{a n}} w(c^{-n} \tau) \,\mathrm{d}\tau,\\ \int_1^\infty t^{-a -1} w(t)\, \,\mathrm{d}t &= \sum_{n \geq 0} \int_{c^n}^{c^{n+1}} t^{-a -1} w(t)\, \,\mathrm{d}t = \int_{1}^{c} \tau^{-a -1} \sum_{n \geq 0} \frac{1}{c^{a n}} w(c^n \tau)\, \,\mathrm{d}\tau. \end{align}\] Since \(f\) is \(b\)decreasing, we have \(b^{-1}f(c^{-n})\leq f(c^{-n} \tau)\leq bf(c^{-n-1})\) for \(\tau \in (\tfrac{1}{c},1)\), while \(b^{-1}f(c^{n+1})\leq f(c^{n} \tau)\leq bf(c^{n})\) for \(\tau \in (1,c)\). Therefore using again monotonicity and the formulas \(\int_{c^{-k}}^{c^{-(k-1)}} t^{a -1} \,\mathrm{d}t = \int_{c^{k-1}}^{c^{k}} t^{-a -1} \,\mathrm{d}t = \frac{c^{a}-1}{a c^{a k}},\) \(k\in \{n,n+1\}\) we find that \[\begin{align} \frac{ab^{-2}}{c^a-1}\int_{c^{-n}}^{c^{-(n-1)}} f(t)t^{a -1} \,\mathrm{d}t & \leq \frac{1}{c^{a n}} f(c^{-n} \tau)\leq \frac{a b^2 c^{2a}}{c^a-1}\int_{c^{-n-2}}^{c^{-n-1}}f(t)t^{a -1} \,\mathrm{d}t \quad \,\, (\tau \in (\tfrac{1}{c},1)),\\ \frac{a b^{-2}c^{2a}}{c^a-1}\int_{c^{n+1}}^{c^{n+2}}f(t)t^{-a -1} \,\mathrm{d}t& \leq \frac{1}{c^{a n}} f(c^{n} \tau) \leq \frac{a b^2}{c^a-1} \int_{c^{n-1}}^{c^{n}} f(t)t^{-a -1} \,\mathrm{d}t \,\,\,\qquad (\tau \in (1,c)). \end{align}\]

Hence it follows that \[\begin{align} &b^{-2}\int_{0}^{c} f(t)t^{a -1} \,\mathrm{d}t \leq \frac{c^a-1}{a}\sum_{n \geq 0}\frac{ f(c^{-n} \tau)}{c^{a n}} \leq c^{2a}b^2\int_{0}^{c^{-1}} f(t)t^{a -1} \,\mathrm{d}t \quad \,\, (\tau \in (\tfrac{1}{c},1)),\\ &c^{2a}b^{-2}\int_{c}^{\infty} f(t)t^{-a -1} \,\mathrm{d}t \leq \frac{c^a-1}{a} \sum_{n \geq 0}\frac{f(c^{n} \tau) }{c^{a n}} \leq b^2\int_{c^{-1}}^{\infty} f(t)t^{-a -1} \,\mathrm{d}t \quad (\tau \in (1,c)). \end{align}\] Since \(f\) is \(b\)-almost decreasing and \(g(ct) \leq g(t)\), if we fix \(\tau>0\), then the sequence \((g(c^{-n} \tau))_n\) is increasing and \((f(c^{-n} \tau))_n\) \(b\)-almost increasing, while the sequence \((g(c^{n} \tau))_n\) is decreasing and \((f(c^{n} \tau))_n\) is \(b\)-almost decreasing. Whence, applying Theorem 8 for discrete sums, with weights \(\alpha_n = \frac{1}{c^{a n}}\) and using \(\sum_{n \geq 0} \frac{1}{c^{a n}} = \frac{c^{a}}{c^{a}-1}\), we obtain \[\begin{align} \sum_{n \geq 0} \frac{1}{c^{a n}} g(c^{-n} \tau)f(c^{-n} \tau) &\geq \frac{c^a-1}{bc^a}\Big( \sum_{n \geq 0} \frac{1}{c^{a n}} g(c^{-n} \tau) \Big)\Big(\sum_{n \geq 0} \frac{1}{c^{a n}} f(c^{-n} \tau)\Big),\\ \sum_{n \geq 0} \frac{1}{c^{a n}} g(c^{n} \tau)f(c^{n} \tau) &\geq \frac{c^a-1}{bc^a} \Big(\sum_{n \geq 0} \frac{1}{c^{a n}} g(c^{n} \tau)\Big) \Big(\sum_{n \geq 0} \frac{1}{c^{a n}} f(c^{n} \tau)\Big). \end{align}\] By exploiting these estimates inn conjunction with the \(c\)-adic decomposition we obtain \[\begin{align} \int_0^1g(t)f(t) t^{a -1} \,\mathrm{d}t &\geq \frac{c^a-1}{bc^a} \int_{\tfrac{1}{c}}^{1} \tau^{a -1} \Big(\sum_{n \geq 0} \frac{1}{c^{a n}} g(c^{-n} \tau) \Big)\Big( \sum_{n \geq 0} \frac{1}{c^{a n}} f(c^{-n} \tau)\Big) \,\mathrm{d}\tau\\ &\geq \frac{a}{b^{3}c^a} \int_{\tfrac{1}{c}}^{1} \tau^{a -1} \Big(\sum_{n \geq 0} \frac{1}{c^{a n}} g(c^{-n} \tau)\Big) \,\mathrm{d}\tau\, \Big(\int_0^c f(t)t^{a -1} \,\mathrm{d}t\Big)\\ &= \frac{a}{b^{3}c^a} \Big(\int_0^1g(t)t^{a -1} \,\mathrm{d}t\Big) \Big(\int_0^c f(t)t^{a -1} \,\mathrm{d}t\Big), \end{align}\] while, similarly, for the second case we get \[\begin{align} \int_1^\infty g(t)t^{-a -1} f(t) \,\mathrm{d}t &\geq \frac{c^a-1}{b^{3}c^a} \int_{1}^{c} \tau^{-a -1} \Big(\sum_{n \geq 0} \frac{1}{c^{a n}} g(c^{n} \tau) \Big) \Big(\sum_{n \geq 0} \frac{1}{c^{a n}} f(c^{n} \tau)\Big) \,\mathrm{d}\tau\\ &\geq \frac{a c^a}{b^{3}} \int_{1}^{c} \tau^{-a -1}\Big( \sum_{n \geq 0} \frac{1}{c^{a n}} g(c^{n} \tau) \Big) \,\mathrm{d}\tau\, \Big(\int_c^\infty f(t)t^{-a -1} \,\mathrm{d}t\Big)\\ &= \frac{a c^a}{b^{3}}\Big(\int_1^\infty g(t)t^{-a -1} \,\mathrm{d}t \Big) \Big(\int_c^\infty f(t)t^{-a -1} \,\mathrm{d}t\Big). \end{align}\] ◻

Proof. Since \(\tilde{g}(t) = g(1/t)\) is \(c\)-adically decreasing, apply Theorem 12 for \(a>0\), \(1/r>0\) and the function \(\tilde{f}(t) = f(1/t)\) is \(b\)-almost monotone oppositely to \(f\). ◻

Proof. The result follows by applying Theorem 10 to the couples \((f,g_{\alpha,\beta})\) and \((g,g_{\alpha,\beta})\) with respect to the measure \(\,\mathrm{d}\mu_\alpha(x)= p_\alpha(x) \,\mathrm{d}\mu(x)\); noting that \[\begin{align} \int_X p_\alpha(x) \,\mathrm{d}\mu(x) = 1= \int_X p_\beta(x) \,\mathrm{d}\mu(x) = \int_X g_{\alpha,\beta}(x) p_\alpha(x) \,\mathrm{d}\mu(x). \end{align}\] ◻

Proof. The claim follows by taking \(b=1\) in Theorem 15. For the special case \(c = 1\), see the different approach in [27]. ◻

Proof. Since \(q\in (1,\infty]\), we put \(q'= \frac{q}{q-1}\in [1,\infty)\). Let us start with the cases where \(g(0)\) and \(g(\infty)\) are finite. Given \(\varepsilon > 0\), there exists \(0 < \delta < r\) such that \(|g(t) - g(0)| < \varepsilon\) for all \(t \in (0, \delta).\) Then for \(a>0\) Hölder inequality implies \[\begin{align} a\Big|\int_0^r g(t)t^{a-1} \,\mathrm{d}t &- \frac{g(0)}{a} r^{a} \Big| = a \Big|\int_0^r (g(t) - g(0)) t^{a-1} \,\mathrm{d}t \Big|\\ &\leq \Big(a \int_0^\delta + a \int_\delta^r \Big)|g(t) - g(0)| t^{a-1} \,\mathrm{d}t \\ &\leq \varepsilon \delta^a + \|g-g(0)\|_{L^q( (\delta,r))} \Big(\frac{a(r^{(a-1)q'+1} - \delta^{(a-1)q'+1})}{(a-1)q'+1}\Big)^{1/q'}. \end{align}\] Letting \(a \to 0^+\), we get \[\begin{align} \limsup_{a \to 0^+} \Big|a\int_0^r g(t)t^{a-1} \,\mathrm{d}t - g(0)r^a \Big|\leq \varepsilon. \end{align}\] Thus, using again the fact that \(r^a \to 1\), we conclude \[\begin{align} \lim_{a\to0^+} a\int_0^r g(t)t^{a-1}\,\mathrm{d}t = g(0). \end{align}\] Analogously, given \(\varepsilon > 0\), if \(g(t) \to g(\infty)\) as \(t\to\infty\), so there exists \(R > r\) such that \(|g(t) - g(\infty)| < \varepsilon\) for all \(t > R.\) Then for \(a>0\) Hölder inequality implies

\[\begin{align} a \Big|\int_r^\infty g(t)t^{-a-1}\,\mathrm{d}t &- \frac{g(\infty)}{a} r^{-a} \Big| = a \Big|\int_r^\infty (g(t) - g(\infty))t^{-a-1}\,\mathrm{d}t \Big|\\ &\leq \Big(a \int_r^R+ a \int_R^\infty \Big) |g(t) - g(\infty)|t^{-a-1}\,\mathrm{d}t\\ &\leq \|g -g(\infty)\|_{L^q( (r,R))}\Big(\frac{a\big( r^{(-a-1)q'+1}-R^{(-a-1)q'+1} \big)}{(a+1)q'-1}\Big)^{1/q'} + \varepsilon R^{-a}. \end{align}\] Letting \(a\to 0^+\), we obtain \[\begin{align} \limsup_{a\to 0^+} \Big| a\int_r^\infty g(t)t^{-a-1}\,\mathrm{d}t - g(\infty) r^{-a} \Big| \leq \varepsilon. \end{align}\] Hence, since \(r^{-a} \to 1\), we conclude \[\begin{align} \lim_{a\to0^+} a\int_r^\infty g(t)t^{-a-1}\,\mathrm{d}t = g(\infty). \end{align}\] Last upon replacing \(g\) with \(-g\), it is sufficient to deal with cases \(g(0)= \infty\) and \(g(\infty)=\infty\). As previously we can choose \(0<\delta<r\) and \(R>r\) such that \[\begin{align} &g(t)\geq 1/\varepsilon\quad\text{for all 0<t<\delta},\quad \text{resp.}\quad g(t)\geq 1/\varepsilon\quad\text{for all t>R}. \end{align}\] By splitting the integrals as previously, the following estimates hold respectively \[\begin{align} a \int_0^r g(t) t^{a-1} \,\,\mathrm{d}t &\geq \frac{\delta^a}{\varepsilon} - a \|g\|_{L^q((\delta,r))} \Big( \frac{r^{(a-1)q'+1} - \delta^{(a-1)q'+1}}{(a-1)q'+1} \Big)^{1/q'}, \\[1.5ex] a \int_r^\infty g(t) t^{-a-1} \,\,\mathrm{d}t &\geq \frac{R^{-a}}{\varepsilon} - a \|g\|_{L^q((r,R))} \Big( \frac{r^{(-a-1)q'+1} - R^{(-a-1)q'+1}}{(-a-1)q'+1} \Big)^{1/q'}. \end{align}\] From these, we deduce \[\begin{align} &\liminf_{a\to 0^+} a\int_0^r g(t)t^{-a-1}\,\mathrm{d}t \geq \frac{1}{\varepsilon},\quad \text{while}\quad \liminf_{a\to 0^+} a\int_r^\infty g(t)t^{-a-1}\,\mathrm{d}t \geq \frac{1}{\varepsilon}. \end{align}\] Since \(\varepsilon>0\) was arbitrary, we conclude: \[\begin{align} &\lim_{a\to0^+} a\int_0^r g(t)t^{a-1}\,\mathrm{d}t = g(0)=\infty \quad \text{and}\quad \lim_{a\to0^+} a\int_r^\infty g(t)t^{-a-1}\,\mathrm{d}t = g(\infty)=\infty. \end{align}\] ◻

Proof. One can either apply Theorem 16 using the probability densities \(p_a(t)= ar^{-a} t^{a-1}\mathbb{1}_{(0,r)}(t)\) and \(p^a(t)= ar^{a} t^{-a-1}\mathbb{1}_{(r,\infty)}(t)\) or apply Theorem 11 with \(f(t)=t^{a-b}\) which is decreasing on \((0,\infty)\) while \(f(t) t^{b-1}= t^{a-1}\) and \(f(t) t^{-a-1}= t^{-b-1}\). ◻

Proof. Both estimates are implied by Theorem 12 since \(f(t)=t^{a-b}\) is decreasing \((0,\infty)\), \(g(t) f(t) t^{b-1}= g(t) t^{a-1}\) while \(g(t) f(t) t^{-a-1}= g(t) t^{-b-1}\) and \[\begin{align} \int_0^{c r} f(t)t^{b-1}\,\mathrm{d}t= \frac{(cr)^{a}}{a} \quad\text{and}\quad \int_{c r}^\infty f(t)t^{-a-1}\,\mathrm{d}t = \frac{(cr)^{-b}}{b}. \end{align}\] The same analogy can applied to \(f(t)=t^{b-a}\) which is increasing on \((0,\infty)\). ◻

Proof. it suffices to prove the claim for indicator functions \(g = \mathbb{1}_A\), where \(A \subset \mathbb{S}^{d-1}\) is a Borel measurable set. This can be done by considering the measure defined by \[\begin{align} \mu(A)= \int_{\mathbb{S}^{d-1}} \sigma_{d-2}(A\cap \mathbb{S}^{d-2}_v ) \frac{ \,\mathrm{d}\sigma_{d-1}(v)}{|\mathbb{S}^{d-2}|}. \end{align}\] Given that the Hausdorff measures \(\sigma_{d-1}\) and \(\sigma_{d-2}\) are respectively rotation invariant, so does the measure \(\mu\) on \(\mathbb{S}^{d-1}\). Indeed, for each \(v\in \mathbb{S}^{d-1}\) and \(O\in \mathcal{O}(d)\), an orthogonal mapping on \(\mathbb{R}^d\), since \(O^\top O= I_d\), one finds that \[\begin{align} O(A) \cap \mathbb{S}^{d-2}_v = O(A \cap \mathbb{S}^{d-2}_{O^\top (v)}). \end{align}\] The rotation invariance of \(\sigma_{d-2}\) implies \[\begin{align} \sigma_{d-2}(O(A)\cap \mathbb{S}^{d-2}_v ) = \sigma_{d-2}(O(A\cap \mathbb{S}^{d-2}_{O^\top (v)})) = \sigma_{d-2}(A\cap \mathbb{S}^{d-2}_{O^\top (v)}). \end{align}\] The rotation invariance of \(\sigma_{d-1}\) yields that \[\begin{align} \int_{\mathbb{S}^{d-1}} \sigma_{d-2}(O(A)\cap \mathbb{S}^{d-2}_v ) \frac{ \,\mathrm{d}\sigma_{d-1}(v)}{|\mathbb{S}^{d-2}|} &= \int_{\mathbb{S}^{d-1}} \sigma_{d-2}(A\cap \mathbb{S}^{d-2}_{O^\top (v)}) \frac{ \,\mathrm{d}\sigma_{d-1}(v)}{|\mathbb{S}^{d-2}|}\\ &= \int_{\mathbb{S}^{d-1}} \sigma_{d-2}(A\cap \mathbb{S}^{d-2}_{v}) \frac{ \,\mathrm{d}\sigma_{d-1}(O(v))}{|\mathbb{S}^{d-2}|}, \end{align}\] in other words \(\mu(O(A)) = \mu(A)\). It follows that, \(\mu= c\sigma_{d-1}\) since rotation invariant Borel measures on \(\mathbb{S}^{d-1}\) are unique up to a multiplicative constant. On the other hand, once again the rotation invariance of \(\sigma_{d-2}\) implies that \(\sigma_{d-2}(\mathbb{S}^{d-1}\cap \mathbb{S}^{d-2}_v) = \sigma_{d-2}(\mathbb{S}^{d-2})\) so that \(\mu(\mathbb{S}^{d-1})= \sigma_{d-1} (\mathbb{S}^{d-1}) =|\mathbb{S}^{d-1}|\), yielding \(\mu= \sigma_{d-1}\). ◻

Proof. For \(u\in L^p(\mathop{\mathrm{\mathbb{R}}}^d)\) define \(U_*(h) = \|u(\cdot+h)-u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\). Thus, we must show \[\begin{align} \int_{\mathbb{S}^{d-1}} U_*(\theta sw)\,\,\mathrm{d}\sigma_{d-1}(w)\leq 2^{p} \int_{\mathbb{S}^{d-1}} U_*(sw)\,\mathrm{d}\sigma_{d-1}(w). \end{align}\] If \(d\) is even then using the block matrices \(\pm\big(\begin{smallmatrix} 0&-1\\1&0 \end{smallmatrix}\big)\) we can construct a rotation \(O\in \mathcal{O}(d)\) on \(\mathop{\mathrm{\mathbb{R}}}^d\), such that \(Ow \cdot w= 0\) for all \(w\in \mathop{\mathrm{\mathbb{R}}}^d\). Equivalently \(O\) is skew symmetric, i.e., \(O^T=-O\). Indeed, since \(w^\top Ow =w^\top O^\top w\) we deduce \(Ow\cdot w= w^\top O w = \frac{1}{2} w^\top (O + O^\top ) w= w^\top O^{\operatorname{sym}}w\). Hence \(Ow\cdot w=0\) if and only if \(w^\top O^{\operatorname{sym}} w=0\). Thus the symmetric part \(O^{\operatorname{sym}}\) of \(O\) vanishes, i.e., \(O^\top =-O\). Thereupon, consider \[\begin{align} O_+ w = \frac{\theta}{2} w+ \frac{\theta'}{2} Ow\quad\text{and}\quad O_- w = \frac{\theta}{2} w- \frac{\theta'}{2}Ow\;\;\text{ with \theta' = \sqrt{4-\theta^2}}. \end{align}\] It appears that \(\theta w= O_-w+O_+w\) and \(O_-,O_+ \in \mathcal{O}(d)\) are rotations. Indeed, since \(|Ow|= |w|\) and \(Ow.\cdot w=0\) we have \[\begin{align} | O_\pm w |^2= \frac{\theta^2}{4}|w|^2+ \frac{\theta'^2}{4}|Ow|^2 \pm \frac{\theta\theta'}{2}O w\cdot w= |w|^2. \end{align}\]

Given that \(\,\mathrm{d}\sigma_{d-1}\) is invariant under rotation, we obtain \[\begin{align} \int_{\mathbb{S}^{d-1}} U_*(\theta sw)\,\mathrm{d}\sigma_{d-1}(w) &\leq 2^{p-1} \int_{\mathbb{S}^{d-1}} U_*(sO_-w)+ U_*(sO_+w)\,\mathrm{d}\sigma_{d-1}(w) \\ &= 2^{p} \int_{\mathbb{S}^{d-1}} U_*(sw)\, \,\mathrm{d}\sigma_{d-1}(w). \end{align}\] If \(d\) is odd then \(d-1\geq 2\) is even thus the previous case reveals that \[\begin{align} \int_{\mathbb{S}^{d-2}_v} U_*(\theta sw)\,\mathrm{d}\sigma_{d-2}(w)\leq 2^{p} \int_{\mathbb{S}^{d-2}_v} U_*(sw) \,\mathrm{d}\sigma_{d-2}(w)\quad\text{for all v\in \mathbb{S}^{d-1}}. \end{align}\] Thus, the slicing integration formula from Theorem 19 yields \[\begin{align} \int_{\mathbb{S}^{d-1}} U_*(\theta sw) \,\mathrm{d}\sigma_{d-1}(w) &= \int_{\mathbb{S}^{d-1}} \Big( \int_{\mathbb{S}^{d-2}_v}U_*(\theta sw) \frac{\,\mathrm{d}\sigma_{d-2}(w)}{|\mathbb{S}^{d-2}|} \Big) \,\mathrm{d}\sigma_{d-1}(v)\\ &\leq 2^{p} \int_{\mathbb{S}^{d-1}} \Big(\int_{\mathbb{S}^{d-2}_v} U_*(sw) \frac{\,\mathrm{d}\sigma_{d-2}(w)}{|\mathbb{S}^{d-2}|} \Big) \,\mathrm{d}\sigma_{d-1}(v)\\ &= 2^{p}\int_{\mathbb{S}^{d-1}} U_*(sw) \,\mathrm{d}\sigma_{d-1}(w). \end{align}\] ◻

Proof. [itme:lp-diff1] The parity of \(U\) and \(AU\) is easy to verify. By Convexity we obtain \[\begin{align} U(t+r) &= \int_{\mathbb{S}^{d-1}} \int_{\mathop{\mathrm{\mathbb{R}}}^d} |u(x + (t+r)\omega) - u(x)|^p \,\mathrm{d}x \,\mathrm{d}\sigma_{d-1}(\omega)\\ &\leq 2^{p-1}\int_{\mathbb{S}^{d-1}} \int_{\mathop{\mathrm{\mathbb{R}}}^d} |u(x + (t+r)\omega) - u(x + r\omega)|^p \,\mathrm{d}x \,\mathrm{d}\sigma_{d-1}(\omega)\\ &+2^{p-1}\int_{\mathbb{S}^{d-1}} \int_{\mathop{\mathrm{\mathbb{R}}}^d} |u(x + r\omega) - u(x)|^p \,\mathrm{d}x \,\mathrm{d}\sigma_{d-1}(\omega)\\ &=2^{p-1}(U(t)+U(r)). \end{align}\] By the same token we obtain the estimate \(U(t)\leq 2^p|\mathbb{S}^{d-1}|\|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\). Furthermore, the above estimate implies that \[\begin{align} U(t)\leq \frac{2^{p-1}}{t} \int_0^t (U(z)+ U(t-z) )\,\mathrm{d}z = \frac{2^p}{t}\int_0^t U(z)\,\mathrm{d}z = 2^p AU(t). \end{align}\] By convexity again for each \(x\in \mathop{\mathrm{\mathbb{R}}}^d\) and \(\omega\in \mathbb{S}^{d-1}\) we have \[\begin{align} |u(x+\ell t\omega )-u(x)|^p &= \ell ^p \big|\frac{1}{\ell}\sum_{i=0}^{\ell-1}u(x+(i+1) t\omega)-u(x+i t\omega) \big|^p\\ &\leq \ell^{p-1} \sum_{i=0}^{\ell-1}|u(x+(i+1) t\omega)-u(x+i t\omega)|^p. \end{align}\] Integrating both sides over \(\mathbb{S}^{d-1}\times \mathop{\mathrm{\mathbb{R}}}^d\) implies \[\begin{align} U(\ell t)\leq \ell^pU(t)\qquad \text{and hence } \qquad g(\ell t) \leq g(t). \end{align}\] [itme:lp-diff2] Since \(t= s\ell+\theta s\) we deduce from the foregoing that \[\begin{align} U(t) \leq 2^{p-1}\big(U(\ell s) +U(\theta s)\big) &\leq 2^{p-1}\big(\ell^p U(s) +U(\theta s)\big)\leq 2^{p-1}\big( (t/s)^p U(s) +U(\theta s)\big). \end{align}\] We conclude that \[\begin{align} \frac{U(t)}{t^p}\leq 2^{p-1}\Big(\frac{U(s)}{s^p}+ \frac{U(\theta s)}{t^p}\Big). \end{align}\] Moreover, we note that \[\begin{align} U(\theta s)\leq \ell^p U\Big(\tfrac{\theta s}{\ell}\Big) \leq 2^{p-1}\ell^p\Big( U(s)+U\big((1-\tfrac{\theta }{\ell})s\big)\Big). \end{align}\] Inserting this in the previous expression gives \[\begin{align} \frac{U(t)}{t^p}\leq 4^{p-1}(1+\ell^p)\Big(\frac{U(s)}{s^p}+ \frac{U\big((1-\tfrac{\theta }{\ell})s\big) }{t^p}\Big). \end{align}\] Using the subadditivity and parity of \(U\) we find that \[\begin{align} U(s) &\leq \frac{2^{p-1}}{t} \int_0^t\big(U(z) + U(s-z)\,\mathrm{d}z \big)= 2^{p-1}\big( AU(t)+ \frac{1}{t} \int^s_{s-t} U(r)\,\mathrm{d}r\big)\\ &\leq 2^{p-1}\big( AU(t)+ \frac{1}{t} \int^t_{-t} U(r)\,\mathrm{d}r\big)= 3.2^{p-1}AU(t)\leq 2^{p+1}AU(t). \end{align}\] That is \(U(r)\leq 2^{p+1} AU(t)\). Using \(U(r)\leq 2^{p+1} AU(t)\) for \(0<r\leq s\leq t\) we get \[\begin{align} AU(s)= \frac{1}{s}\int_0^s U(r)\,\mathrm{d}r \leq 2^{p+1}AU(t). \end{align}\] [itme:lp-diff2-bis] Now assuming \(d\geq 2\), by Lemma 2, we implies that \(U(s)\leq 2^p U(t)\) for \(0<s\leq t\). Moreover, while \(U(t)\leq 2^{p} AU(t)\)has already been proved, combining the latter with the previously established estimates gives \[\begin{align} \frac{U(t)}{t^p}&\leq 2^{p-1}\Big(\frac{U(s)}{s^p}+ \frac{U(\theta s)}{t^p}\Big)\leq 4^{p}\frac{U(s)}{s^p},\\ U(t+s)&\leq 2^{p-1}(U(t)+U(s))\leq 4^p U(t),\\ AU(t)&= \frac{1}{t}\int_0^t U(z)\,\mathrm{d}z \leq 2^p U(t). \end{align}\]

[itme:lp-diff3] The continuity of the shift, i.e., \(\|u(\cdot+h)-u\|_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\to 0\) as \(|h|\to0\) implies \(\lim_{t \to 0^+} U(t)= \lim_{t \to 0^+} AU(t)=0\) and the continuity of \(U\) and hence of \(AU\), \(\frac{U(t)}{t^p}\). To compute the limit of \(U(t)\) as \(t\to\infty\) let us consider \((u_n)_n\subset C_c^\infty(\mathop{\mathrm{\mathbb{R}}}^d)\) be a sequence converging to \(u\) in \(L^p(\mathop{\mathrm{\mathbb{R}}}^d)\) and let \(U_n\) be defined for \(u_n\) in the same manner as \(U\) is defined for \(u\). For fixed \(u_n\), suppose that \(\mathop{\mathrm{supp}}u_n \subset B_R(0)\) for some \(R>0\) so that \(\mathop{\mathrm{supp}}~u_n(\cdot+t\omega) \subset B_R(-t\omega)\). Therefore, for \(\omega \in \mathbb{S}^{d-1}\) and \(t>2R\) \[\begin{align} \mathop{\mathrm{supp}}u_n \;\cap\; \mathop{\mathrm{supp}}\big(u_n(\cdot+t\omega)\big) = \emptyset. \end{align}\] Consequently, since for \(t>2R\), \(x\in\mathop{\mathrm{\mathbb{R}}}^d\) and \(\omega\in \mathbb{S}^{d-1}\) we have \[\begin{align} |u_n(x+t\omega)\pm u_n(x)|^p = |u_n(x+t\omega)|^p + |u_n(x)|^p. \end{align}\] Hence we deduce that for all \(t\geq 2R\) we have \[\begin{align} \int_{\mathbb{S}^{d-1}}\|u_n(\cdot+t\omega)\pm u_n\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\,\mathrm{d}\sigma_{d-1}(\omega) = 2|\mathbb{S}^{d-1}| \|u_n\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}. \end{align}\] This shows that \(\|u_n(\cdot+ t\omega)\pm u_n\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\to 2\|u_n\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\) and \(U_n(t)\to 2|\mathbb{S}^{d-1}|\|u_n\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\) as \(t\to \infty\). Next, by reverse triangular inequality, we obtain \[\begin{align} \big|U^{\frac{1}{p}}_n(t)-U^{\frac{1}{p}}(t)\big|&\leq 2|\mathbb{S}^{d-1}|^{\frac{1}{p}}\| u_n-u\|_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}. \end{align}\] Therefore, for each \(u_n\), we find that \[\begin{align} \limsup_{t\to\infty}|U^{\frac{1}{p}}(t)&- 2|\mathbb{S}^{d-1}|^{\frac{1}{p}}\|u\|_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}|\\ &\leq \limsup_{t\to\infty}\Big(|U^{\frac{1}{p}}_n(t)- 2|\mathbb{S}^{d-1}|^{\frac{1}{p}}\|u\|_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}|+\big|U^{\frac{1}{p}}_n(t)- U^{\frac{1}{p}}(t)\big|\Big)\\ &=2|\mathbb{S}^{d-1}|^{\frac{1}{p}} | \|u_n\|_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}-\|u\|_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}|+ \limsup_{t\to\infty}\big|U^{\frac{1}{p}}_n(t)- U^{\frac{1}{p}}(t)\big|\\ &\leq 4|\mathbb{S}^{d-1}|^{\frac{1}{p}}\| u_n-u\|_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}. \end{align}\] Letting \(n\to \infty\) so that \(\| u_n-u\|_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\to 0\) thereby yielding \[\begin{align} \lim_{t\to\infty}U(t)=U(\infty):= 2|\mathbb{S}^{d-1}|\|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\quad\text{and hence}\quad \lim_{t\to\infty}\frac{U(t)}{t^p}=0. \end{align}\] Analogously, one establishes that that for almost all \(\omega\in \mathbb{S}^{d-1}\), \[\begin{align} \lim_{t\to\infty}\|u(\cdot+ t\omega)\pm u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}= 2\|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}. \end{align}\] For the case \(AU\), let \(\varepsilon>0\) and \(t_0\) such that \(|U(t) -U(\infty)|<\varepsilon\) for all \(t>t_0\). \[\begin{align} |AU(t) -U(\infty)|= \Big|\frac{1}{t}\int_0^t(U(z) -U(\infty))\,\mathrm{d}z\Big|\leq \frac{\varepsilon(t-t_0) }{t}+ \frac{1}{t}\int_{0}^{t_0} |U(z) -U(\infty)|\,\mathrm{d}z. \end{align}\] It follows that \(\limsup\limits_{t\to\infty} |AU(t) -U(\infty)|\leq \varepsilon\). We deduce \(AU(t)\xrightarrow{t\to\infty} U(\infty)\).

[itme:lp-diff4] To show that \(\frac{U(t)}{t^p}\to g(\infty):=K_{d,p}|\mathbb{S}^{d-1}|\|\nabla u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\) as \(t\to 0\), note that \[\begin{align} \int_{ \mathbb{S}^{d-1}}\int_{\mathop{\mathrm{\mathbb{R}}}^d}|\nabla u(x)\cdot \omega|^p\,\mathrm{d}x \,\mathrm{d}\sigma_{d-1}(\omega)= |\mathbb{S}^{d-1}|K_{d,p}\|\nabla u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}= g(\infty). \end{align}\] The fundamental theorem of calculus implies \(U(t)\leq |\mathbb{S}^{d-1}| K_{d,p}\|\nabla u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}t^p\) and \[\begin{align} \Big|(|\mathbb{S}^{d-1}|&K_{d,p})^{\frac{1}{p}} \|\nabla u\|_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}-\frac{U^{\frac{1}{p}}(t)}{t} \Big| =\left|\Big( \int_{\mathbb{S}^{d-1}} \int_{\mathop{\mathrm{\mathbb{R}}}^d} \Big| \int_0^1\nabla u(x)\cdot\omega\;d \tau \Big|^p \,\mathrm{d}x \,\mathrm{d}\sigma_{d-1}(\omega)\Big)^{\frac{1}{p}}\right. \\ &\qquad\left.-\Big( \int_{\mathbb{S}^{d-1}} \int_{\mathop{\mathrm{\mathbb{R}}}^d} \Big| \int_0^1\nabla u(x+ \tau t\omega)\cdot \omega\;d \tau \Big|^p \,\mathrm{d}x \,\mathrm{d}\sigma_{d-1}(\omega)\Big)^{\frac{1}{p}}\right|. \end{align}\] The reverse triangular inequality, Jensen’s inequality and Fubini’s imply

\[\begin{align} \Big|[g(\infty)]^{\frac{1}{p}} -\frac{U^{\frac{1}{p}}(t)}{t} \Big| \leq \Big( \int_0^1 \int_{\mathbb{S}^{d-1}} \|\nabla u(\cdot + \tau t\omega)-\nabla u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)} \,\mathrm{d}\sigma_{d-1}(\omega) \,\mathrm{d}\tau\Big)^{\frac{1}{p}}\xrightarrow{t\to0}0, \end{align}\] where the conclusion is implied by the dominated convergence theorem, since \(\|\nabla u(\cdot + \tau t\omega)-\nabla u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)} \leq 2^p\|\nabla u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\) and \(\|\nabla u(\cdot + \tau t\omega) - \nabla u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)} \to 0\) as \(t \to 0\) by the continuity of the shift operator in \(L^p(\mathop{\mathrm{\mathbb{R}}}^d)\). Next let \(\varepsilon>0\) and consider \(t_0>0\) such that \(\big|\frac{U(t)}{t^p} -g(\infty)\big|<\varepsilon\) for all \(0<t<t_0\). For all \(0<t<t_0\) we find that \[\begin{align} \Big |\frac{(p+1)}{t^p}AU(t) -g(\infty)\Big|= \frac{(p+1)}{t^{p+1}}\int_0^t z^p\Big|\frac{U(z) }{z^p}-g(\infty)\Big|\,\mathrm{d}z \leq (p+1)\varepsilon. \end{align}\] We conclude that \((p+1)\frac{AU(t)}{t^p}\xrightarrow{t\to\infty} g(\infty)\). ◻

Proof. According to Theorem 20 we know that \(U(t)\to |\mathbb{S}^{d-1}| \|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\) as \(t\to\infty\) and \(U(t)\leq 2^p |\mathbb{S}^{d-1}| \|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\). Therefore, we deduce from Lemma 1 that \[\begin{align} &\lim_{s \to 0^+} \frac{sp}{p} \int_r^\infty \frac{U(t)}{t^{1+s p}} \,\mathrm{d}t = \frac{|\mathbb{S}^{d-1}|}{p} \|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}. \end{align}\] Furthermore, since \(U(t)\leq 2^p|\mathbb{S}^{d-1}|\|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\) we have \[\begin{align} \lim_{s\to1^-} (1-s)\int_r^\infty \frac{U(t)}{t^{1+s p}}\leq 2^p|\mathbb{S}^{d-1}|\|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\lim_{s\to1^-} \frac{(1-s) r^{-sp}}{sp}=0. \end{align}\] Assume \(u\in W^{s_0, p}(\mathop{\mathrm{\mathbb{R}}}^d)\) for some \(0<s_0<1\) then, for \(0<s<s_0\) we have \[\begin{align} s\int_0^r \frac{U(t)}{t^{1+s p} }\leq sr^{(s_0-s)p} \int_0^r \frac{U(t)}{t^{s_0p+1} } \leq sr^{(s_0-s)p} |u|^p_{W^{s_0, p}(\mathop{\mathrm{\mathbb{R}}}^d)}\xrightarrow{s\to 0}0. \end{align}\] Now if \(u \in W^{1,p}(\mathop{\mathrm{\mathbb{R}}}^d)\), then \(g(t)=\frac{U(t)}{t^p}\) satisfies \(g(t)\leq |\mathbb{S}^{d-1}| K_{d,p} \|\nabla u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\) and \[\begin{align} \lim_{t \to 0^+} \frac{U(t)}{t^p} = |\mathbb{S}^{d-1}| K_{d,p} \|\nabla u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)} = g(0). \end{align}\] Therefore, putting \(a = (1 - s)p\) we deduce from Lemma 1 that \[\begin{align} \lim_{s \to 1^-}(1 - s) \int_0^r \frac{U(t)}{t^{1+s p}} \,\mathrm{d}t = \lim_{a \to 0^+} \frac{a}{p} \int_0^r g(t)t^{a-1}\,\mathrm{d}t=\frac{g(0)}{p}= \frac{|\mathbb{S}^{d-1}|}{p} K_{d,p} \|\nabla u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}. \end{align}\] ◻

Proof. For \(u\in \bigcup_{s\in (0,1)} W^{s, p}(\mathop{\mathrm{\mathbb{R}}}^d)\) Theorem 22 implies \[\begin{align} \lim_{s\to0^+} [u]^p_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)}&= \lim_{s\to0^+}\Big( s\int_1^\infty \frac{U(t)}{t^{1+s p}} +s\int_0^1 \frac{U(t) }{t^{1+s p}}\,\mathrm{d}t\Big) =\frac{2|\mathbb{S}^{d-1}|}{p}\|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}, \end{align}\] where we recall that \[\begin{align} [u]^p_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)}= s(1-s)\iint_{\mathop{\mathrm{\mathbb{R}}}^d \times \mathop{\mathrm{\mathbb{R}}}^d} \frac{ |u(x) - u(y)|^p}{|x-y|^{d+s p}}\,\mathrm{d}y\,\mathrm{d}x. \end{align}\] Conversely, if the above limit holds then \(u\in \bigcup_{0 < s < 1} W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)\) since \[\begin{align} s\iint_{\mathop{\mathrm{\mathbb{R}}}^d \times \mathop{\mathrm{\mathbb{R}}}^d} \frac{ |u(x) - u(y)|^p}{|x-y|^{d+s p}}\,\mathrm{d}y\,\mathrm{d}x\leq \big(1+\tfrac{2|\mathbb{S}^{d-1}|}{p}\|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\big)\quad\text{for all 0<s<s_0}, \end{align}\] for some \(s_0\in (0,1)\). Next for \(u\in W^{1,p}(\mathop{\mathrm{\mathbb{R}}}^d)\), \(p\geq1\), Theorem 22 implies \[\begin{align} \lim_{s\to0^+} [u]^p_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)} &= \lim_{s\to1^-} (1-s) \Big( \int_0^1 \frac{U(t) }{t^{1+s p}}\,\mathrm{d}t+ \int_1^\infty \frac{U(t) }{t^{1+s p}}\,\mathrm{d}t\Big) = K_{d,p}\frac{|\mathbb{S}^{d-1}|}{p}\|\nabla u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}. \end{align}\] The general case \(u\in BV(\mathop{\mathrm{\mathbb{R}}}^d)\) when \(p=1\) follows analogously, see [6] and [33]. Recall that by [4], see also [6], we find that \[\begin{align} &\liminf_{s\to1^-} [u]^p_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)}<\infty \implies \begin{cases} u\in W^{1,p}(\mathop{\mathrm{\mathbb{R}}}^d), & \text{if } 1<p<\infty\\ u\in BV(\mathop{\mathrm{\mathbb{R}}}^d),&\text{if } p=1. \end{cases} \end{align}\] In particular, if \(u\notin W^{1,p}(\mathop{\mathrm{\mathbb{R}}}^d)\) for \(1<p<\infty\) (resp. \(u\notin BV(\mathop{\mathrm{\mathbb{R}}}^d)\) for p=1) then \[\begin{align} \liminf_{s\to1^-}s(1-s)\iint_{\mathop{\mathrm{\mathbb{R}}}^d \times \mathop{\mathrm{\mathbb{R}}}^d} \frac{ |u(x) - u(y)|^p}{|x-y|^{d+s p}}\,\mathrm{d}y \,\mathrm{d}x=\infty, \end{align}\] that is we have \(\|\nabla u\|^p_{L^{p}(\mathop{\mathrm{\mathbb{R}}}^d)}=\infty\) (resp. \(|u|_{BV(\mathop{\mathrm{\mathbb{R}}}^d)}=\infty\)). ◻

Proof. This is directly implied by Theorem 23 and the fact that \[\begin{align} \lim_{s\to 1^-} \frac{K_{d,p}C_{d,p,s} }{2s(1-s)} =\lim_{s\to 0^+} \frac{C_{d,2,\frac{sp}{2}}}{s(1-s)}=\frac{p}{|\mathbb{S}^{d-1}|}. \end{align}\] ◻

Proof. Since \(x\in \Omega\) we have \(\delta_x=\operatorname{dist}(x,\partial\Omega)>0\). Let \(c_1= \|f\|_{L^\infty(B_\rho(x))}\) and \(c_2= \|g\|_{L^\infty(\Omega)}\). For arbitrarily large \(\rho>\delta_x\) and arbitrarily small \(0<\delta<\delta_x\) we have \[\begin{align} &s \int_{ \Omega^c\cap B_\rho(x)} \frac{|f(y)|\, \,\mathrm{d}y}{|x-y|^{d+sp}} \leq \int_{B_\rho(x)\setminus B_{\delta_x}(x)}\frac{s c_1\,\mathrm{d}y}{|x-y|^{d+sp}} = c_1\frac{|\mathbb{S}^{d-1}|}{p} (\delta_x^{-sp} -\rho^{-sp})\xrightarrow{s\to 0^+}0,\\ &(1-s) \int_{\Omega\cap B^c_{\delta} (x)} \frac{ |g(y)|\,\mathrm{d}y}{|x-y|^{d-(1-s)p}} \leq \int_{B^c_\delta(x)} \frac{c_2(1-s)\,\mathrm{d}y}{|x-y|^{d-(1-s)p}} = c_2\frac{ (1-s) |\mathbb{S}^{d-1}|}{sp \delta^{(s-1)p}} \xrightarrow{s\to 1^-}0. \end{align}\] Using the previous remark, Lemma 1 and dominated convergence we get \[\begin{align} \lim_{s \to 0^+}s \int_{\mathop{\mathrm{\mathbb{R}}}^d \setminus \Omega} \frac{ f(y)\, \,\mathrm{d}y}{|x-y|^{d+sp}} &=\lim_{s \to 0^+}s \int_{B^c_\rho(x)} \frac{ f(y)\mathbb{1}_{\Omega^c}(y)\, \,\mathrm{d}y}{|x-y|^{d+sp}} \\ &= \int_{\mathbb{S}^{d-1}} \lim_{s\to 0^+}s\int_\rho^\infty f(x + rw)\mathbf{1}_{\Omega^c}(x + rw) r^{-sp-1}\,\mathrm{d}r \, \,\mathrm{d}\sigma_{d-1}(w)\\ &= \frac{1}{p} \int_{\mathbb{S}^{d-1}}\lim_{r\to \infty} f(x+rw) \mathbb{1}_{\Omega^c}(x+rw)\, \,\mathrm{d}\sigma_{d-1}(w)\\ &=\frac{1}{p} \int_{\mathbb{S}^{d-1}} f_\infty(x, w) \mathbb{1}_{S_\Omega(x)}(w)\, \,\mathrm{d}\sigma_{d-1}(w), \end{align}\] where we note that, for each \(w \in \mathbb{S}^{d-1}\), \[\begin{align} \lim_{r\to \infty} \mathbb{1}_{\Omega^c}(x+rw)&= \mathbb{1}_{S_\Omega(x)} (w)= \begin{cases} 1 & \text{if } x + rw \in \Omega^c \text{ for all sufficiently large } r, \\ 0 & \text{otherwise}. \end{cases} \end{align}\] Likewise, since \(B_\delta(x)\subset B_{\delta_x}(x)\subset \Omega\), dominated convergence and Lemma 1 yield \[\begin{align} \lim_{s\to 1^-}(1-s) \int_{\Omega} \frac{ g(y)\,\mathrm{d}y}{|x-y|^{d-(1-s)p}} &= \lim_{a \to 0^+}\frac{a}{p} \int_{B_{\delta} (x)} \frac{ g(y)\,\mathrm{d}y}{|x-y|^{d-a}} \\ &= \int_{\mathbb{S}^{d-1}} \lim_{a \to 0^+}\frac{a}{p} \int_0^\delta g(x+r w) r^{a-1}\,\mathrm{d}r \,\mathrm{d}\sigma_{d-1}(w) \\ &= \frac{1}{p} \int_{\mathbb{S}^{d-1}} g_0(x,w)\,\mathrm{d}\sigma_{d-1}(w). \end{align}\] Finally, if \(\Omega\) is bounded then we have \(\Omega\subset B_R(x)\) for \(R>0\) large enough, so that \(x+rw\in B_R^c(x)\subset \Omega^c\) for all \(w\in \mathbb{S}^{d-1}\) and \(r>R.\) Hence one readily gets that \(S_\Omega(x) =\mathbb{S}^{d-1}\). Likewise if \(\Omega^c\) is bounded say \(\Omega^c \subset B_R(x)\) for some \(R>0\), then for all \(r>R\) and \(w\in \mathbb{S}^{d-1}\) we have \(x+rw\in \Omega\subset B_R^c(x)\), that is, \(\mathbb{1}_{\Omega^c}(x+rw)=0\) and hence \(S_\Omega(x)= \emptyset\). Analogously it is not difficult to verify that \(S_{\mathop{\mathrm{\mathbb{R}}}_+}(x)=\mathbb{S}^{d-1}_+\). ◻

Proof. \(\eqref{item:mazya-dom0}\) Let \(u\in L^p(\Omega, \delta_x^{-s_0p})\) for some \(0<s_0<1\). Lemma 3 implies \[\begin{align} s \int_{\mathop{\mathrm{\mathbb{R}}}^d \setminus \Omega} \frac{\,\mathrm{d}y}{|x-y|^{d+sp}} \xrightarrow{s\to0^+} \frac{1}{p}|S_\Omega(x)|\qquad x\in \Omega. \end{align}\] Moreover, for all \(0<s\leq s_0\) we have \[\begin{align} s \int_{\mathop{\mathrm{\mathbb{R}}}^d \setminus \Omega} \frac{\,\mathrm{d}y}{|x-y|^{d+sp}}\leq s \int_{\mathop{\mathrm{\mathbb{R}}}^d \setminus B_{\delta_x}(x)} \frac{\,\mathrm{d}y}{|x-y|^{d+sp}}= \frac{|\mathbb{S}^{d-1}|}{p}\delta_x^{-sp} \leq \frac{|\mathbb{S}^{d-1}|}{p}(1+\delta_x^{-s_0p}). \end{align}\] Since \(|u(x)|^p(1+\delta_x^{-s_0p})\in L^1(\Omega)\), the dominated convergence theorem yields \[\begin{align} \lim_{s\to0^+} s \int_{\Omega} |u(x)|^p \int_{\mathop{\mathrm{\mathbb{R}}}^d \setminus \Omega} \frac{\,\mathrm{d}y }{|x-y|^{d+sp}} \,\mathrm{d}x=\frac{|\mathbb{S}^{d-1}|}{p} \int_{\Omega} |u(x)|^p\, |S_\Omega(x)| \,\mathrm{d}x. \end{align}\] \(\eqref{item:mazya-dom}\) Assume \(u\in W^{s_0,p}(\Omega)\cap L^p(\Omega, \delta_x^{-s_0p})\) for some \(0<s_0<1\). It appears that \(\widetilde{u}\in W^{s_0,p}(\mathop{\mathrm{\mathbb{R}}}^d)\) where \(\widetilde{u}\) is the zero-extension of \(u\), i.e., \(\widetilde{u}(x)= u(x)\) for \(x\in \Omega\) and \(\widetilde{u}(x)=0\) for \(x\in \Omega^c\). Indeed, we have \[\begin{align} \iint_{\mathop{\mathrm{\mathbb{R}}}^d \times \mathop{\mathrm{\mathbb{R}}}^d} \frac{|\widetilde{u}(x) - \widetilde{u}(y)|^p}{|x-y|^{d+s_0p}} \,\mathrm{d}y \,\mathrm{d}x &\leq \iint_{\Omega \times \Omega} \frac{|u(x) - u(y)|^p}{|x-y|^{d+s_0p}} \,\mathrm{d}y \,\mathrm{d}x + \frac{2 |\mathbb{S}^{d-1}|}{s_0p}\int_{\Omega} \frac{|u(x)|^p}{\delta_x^{s_0p}}\,\mathrm{d}x. \end{align}\] Therefore, since \(\widetilde{u}\in W^{s_0,p}(\mathop{\mathrm{\mathbb{R}}}^d)\), Theorem 23 implies \[\begin{align} \lim_{s \to 0^+} s [\widetilde{u}]^p_{W^{s,p}(\mathbb{R}^d)} = \frac{2}{p}|\mathbb{S}^{d-1}| \int_{\mathop{\mathrm{\mathbb{R}}}^d} |\widetilde{u}(x)|^p \,\mathrm{d}x= \frac{2}{p}|\mathbb{S}^{d-1}| \int_{\Omega} |u(x)|^p \,\mathrm{d}x. \end{align}\] This in conjunction with the foregoing gives \[\begin{align} \lim_{s\to 0^+} [u]^p_{W^{s,p}(\Omega)} &= \lim_{s\to 0^+} [\widetilde{u}]^p_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)}-2\lim_{s\to 0^+}s \int_{\Omega} |u(x)|^p \Big( \int_{\mathop{\mathrm{\mathbb{R}}}^d \setminus \Omega} \frac{\,\mathrm{d}y}{|x-y|^{d+sp}} \Big) \,\mathrm{d}x\\ &= \frac{2}{p}\int_\Omega (|\mathbb{S}^{d-1}|- |S_\Omega(x)| )|u(x)|^p\,\mathrm{d}x. \end{align}\] \(\eqref{item:mazya-dom1}\) If \(\partial\Omega\) is compact Lipschitz or \(\Omega=\{x= (x',x_d): x_d>\zeta(x')\}\), then the fractional Hardy inequality [34], [35] entails the continuous embedding \(W^{s,p}(\Omega)\hookrightarrow L^p(\Omega, \delta_x^{-sp})\) for \(0<s<\frac{1}{p}\). The desired results follow from the previous assertion.

[item:mazya-dom2] This follows from Lemma 3 since for all \(x\in \mathop{\mathrm{\mathbb{R}}}^d_+\) we have \(S_{\mathop{\mathrm{\mathbb{R}}}^d_+}(x)=\mathbb{S}^{d-1}_+\), whereas for all \(x\in \Omega\) we have \(S_{\Omega}(x)=\mathbb{S}^{d-1}\) if \(\Omega\) is bounded and \(S_{\Omega}(x)=\emptyset\) if \(\Omega^c\) is bounded. ◻

Proof. Since \(B_R(0)\) is Lipschitz and \(\Omega\subset B_R(0)\) for some \(R>0\), Theorem 24 implies \[\begin{align} &\limsup_{s\to 0^+} s \iint_{\Omega \times \Omega} \frac{|u(x) - u(y)|^p}{|x-y|^{d+sp}} \,\mathrm{d}y \,\mathrm{d}x\leq \limsup_{s\to 0^+} s \iint_{B_R(0) \times B_R(0)} \frac{|u(x) - u(y)|^p}{|x-y|^{d+sp}} \,\mathrm{d}y \,\mathrm{d}x=0. \end{align}\] Next, combining Theorem 23 and Theorem 24 we find that \[\begin{align} \lim_{s\to 0^+} s\iint_{\mathop{\mathrm{\mathbb{R}}}^d_- \times \mathop{\mathrm{\mathbb{R}}}^d_+} \frac{|u(x) - u(y)|^p}{|x-y|^{d+sp}} \,\mathrm{d}y \,\mathrm{d}x &=\lim_{s\to 0^+} \frac{s}{2}\Big(\iint_{\mathop{\mathrm{\mathbb{R}}}^d \times \mathop{\mathrm{\mathbb{R}}}^d}- \iint_{\mathop{\mathrm{\mathbb{R}}}^d_+ \times \mathop{\mathrm{\mathbb{R}}}^d_+}- \iint_{\mathop{\mathrm{\mathbb{R}}}^d_- \times \mathop{\mathrm{\mathbb{R}}}^d_-}\Big)[\cdots]\\ &=\frac{|\mathbb{S}^{d-1}|}{2p}\Big( 2\|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}- \|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d_+)}-\|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d_-)}\Big)\\ &=\frac{|\mathbb{S}^{d-1}|}{2p}\|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}. \end{align}\] Analogously if \(\Omega\) is bounded Lipschitz so that \(|\partial \Omega|=0\), then Theorem 24 implies \[\begin{align} &\lim_{s\to 0^+} s(1-s) \iint_{\Omega^c \times \Omega^c} \frac{|u(x) - u(y)|^p}{|x-y|^{d+sp}} \,\mathrm{d}y \,\mathrm{d}x= \frac{2|\mathbb{S}^{d-1}|}{p}\|u\|^p_{L^p(\Omega^c)}, \end{align}\] while, splitting the integral, we deduce from the foregoing that \[\begin{align} \lim_{s\to 0^+} s\iint_{\Omega^c \times \Omega}\frac{|u(x) - u(y)|^p}{|x-y|^{d+sp}} &\,\mathrm{d}y \,\mathrm{d}x = \lim_{s\to 0^+} \frac{s}{2}\Big(\iint_{\mathop{\mathrm{\mathbb{R}}}^d \times \mathop{\mathrm{\mathbb{R}}}^d} - \iint_{\Omega \times \Omega}- \iint_{\Omega^c \times \Omega^c}\Big)[\cdots]\\ &=\frac{|\mathbb{S}^{d-1}|}{2p} \Big( \|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}-\|u\|^p_{L^p(\Omega^c)}\Big)=\frac{|\mathbb{S}^{d-1}|}{p}\|u\|^p_{L^p(\Omega)}. \end{align}\] ◻

Proof. The first limit near \(s=1\) can be found in [8]. If \(u(\infty)\) exists and \(u\in L^q( \mathop{\mathrm{\mathbb{R}}}^d)\) with \(1<q<\infty\), necessarily \(u(\infty)=0.\) Moreover, note that \(\lim_{s\to0}\frac{ \widetilde{C}_{d,p,s}}{s(1-s)} =\frac{p}{|\mathbb{S}^{d-1}|}.\) Since \(u\in C^\varepsilon(B_1(x))\), there exists \(C>0\) such that \(|u(x)-u(y)| \leq C |x-y|^\varepsilon\), \(y\in B_1(x)\). Therefore, for \(0<s<\varepsilon(p-1)\), we find that \[\begin{align} \left| s\int_{B_1(x)} \frac{|u(x)-u(y)|^{p-2}(u(x)-u(y))}{|x-y|^{d+sp}}dy \right| &\leq C \int_{B_1(x)} |x-y|^{-d-sp+\varepsilon(p-1)}dy \\ &=C\frac{ s|\mathbb{S}^{d-1}|}{ -sp+\varepsilon(p-1)} \to 0 \quad \text{as } s\to0^+. \end{align}\] Next, define \(u_\infty(x)= u(x)-u(\infty)\) and \[g_{x, w}(t)= |u(x)-u(x+t w)|^{p-2}(u(x)-u(x+t w))\] so that \(g_{x,w}(\infty) = |u_\infty(x)|^{p-2}u_\infty(x)\) and \(g_{x, w}\in L^{(p-1)q}_{\mathrm{loc}}((0,\infty))\) with \(1<(p-1)q\leq \infty\). In view of Lemma 1 we find that \[\begin{align} \lim_{s\to0^+}s\int_{B^c_1(x)} \frac{|u(x)-u(y)|^{p-2}(u(x)-u(y))}{|x-y|^{d+sp}}\,\mathrm{d}y&= \lim_{s\to0^+}s\int_{\mathbb{S}^{d-1}} \int_1^\infty \frac{g_{x, w}(t)}{ t^{1+sp}} \,\mathrm{d}t\,\mathrm{d}\sigma_{d-1}(w)\\ &=\frac{|\mathbb{S}^{d-1}|}{p}g_{x, w}(\infty)= \frac{|\mathbb{S}^{d-1}|}{p}|u_\infty(x)|^{p-2}u_\infty(x), \end{align}\] with \(u_\infty(x)= u(x)-u(\infty)\). Altogether we obtain \[\begin{align} \lim_{s\to0^+} \frac{sp}{|\mathbb{S}^{d-1}|}\int_{\mathop{\mathrm{\mathbb{R}}}^d} &\frac{|u(x)-u(y)|^{p-2}(u(x)-u(y))}{|x-y|^{d+sp}}\,\mathrm{d}y= |u_\infty(x)|^{p-2}u_\infty(x). \end{align}\] ◻

Proof. For \(\xi\in \mathop{\mathrm{\mathbb{R}}}\) the mapping \(t\mapsto g_\xi(t) =\frac{1-\cos (\xi t)}{t^2}\) with \(g_\xi(0)= \frac{\xi^2}{ 2}\) is continuous and dyadically decreasing, i.e., \(g_\xi(2t)\leq g_\xi(t)\) for all \(t\geq 0\). Applying the Chebyshev type inequality in Theorem 18 with \(a=2(1-s)\leq b=2(1-\eta)\) implies that \[\begin{align} 2(1-\eta)(2 r)^{-2(1-\eta)} \int_0^{r} g_\xi(t) t^{2(1-\eta)-1}\,\mathrm{d}t\leq 2(1-s)(2 r)^{-2(1-s)}\int_0^r g_\xi(t)t^{2(1-s)-1} \,\mathrm{d}t, \end{align}\] yielding the sought inequality. ◻

Proof. Up to replacing \(\xi\) by \(r\xi\) we can assume by scaling argument that \(r=1\) and by parity that \(\xi\geq 0\). Define the function \[\begin{align} f(\eta,\xi):=\eta \int_1^\infty \frac{(1-\cos (\xi t))}{t^{1+2\eta}}\,\mathrm{d}t&= \eta \xi^{2\eta} \int_{\xi}^\infty \frac{1-\cos (t)}{t^{1+2\eta}}\,\mathrm{d}t. \end{align}\] In most cases we will only show that \(f(\eta,\xi)\leq 3 f(\sigma,\xi)\) as this readily implies the second desired estimate \((1-\eta)f(\eta,\xi)\leq 3 (1-\sigma)f(\sigma,\xi)\), since \(1-\eta\leq 1-\sigma\). We begin by performing integration by parts, yielding \[\begin{align} \label{eq:xexpansion-away-zero} f(\eta,\xi)= \frac{1}{2}- \eta \int_1^\infty \frac{\cos (\xi t)}{t^{1+2\eta}}\,\mathrm{d}t &=\frac{1}{2} + \frac{\eta\sin (\xi)}{\xi} -\frac{\eta(1+2\eta)}{\xi} \int_1^\infty \frac{\sin (\xi t)}{t^{2+2\eta}}\,\mathrm{d}t. \end{align}\tag{5}\] For fixed \(\xi_0>0\) then we have \[\begin{align} \label{eq:xuniform-away-zero} \sup_{\xi\geq \xi_0}\Big| f(\eta,\xi)-\frac{1}{2}\Big| \leq \frac{\eta}{\xi_0}+ \frac{\eta (1+2\eta)}{\xi_0}\int_1^\infty\frac{\,\mathrm{d}t}{t^{2+2\eta}}= \frac{2\eta}{\xi_0}. \end{align}\tag{6}\] We note in passing that \(f(\sigma, 0)=0\) while for \(\xi>0\) the estimate 6 readily implies \[\begin{align} f(0,\xi)=& \lim_{\sigma\to 0^+}\sigma \int_1^\infty \frac{(1-\cos (\xi t))}{t^{1+2\sigma}}\,\mathrm{d}t =\frac{1}{2}, \\ f(\sigma, \infty)=&\lim_{\xi\to \infty}\sigma \int_1^\infty \frac{(1-\cos (\xi t))}{t^{1+2\sigma}}\,\mathrm{d}t = \frac{1}{2}. \end{align}\] From now on, we take \(\xi_0=8\) hence we arrive at \[\begin{align} \label{eq:xgen-bound} \frac{1}{2}- \frac{\eta}{4}\leq f(\eta,\xi)\leq \frac{1}{2}+ \frac{\eta}{4}&& \text{for all \eta\in (0,1)and \xi\geq 8}. \end{align}\tag{7}\] In particular, since \(0\leq\eta\leq 1\) we find that \[\begin{align} \frac{1}{4}\leq f(\eta,\xi)\leq \frac{3}{4}&& \text{for all \eta\in (0,1)and \xi\geq 8}. \end{align}\] The latter estimate clearly implies \[\begin{align} f(\eta,\xi)\leq 3f(\sigma,\xi) && \text{for all \sigma, \eta\in (0,1)and \xi\geq 8}. \end{align}\] Therefore, it remains to deal with the case \(0\leq \xi\leq 8\) distinguished in tree cases.

Case 1: Assume \(0\leq \sigma\leq \eta\leq \frac{1}{4}\) and \(0\leq \xi\leq 8\). Let us obverse that \[\begin{align} f(\eta,\xi)=\eta \xi^{2\eta} \int_{\xi}^\infty \frac{1-\cos (t)}{t^{1+2\eta}}\,\mathrm{d}t\leq \frac{\eta \xi^{2\eta}}{2}\int_{\mathop{\mathrm{\mathbb{R}}}} \frac{1-\cos (t)}{t^{1+2\eta}}\,\mathrm{d}t= \frac{\eta \xi^{2\eta} }{2 C_{1,\eta}} , \end{align}\] where \(C_{1,\eta}\), is normalization constant of the one dimensional fractional Laplacian, see [12] or [36], given by \[\begin{align} \frac{1}{C_{1,\eta}} = \int_{\mathop{\mathrm{\mathbb{R}}}} \frac{1-\cos (t)}{t^{1+2\eta}}\,\mathrm{d}t =\frac{\pi^{1/2} \Gamma(1 - \eta)}{2^{2\eta} \eta\Gamma\left( \frac{1 + 2\eta}{2} \right)}= \frac{ \pi}{ \Gamma(1+2\eta) \sin(\pi \eta)}. \end{align}\] The last expression follows from the Schwartz’s reflection and duplication formulas \[\begin{align} \Gamma(1-\eta)\Gamma (\eta)= \frac{\pi}{\sin (\pi\eta)}\qquad \text{ and }\qquad \Gamma(2\eta)= \frac{2^{2\eta} \Gamma(\eta)\Gamma(\tfrac{1+2\eta}{2})}{2\pi^{1/2}}. \end{align}\] Note that \(\frac{\sqrt{\pi}}{2}= \Gamma(1+\tfrac{1}{2})\leq \Gamma(1+2x)\leq 1\) for \(x\in (0,\frac{1}{4})\), while since \(x\mapsto \frac{\sin (x)}{x}\) is decreasing on \([0, \frac{\pi}{4}]\) we have, \[\begin{align} 1\leq \frac{x}{\sin (x)}\leq \frac{\pi\sqrt{2}}{4},\qquad x\in [0, \frac{\pi}{4}]. \end{align}\] Inserting this in the previous expression, implies that \[\begin{align} \label{eq:eta-less-quart-upb} f(\eta,\xi)\leq \frac{ \pi\eta \xi^{2\eta}}{ 2\Gamma(1+2\eta) \sin(\pi \eta)}\leq \frac{\sqrt{2\pi}}{4}\xi^{2\eta} &&\quad \text{ for all \,\, 0\leq \eta\leq \frac{1}{4} and \xi\geq 0}. \end{align}\tag{8}\] Now, since \(0\leq \xi\leq \xi_0=8\) and \(0\leq \sigma\leq \frac{1}{4}\), the lower estimate in 7 implies \[\begin{align} f(\sigma,\xi)\geq \sigma \xi^{2\sigma} \int_{\xi_0}^\infty \frac{1-\cos (t)}{t^{1+2\sigma}}\,\mathrm{d}t= \big(\frac{\xi}{\xi_0}\big)^{2\sigma}f(\sigma,\xi_0)\geq\big(\frac{\xi}{\xi_0}\big)^{2\sigma} \frac{2-\sigma}{4}\geq \frac{7}{16} \big(\frac{\xi}{\xi_0}\big)^{2\sigma}. \end{align}\] In other words, we have \[\begin{align} \label{eq:sigma-less-quart-lob} f(\sigma,\xi)\geq\big(\frac{\xi}{\xi_0}\big)^{2\sigma} \frac{2-\sigma}{4}\geq \frac{7}{16} \big(\frac{\xi}{\xi_0}\big)^{2\sigma}&&\text{0\leq \sigma\leq \frac{1}{4} and 0\leq \xi\leq \xi_0=8 }. \end{align}\tag{9}\]

• If \(0\leq \sigma\leq \eta\leq \frac{1}{4}\) and \(0\leq \xi\leq 3\), the estimates 8 and 9 imply \[\begin{align} \frac{f(\eta,\xi) }{f(\sigma,\xi)}\leq\xi^{2\eta} \big(\frac{\xi}{\xi_0}\big)^{2(\eta-\sigma)} \frac{4\sqrt{2\pi}}{7}\leq \xi^{1/2} \frac{4\sqrt{2\pi}}{7}\leq \frac{4\sqrt{6\pi}}{7}<\frac{5}{2}<3. \end{align}\]

• If \(0\leq \sigma\leq \eta\leq\frac{1}{4}\) and \(3\leq \xi\leq 8\), the lower estimate in 6 implies that \[\begin{align} \label{eq:sigma-less-quart-lob-bis} f(\sigma,\xi)\geq\big( \frac{1}{2}-\frac{2\sigma}{\xi}\big)\geq \big( \frac{1}{2}-\frac{1}{6}\big)=\frac{1}{3}&& \text{for all 0\leq \sigma\leq \frac{1}{4} and \xi\geq 3}. \end{align}\tag{10}\] This together with the fact that \(f(\eta,\xi) \leq 1\) implies \[\begin{align} f(\eta,\xi) \leq 1\leq3f(\sigma,\xi). \end{align}\]

Case 2: Assume that \(\frac{1}{4}\leq\sigma\leq \eta\leq 1\) and \(0\leq \xi\leq 8\).

• Assume \(\frac{1}{4}\leq\sigma\leq \eta\leq \frac{3}{4}\) or \(\frac{3}{4}\leq\sigma\leq \eta\leq 1\) and \(0\leq \xi\leq 8\). Since \(t^{-2\eta}\leq t^{-2\sigma}\), \(t\geq 1\) and \(\frac{1}{3}\leq \frac{\eta}{\sigma} \leq 3\), it follows that \[\begin{align} f(\eta,\xi)\leq 3f(\sigma,\xi) && \text{for all \xi\geq 0}. \end{align}\]

• Assume \(\frac{1}{3}\leq\sigma\leq \frac{3}{4}\leq \eta\leq 1\) and \(0\leq \xi\leq 8\). Since \(\frac{1}{3}\leq \frac{\eta}{\sigma} \leq 3\) and \(t^{-2\eta}\leq t^{-2\sigma}\), \(t\geq 1\), it follows that \[\begin{align} f(\eta,\xi)\leq 3f(\sigma,\xi) && \text{for all \xi\geq 0}. \end{align}\]

• Assume \(\frac{1}{4}\leq\sigma\leq \frac{1}{3}\) and \(\frac{3}{4}\leq \eta\leq 1\) and \(0\leq \xi\leq 8\). Analogously, it follows that \[\begin{align} f(\eta,\xi)\leq 4f(\sigma,\xi) && \text{for all\xi\geq 0}. \end{align}\] On the other hand, using again \(t^{-2\eta}\leq t^{-2\sigma}\), \(t\geq1\), we immediately have \[\begin{align} (1-\eta) f(\eta, \xi)\leq (1-\sigma) f(\sigma,\xi),&& \text{for all\xi\geq 0}, \end{align}\] where we note that \(\frac{\eta(1-\eta)}{\sigma(1-\sigma)}\leq 1\) since \[\begin{align} \max_{\tau\in\left[\frac{3}{4},\,1\right]} \tau(1-\tau) = \frac{3}{16} = \min_{\tau\in\left[\frac{1}{4},\,\frac{1}{3}\right]} \tau(1-\tau). \end{align}\]

Case 3: Assume \(0\leq \sigma\leq \frac{1}{4}\leq \eta\leq 1\) and \(0\leq \xi\leq 8\).

• If \(0\leq \sigma\leq \frac{1}{4}\leq \eta\leq 1\) and \(3\leq \xi\leq 8\), the estimate 6 yields \[\begin{align} \label{usifzawp} f(\sigma,\xi)\geq\big( \frac{1}{2}-\frac{2\sigma}{\xi}\big)\geq \big( \frac{1}{2}-\frac{1}{6}\big)=\frac{1}{3}&& \text{for all 0\leq \sigma\leq \frac{1}{4} and \xi\geq 3}. \end{align}\tag{11}\] This together with the fact that \(f(\eta,\xi) \leq 1\) implies \[\begin{align} f(\eta,\xi) \leq 1\leq 3f(\sigma,\xi). \end{align}\]

• If \(0\leq \sigma\leq \frac{1}{4}\leq \eta\leq 1\) and \(0\leq \xi\leq 3\). We will show that \(f(\eta,\xi)\leq 4f(\sigma,\xi)\) which implies \((1-\eta)f(\eta,\xi)\leq 3(1-\sigma)f(\sigma,\xi)\) since \((1-\eta)\leq \frac{3}{4}\leq (1-\sigma)\). Let us observe that

\[\begin{align} f(\eta, \xi) &= \eta \int_{1}^{\infty} (1-\cos(\xi t)) \left( \frac{1}{\Gamma(1+2\eta)} \int_{0}^{\infty} x^{2\eta} e^{-xt} \, \,\mathrm{d}x \right) \,\mathrm{d}t \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{x^{2\eta-1} }{\Gamma(2\eta)} \left( \int_{1}^{\infty} (1-\cos(\xi t)) xe^{-xt} \, \,\mathrm{d}t \right) \,\mathrm{d}x \\ &= \frac{1}{2} \int_{0}^{\infty} K(x,\xi) \frac{x^{2\eta-1}}{\Gamma(2\eta)} e^{-x} \,\mathrm{d}x \end{align}\] where we consider the kernel \[\begin{align} K(x,\xi)=e^x \operatorname{Re}\left( \int_{1}^{\infty} (1-e^{i\xi t}) xe^{-xt} \, \,\mathrm{d}t \right) = 1- \frac{x^2\cos (\xi) -x\xi \sin(\xi)}{x^2+\xi^2 }. \end{align}\] Cauchy-Schwartz inequality implies \(-\xi\sin\xi + x\cos\xi \leq \sqrt{x^2+\xi^2}\), so that \[\begin{align} \frac{x\xi\sin\xi}{x^2+\xi^2} \leq \frac{x}{\sqrt{x^2+\xi^2}}- \frac{x^2\cos\xi}{x^2+\xi^2} \leq 1-\frac{x^2\cos\xi}{x^2+\xi^2}. \end{align}\] Since \(\sin\xi\geq 0\) as \(0\leq \xi\leq 3<\pi\) if follows that \[\begin{align} g(x, \xi) \leq K(x,\xi)= g(x,\xi) + \frac{x\xi\sin\xi}{x^2+\xi^2}\leq 2g(x, \xi) \end{align}\] where \(g(\cdot,\xi)\) is given by \[\begin{align} g(x,\xi)= 1-\frac{x^2\cos\xi}{x^2+\xi^2}= 1-\cos\xi + \frac{\xi^2\cos\xi}{x^2+\xi^2}. \end{align}\] If \(0\leq \xi\leq \frac{\pi}{2}\) so that \(\cos \xi\geq 0\) then clearly \(g(x,\xi)\) is decreasing. Hence \[\begin{align} K(y,\xi)\leq 2K(x,\xi)\qquad \text{x\leq y and 0\leq \xi\leq \frac{\pi}{2}}. \end{align}\] If \(\frac{\pi}{2}<\xi\leq \pi,\) then \(g(x,\xi)\) is increasing, \(g(\infty,\xi) =2(1-\cos\xi ) g(0, \xi) \leq 4 g(x,\xi)\). Then \[\begin{align} K(y,\xi)\leq 2 g(y,\xi)\leq 2 g(\infty,\xi) \leq 4 g(x,\xi) \leq 4 K(x,\xi) \end{align}\] for all \(x,y\in (0,\infty)\). Altogether, we deduce that \(K(\cdot,\xi)\) is \(4\)-almost decreasing for \(0\leq \xi\leq 3\), i.e., we have \(K(y,\xi)\leq 4K(x,\xi)\) for \(0<x<y\). The family \((\frac{x^{2\eta-1}}{\Gamma(2\eta)} e^{-x})_\eta\) of probability densities function of \((2\eta,1)\)-Gamma distributions on \((0,\infty)\) and possesses the monotone likelihood ratio property in \(x\), i.e., for \(\sigma<\eta\), the mapping \[\begin{align} x\mapsto \frac{\Gamma(2\sigma)}{\Gamma(2\eta)}x^{2(\eta-\sigma)} \quad\text{is increasing on (0,\infty)}. \end{align}\] Accordingly, Theorem 16 implies that \(f(\eta,\xi)\leq 4f(\sigma,\xi)\). ◻

Proof. Using the Fourier transform, Plancherel’s identity implies \[\begin{align} U(t)&= \int_{ \mathbb{S}^{d-1}} \int_{\mathop{\mathrm{\mathbb{R}}}^d} | u(x+\omega t)-u(x)|^2\,\mathrm{d}x\,\mathrm{d}\sigma_{d-1}(\omega)\\ &= 2\int_{\mathbb{S}^{d-1}} \int_{\mathop{\mathrm{\mathbb{R}}}^d} |\widehat{u}(\xi)|^2(1-\cos (|\omega \cdot \xi| t))\,\mathrm{d}\xi\,\mathrm{d}\sigma_{d-1}(\omega). \end{align}\] Accordingly, Lemma 4 and Lemma 5 imply all desired estimates since \[\begin{align} T^2_2 (\eta, u,r) &=2 \eta \int_{\mathbb{S}^{d-1}} \int_{\mathop{\mathrm{\mathbb{R}}}^d} |\widehat{u}(\xi)|^2 r^{2\eta}\int_r^\infty \frac{(1-\cos (|\omega\cdot \xi| t))}{t^{1+2\eta}}\,\mathrm{d}t \,\mathrm{d}\xi \,\mathrm{d}\sigma_{d-1}(\omega),\\ L^2_2 (\eta, u,r) &=2 (1-\eta )\int_{\mathbb{S}^{d-1}} \int_{\mathop{\mathrm{\mathbb{R}}}^d} |\widehat{u}(\xi)|^2 (2r)^{2\eta}\int_0^r \frac{(1-\cos (|\omega\cdot \xi| t))}{t^{1+2\eta}}\,\mathrm{d}t \,\mathrm{d}\xi \,\mathrm{d}\sigma_{d-1}(\omega). \end{align}\] ◻

Proof of Theorem 2. The extreme cases \(\sigma=0\) and \(s=1\) follow by letting \(\sigma\to 0^+\) and \(s\to1^-\) using Theorem 23, thus we only consider \(0<\sigma<\eta<s<1\). The continuities are not difficult to establish. By Theorem 20 we know that \(t \mapsto g(t) = \frac{U(t)}{t^p}\) is dyadically decreasing, i.e., \(g(2t) \leq g(t)\) and satisfies \[\begin{align} g(0)=\lim_{t\to0^+} g(t)= \frac{|\mathbb{S}^{d-1}|}{p}K_{d,p} \|\nabla u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)} =[u]^p_{W^{1,p}(\mathop{\mathrm{\mathbb{R}}}^d)}. \end{align}\] Therefore, for \(0<\eta<s\leq 1\) and \(1\leq p<\infty\), Theorem 18 implies that \[\begin{align} (1-\eta)(2r)^{-(1-\eta )p}\int_0^r g(t)t^{(1-\eta) p-1} \,\mathrm{d}t &\leq (1-s) (2r)^{-(1-s) p}\int_0^r g(t)t^{(1-s) p-1}\,\mathrm{d}t. \end{align}\] In other words we have \[\begin{align} (1-\eta)(2r)^{-(1-\eta) p}\int_0^r \frac{U(t)}{t^{1+\eta p}} \,\mathrm{d}t &\leq (1-s) (2r)^{-(1-s) p}\int_0^r \frac{U(t)}{t^{1+s p}}\,\mathrm{d}t. \end{align}\] That is, \(L_p(\eta, u,r) \leq L_p(s, u,r)\). Now if \(p=\infty\) and \(0<\eta<s\leq 1\), we get \[\begin{align} L_\infty(\eta, u,r)&= (2r)^{\eta-s} (2r)^{-(1-s)} \sup_{0<|x-y|\leq r} \frac{|u(x)-u(y)|}{|x-y|^s} |x-y|^{s-\eta}\\ &\leq 2^{\eta-s}L_\infty(s, u,r)\leq L_\infty(s, u,r). \end{align}\] In particular, \(2^{-\eta}L_\infty(\eta, u,r) \leq 2^{-s}L_\infty(s, u,r)\). Next, we turn our attention to long range. Since \(U(t)\leq 2^p |\mathbb{S}^{d-1}| \|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\) by Theorem 20, for \(\sigma=0\) we get \[\begin{align} T_p(\eta, u,r)\leq \Big(2^{p-1}\frac{2|\mathbb{S}^{d-1}|}{p}\|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)} \Big)^{\frac{1}{p}}=2^{1-\frac{1}{p}}T_p(0, u,r)\leq 2T_p(0, u,r). \end{align}\] For \(p=\infty\), we get \(T_\infty(\eta, u, r)\leq 2T_\infty(0, u,r)\), while for \(0<\sigma<\eta<1\) we have \[\begin{align} T_\infty(\eta, u,r)&= r^{\sigma}r^{\eta-\sigma}\sup_{|x-y|\geq r} \frac{|u(x)-u(y)|}{|x-y|^{\sigma}}|x-y|^{\sigma-\eta} \leq T_\infty(\sigma, u,r). \end{align}\] The case \(p=2\) and \(d\geq 1\) follows immediately from Lemma 5. Now assume \(d \geq 2\). In view of Theorem 20 we know that \[\begin{align} U(\infty)=\lim_{t\to^\infty} U(t)= \frac{2|\mathbb{S}^{d-1}|}{p} \|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)} =[u]^p_{W^{0,p}(\mathop{\mathrm{\mathbb{R}}}^d)}, \end{align}\] and \(U\) is \(2^p\)-almost increasing, i.e., \(U(s) \leq 2^p U(t)\) for all \(s < t\). Accordingly, Theorem 17 implies that for \(0\leq \sigma<\eta\) we have \(T_p(\eta, u,r) \leq 2 T_p(\sigma, u,r)\), that is, \[\begin{align} \eta r^{\eta p}\int_r^\infty \frac{U(t)}{t^{1+\eta p}}\,\mathrm{d}t &\leq 2^p\sigma r^{ \sigma p}\int_r^\infty \frac{U(t)}{t^{ 1+\sigma p}} \,\mathrm{d}t. \end{align}\]

Finally, it remains to show that \(T_p(\eta, u,r) \leq 2^{1+\frac{1}{p}} r^\sigma(1-\sigma)^{-\frac{1}{p}} [u]_{W^{\sigma,p}(\mathop{\mathrm{\mathbb{R}}}^d)}\) for \(d\geq1\), \(\sigma>0\) and \(p\neq \infty\). Since \(z^{-\frac{1}{\eta p}} \leq z^{-\frac{1}{\sigma p}}\) for \(z\in (0,1)\), Theorem 20 implies \(U(rz^{-\frac{1}{\eta p}})\leq 2^{p+1}AU(rz^{-\frac{1}{\sigma p}})\), yielding \[\begin{align} \int_0^1 U(rz^{-\frac{1}{\eta p}}) \,\mathrm{d}z \leq 2^{p+1} \int_0^1 AU(rz^{-\frac{1}{\sigma p}})\,\mathrm{d}z . \end{align}\] The latter estimate is equivalent to saying \[\begin{align} \eta r^{\eta p}\int_r^\infty \frac{U(t)}{t^{1+\eta p}}\,\mathrm{d}t &\leq 2^{p+1}\sigma r^{\sigma p}\int_r^\infty \frac{AU(t)}{t^{ 1+\sigma p}} \,\mathrm{d}t. \end{align}\] Now, applying Fubini’s theorem to interchange the order of integration gives \[\begin{align} \int_r^\infty \frac{AU(t)}{t^{1+\sigma p}}\,\mathrm{d}t &= \int_r^\infty \Big(\frac{1}{t}\int_0^t U(z)\,\mathrm{d}z\Big) \frac{\,\mathrm{d}t}{ t^{1+\sigma p}}= \int_0^\infty U(z) \Big( \int_{\max(r,z)}^\infty \frac{\,\mathrm{d}t}{ t^{2+\sigma p}}\Big) \,\mathrm{d}z\\ &\leq \int_0^\infty U(z) \Big(\int_{z}^\infty t^{-(2+\sigma p)}\,\mathrm{d}t\Big) \,\mathrm{d}z = \frac{1}{1+\sigma p} \int_0^\infty \frac{U(z)}{z^{1+\sigma p}} \,\mathrm{d}z. \end{align}\] Altogether yields the estimate \(T_p(\eta, u,r) \leq 2^{1+\frac{1}{p}} r^\sigma(1-\sigma)^{-\frac{1}{p}} [u]_{W^{\sigma,p}(\mathop{\mathrm{\mathbb{R}}}^d)}\) since \[\begin{align} \eta r^{\eta p}\int_r^\infty \frac{U(t)}{t^{1+\eta p}}\,\mathrm{d}t \leq2^{p+1} \sigma r^{\sigma p} \int_r^\infty \frac{AU(t)}{t^{1+\sigma p}}\,\mathrm{d}t\leq 2^{p+1} \sigma r^{\sigma p} \int_0^\infty \frac{U(t)}{t^{1+\sigma p}}\,\mathrm{d}t. \end{align}\] ◻

Proof. Assume \(0<s<1\). By Theorem 2, \(s\mapsto L_p(s,u,r)\) is increasing. Therefore, using this and Fatou’s Lemma imply that \[\begin{align} L_p(s,u,r)\leq \liminf_{n\to\infty} L_p(s-\tfrac{1}{n},u,r)\leq \sup_{0\leq \eta< s} L_p(\eta,u,r)\leq L_p(s,u,r). \end{align}\] That is, \(L_p(s,u,r) = \sup_{0\leq \eta< s} L_p(\eta,u,r).\) In particular if \(u\in W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)\) then \[\begin{align} \sup_{0\leq \eta< s} L_p(\eta,u,r)=L_p(s,u,r) <\infty. \end{align}\] Conversely if \(\sup_{0\leq \eta< s} L_p(\eta,u,r)<\infty\) then \(u\in W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)\). Indeed, since \(T_p^p(s, u,r)\leq 2^pT_p^p(0, u,r)\) with \(T_p^p(0, u,r) =\frac{2}{p}|\mathbb{S}^{d-1}|\|u\|^p_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)}\) we have \[\begin{align} [u]^p_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)} &= (2r)^{(1-s)p} s L_p^p(s, u,r) + r^{-sp} (1-s)T_p^p(s, u,r)\\ &\leq (2r)^{(1-s)p} s \sup_{0\leq \eta< s} L^p_p(\eta,u,r) + r^{-sp} 2^pT_p^p(0, u,r)<\infty. \end{align}\] The case \(s = 1\), which provides a standard nonlocal characterization of \(W^{1,p}(\mathop{\mathrm{\mathbb{R}}}^d)\), is less obvious. It follows from [6]; see also [4]. ◻

Proof. The unique critical point is \(r_* = \big(\frac{b \beta}{a \alpha}\big)^{\frac{1}{\alpha + \beta}}\) satisfying \(F'(r_*) = 0\) with \(F'(r) = a \alpha r^{\alpha - 1} - b \beta r^{-\beta - 1}.\) Moreover, evaluating the second derivative \(F''(r)\) at \(r=r_*\) reveals that \(r_*\) is a global minimum since \[\begin{align} F''(r_*) &= a \alpha (\alpha - 1) r_*^{\alpha - 2} + b \beta (\beta + 1) r_*^{-\beta - 2} = (a\alpha)^{\frac{\alpha+2}{\alpha + \beta} } (b\beta)^{\frac{\beta-2}{\alpha + \beta} } (\alpha+\beta)>0 \end{align}\] It follows that, the minimal value is given by \[\begin{align} F(r_*) &= a \Big(\frac{b \beta}{a \alpha}\Big)^{\frac{\alpha}{\alpha + \beta}} + b \Big(\frac{b \beta}{a \alpha}\Big)^{-\frac{\beta}{\alpha + \beta}} =(\alpha + \beta) \Big(\frac{a}{\alpha}\Big)^{\frac{\beta}{\alpha + \beta}} \Big(\frac{b }{\beta}\Big)^{\frac{\alpha}{\alpha + \beta}}= \kappa(\theta)\, a^{\theta} b^{1 - \theta}. \end{align}\] Next, taking \(\rho(\theta)= \log(\theta^{\theta} (1-\theta)^{(1-\theta)})\) we find that \[\begin{align} \rho'(\theta) &= \log \theta - \log(1 - \theta) \quad\text{and}\quad \rho''(\theta) = \frac{1}{\theta} + \frac{1}{1 - \theta}>0. \end{align}\] This implies that \(\rho'\) is strictly increasing and for all \(\theta\in (0, \frac{1}{2})\) we have \[\begin{align} -\infty=\lim_{\theta\to 0^+} \rho' (\theta)< \rho'(\theta)< \rho'(1/2)=0< \rho'(1-\theta)<\lim_{\theta\to 0} \rho' (1-\theta)=\infty. \end{align}\] Thus, \(\rho\) is strictly decreasing on \((0, \tfrac{1}{2})\) and strictly increasing on \((\tfrac{1}{2}, 1)\), yielding that the minimum is achieved at \(\theta = \tfrac{1}{2}\) and the maximum at \(\theta \in \{0,1\}\). We find that \[\begin{align} \frac{1}{2}\leq \theta^{\theta} (1-\theta)^{(1-\theta)} \leq 1 \quad \text{for all } \theta \in (0,1). \end{align}\] Therefore, since \(1/\kappa(\theta) = \theta^{\theta}(1 - \theta)^{1-\theta}\) we deduce \[\begin{align} \kappa(0^+) =\kappa(1^-) =1\leq \kappa(\theta) \leq 2=\kappa(1/2) \quad \text{ for all } \theta \in (0,1). \end{align}\] The claim for \(G\) follows since \(G\) is decreasing on \((0, r_0)\) and increasing on \((r_0,\infty)\). ◻

Proof of Theorem 1. Let \(0\leq \sigma<\eta<s\leq 1\) and \(\theta=\tfrac{\eta-\sigma}{s-\sigma}\in(0,1)\). We only prove the estimates for \(\mathop{\mathrm{\mathbb{R}}}^d\) as those of \(\mathop{\mathrm{\mathbb{R}}}^d_+\) follow by even reflection; see Lemma 8. For \(1 \leq p < \infty\) and \(r > 0\), setting \(\rho = r^{s-\sigma}\) in Theorem 2 implies that \[\begin{align} (1-\eta)^{\frac{1}{p}} T_p(\eta, u, r) \leq 2\varkappa r^{\sigma} [u]_{W^{\sigma,p}(\mathop{\mathrm{\mathbb{R}}}^d)} \quad \text{and} \quad s^{\frac{1}{p}} L_p(\eta, u, r) \leq (2r)^{-(1-s)} [u]_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)}, \end{align}\] where we recall \(\tiny{\varkappa = \begin{cases} 2^{\frac{1}{p}}& \text{if } d = 1, \\ 1 & \text{if } d \geq 2. \end{cases}}\). Therefore, we deduce \[\begin{align} [u]^p_{W^{\eta,p}(\mathop{\mathrm{\mathbb{R}}}^d)} &=\eta(1-\eta)\int_0^r \frac{U(t)}{t^{1+\eta p}} \,\mathrm{d}t + \eta(1-\eta)\int_r^\infty \frac{U(t)}{t^{1+\eta p}} \,\mathrm{d}t\\ &= (2r)^{(1-\eta)p} \eta L_p^p(\eta, u,r) + r^{-\eta p} (1-\eta)T_p^p(\eta, u,r) \\ &\leq (2r)^{(1-\eta)p} sL_p^p(s, u,r) + 2^pr^{-\eta p} (1-\sigma)T_p^p(\sigma, u,r) \\ & \leq 2^{(s-\eta) p}r^{(s-\eta) p} [u]^p_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)}+ 2^p\varkappa^p r^{-(\eta-\sigma) p} [u]^p_{W^{\sigma,p}(\mathop{\mathrm{\mathbb{R}}}^d)}\\ &=2^{(s-\sigma)(1-\theta)p} \rho^{(1-\theta)p } [u]^p_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)}+ 2^p\varkappa^p\rho^{-\theta p} [u]^p_{W^{\sigma,p}(\mathop{\mathrm{\mathbb{R}}}^d)}, \end{align}\] According to Lemma 6 the value \(\rho_*\) that minimizes the right-hand side is \[\begin{align} \rho_* & = \Big( \frac{2^{p} \varkappa^p \theta }{2^{(s-\sigma)(1-\theta)p}(1-\theta) } \frac{[u]^p_{W^{\sigma,p}(\mathop{\mathrm{\mathbb{R}}}^d)}}{[u]^p_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)}} \Big)^{\frac{1}{p}}. \end{align}\] Moreover, for this choice \(\rho_*\), the minimal value of the right-hand side becomes: \[\begin{align} [u]_{W^{\eta,p}(\mathop{\mathrm{\mathbb{R}}}^d)} &\leq \min_{\rho>0}\Big(2^{(s-\sigma)(1-\theta)p} \rho^{(1-\theta)p } [u]^p_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)}+ 2^p\rho^{-\theta p } [u]^p_{W^{\sigma,p}(\mathop{\mathrm{\mathbb{R}}}^d)}\Big)^{\frac{1}{p}} \\ &= \varkappa 2^{(s-\eta)\theta+(1-\theta)}\kappa^{\frac{1}{p}}(\theta) [u]_{W^{s,p} (\mathop{\mathrm{\mathbb{R}}}^d)}^{\theta } \cdot [u]_{W^{\sigma,p} (\mathop{\mathrm{\mathbb{R}}}^d)}^{1 - \theta}\\ &=\varkappa N_p(\sigma, \eta, s) [u]_{W^{s,p} (\mathop{\mathrm{\mathbb{R}}}^d)}^{\theta } \cdot [u]_{W^{\sigma,p} (\mathop{\mathrm{\mathbb{R}}}^d)}^{(1 - \theta)}. \end{align}\] where we note that \(\kappa(\theta)=\theta^{-\theta}(1-\theta)^{-(1-\theta)}\in [1,2]\) and define \[\begin{align} N_p(\sigma,\eta,s) =2^{(s-\eta)\theta+(1-\theta)} \kappa(\theta)^{\frac{1}{p}} = 2^{\frac{(s-\eta)}{s-\sigma}(\eta+1-\sigma)} \left(\frac{s-\sigma}{\eta-\sigma}\right)^{\frac{\eta-\sigma}{(s-\sigma)p}} \left(\frac{s-\sigma}{s-\eta}\right)^{\frac{s-\eta}{(s-\sigma)p}}, \end{align}\] which satisfies \(1\leq N_p(\sigma, \eta, s)\leq 2^{1+\frac{1}{p}}\). Analogously for the case \(p=\infty\) we have \[\begin{align} |u|_{W^{\eta,\infty}(\mathop{\mathrm{\mathbb{R}}}^d)} &= \max\Big(\sup_{|x-y|< r}\frac{| u(x)-u(y)|}{|x-y|^\eta}, \,\, \sup_{|x-y|\geq r}\frac{| u(x)-u(y)|}{|x-y|^\eta}\Big)\\ &= \max \Big((2r)^{(1-\eta)} L_\infty(\eta, u,r) ,\,\, r^{-\eta} T_\infty(\eta, u,r)\Big) \\ &\leq \max\Big((2r)^{(1-\eta)} L_\infty(s, u,r) , \,\, 2r^{-\eta} T_\infty(\sigma, u,r)\Big) \\ &\leq \max \Big((2r)^{s-\eta}[u]_{W^{1,\infty}(\mathop{\mathrm{\mathbb{R}}}^d)}, \,\, 2r^{\sigma-\eta}[u]_{W^{0,\infty}(\mathop{\mathrm{\mathbb{R}}}^d)}\Big)\\ &=\max\Big(2^{(s-\sigma)(1-\theta)} \rho^{1-\theta} [u]_{W^{s,\infty}(\mathop{\mathrm{\mathbb{R}}}^d)}, \,\, 2\rho^{-\theta } [u]_{W^{\sigma,\infty}(\mathop{\mathrm{\mathbb{R}}}^d)}\Big). \end{align}\] By Lemma 6 taking \(\rho= \rho_*= 2^{\eta+ 1-s}\frac{ [u]_{W^{\sigma,\infty}(\mathop{\mathrm{\mathbb{R}}}^d)}}{ [u]_{W^{s,\infty}(\mathop{\mathrm{\mathbb{R}}}^d)}}\) yields that \[\begin{align} |u|_{W^{\eta,\infty}(\mathop{\mathrm{\mathbb{R}}}^d)} &\leq \min_{\rho>0} \max\Big(2^{(s-\sigma)(1-\theta)} \rho^{1-\theta} [u]_{W^{s,\infty}(\mathop{\mathrm{\mathbb{R}}}^d)}, \,\, 2\rho^{-\theta } [u]_{W^{\sigma,\infty}(\mathop{\mathrm{\mathbb{R}}}^d)}\Big)\\ &=2^{(s-\eta)\theta+(1-\theta)}[u]_{W^{s,\infty} (\mathop{\mathrm{\mathbb{R}}}^d)}^{\theta } \cdot [u]_{W^{\sigma,\infty} (\mathop{\mathrm{\mathbb{R}}}^d)}^{1 - \theta}\\ &= N_\infty(\sigma, \eta, s)[u]_{W^{s,\infty} (\mathop{\mathrm{\mathbb{R}}}^d)}^{\theta } \cdot [u]_{W^{\sigma,\infty} (\mathop{\mathrm{\mathbb{R}}}^d)}^{1 - \theta}. \end{align}\] On the other side, given that \([u]^{*\,p}_{W^{t,p}(\mathop{\mathrm{\mathbb{R}}}^d)}= B_{d,p,t} [u]^p_{W^{t,p}(\mathop{\mathrm{\mathbb{R}}}^d)}\), it merely follows that \[\begin{align} [u]^*_{W^{\eta,p} (\mathop{\mathrm{\mathbb{R}}}^d)} &\leq \Upsilon^*_{d,p}(\sigma, \eta, s) [u]^{*\, \theta }_{W^{s,p} (\mathop{\mathrm{\mathbb{R}}}^d)} \cdot [u]^{*\,(1 - \theta)}_{W^{\sigma,p} (\mathop{\mathrm{\mathbb{R}}}^d)}, \end{align}\] with \(\Upsilon^*_{d,p}(\sigma, \eta, s)= N_p(\sigma, \eta, s) \Big(\frac{B_{d,p,\eta}}{B_{d,p,\sigma}^{\theta}B_{d,p,s}^{1-\theta}} \Big)^{\frac{1}{p}}\). Likewise, by substituting the expression of \([\cdot]_{W^{t,p}(\mathop{\mathrm{\mathbb{R}}}^d)}\) in term of \(|\cdot|_{W^{t,p}(\mathop{\mathrm{\mathbb{R}}}^d)}\) yields:

Case \(\sigma\neq 0,\,\, s\neq 1\). \[\begin{align} |u|_{W^{\eta,p}(\mathop{\mathrm{\mathbb{R}}}^d)} \leq \frac{ N_p(\sigma,\eta,s)}{[\eta(1-\eta)]^{\frac{1}{p}}} \Big[ (s(1-s))^\theta (\sigma(1-\sigma))^{1-\theta}\Big]^{\frac{1}{p}} |u|_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)}^{\theta} |u|_{W^{\sigma,p}(\mathop{\mathrm{\mathbb{R}}}^d)}^{1-\theta}. \end{align}\] Case \(\sigma=0, s\neq 1\). \[\begin{align} |u|_{W^{\eta,p}(/R^d)} \leq\frac{ N_p(\sigma,\eta,s)}{[\eta(1-\eta)]^{\frac{1}{p}}} \Big[ (s(1-s))^\theta (2|\mathbb{S}^{d-1}|/p)^{1-\theta} \Big]^{\frac{1}{p}} |u|_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)}^\theta |u|_{W^{0,p}(\mathop{\mathrm{\mathbb{R}}}^d)}^{1-\theta}. \end{align}\] Case \(\sigma\neq 0,\,\, s=1\). \[\begin{align} |u|_{W^{\eta,p}(\mathop{\mathrm{\mathbb{R}}}^d)} \leq \frac{ N_p(\sigma,\eta,s)}{[\eta(1-\eta)]^{\frac{1}{p}}} \Big[ (K_{d,p} |\mathbb{S}^{d-1}|/p)^\theta (\sigma(1-\sigma))^{1-\theta}\Big]^{\frac{1}{p}} |u|_{W^{1,p}(\mathop{\mathrm{\mathbb{R}}}^d)}^\theta |u|_{W^{\sigma,p}(\mathop{\mathrm{\mathbb{R}}}^d)}^{1-\theta}. \end{align}\] Case \(\sigma=0, s=1\). \[\begin{align} |u|_{W^{\eta,p}(\mathop{\mathrm{\mathbb{R}}}^d)} \leq \frac{ N_p(\sigma,\eta,s)}{[\eta(1-\eta)]^{\frac{1}{p}}} \Big[ \frac{|\mathbb{S}^{d-1}|}{p}K_{d,p}^\theta 2^{1-\theta} \Big]^{\frac{1}{p}} |u|_{W^{1,p}(\mathop{\mathrm{\mathbb{R}}}^d)}^\theta |u|_{W^{0,p}(\mathop{\mathrm{\mathbb{R}}}^d)}^{1-\theta}. \end{align}\] ◻

Proof. By Theorem 32 (see Appendix 7), there exist a \(C=C(d,p,\Omega)>0\) and \(\overline{u}\in L^p(\mathop{\mathrm{\mathbb{R}}}^d)\) a suitable extension of \(u\in L^p(\Omega)\) such that \(\overline{u}|_\Omega=u\) a.e. in \(\Omega\) and \(\|\overline{u}\|_{W^{t, p}(\mathop{\mathrm{\mathbb{R}}}^d)}\leq C \|u\|_{W^{t, p}(\Omega)}\) for all \(t\in [0,1]\). In view of Theorem 1 we find that \[\begin{align} \|u\|_{W^{\eta, p}(\Omega)} &=\Big(\|u\|^p_{L^p(\Omega)}+ [u]^p_{W^{\eta,p}(\Omega)} \Big)^{1/p}\leq \Big(\|u\|^p_{L^p(\Omega)}+ [\overline{u}]^p_{W^{\eta,p}(\mathop{\mathrm{\mathbb{R}}}^d)} \Big)^{1/p} \\ &\leq \Big(\|u\|^p_{L^p(\Omega)}+N_p(\sigma, \eta, s) [\overline{u}]^\theta_{W^{s, p}(\mathop{\mathrm{\mathbb{R}}}^d)} [\overline{u}]^{1-\theta}_{W^{\sigma, p}(\mathop{\mathrm{\mathbb{R}}}^d)}\Big)^{1/p}\\ &\leq \Big(\|u\|^{\theta p}_{L^p(\Omega)}\|u\|^{(1-\theta)p}_{L^p(\Omega)}+ CN^p_p(\sigma, \eta, s)\|u\|^{\theta p}_{W^{s, p}(\Omega)}\|u\|^{(1-\theta)p}_{W^{\sigma, p}(\Omega)} \Big)^{1/p}\\ &\leq C(N_p(\sigma, \eta, s)+1) \|u\|^{\theta }_{W^{s, p}(\Omega)}\|u\|^{1-\theta}_{W^{\sigma, p}(\Omega)} \leq 9C \|u\|^{\theta }_{W^{s, p}(\Omega)}\|u\|^{1-\theta}_{W^{\sigma, p}(\Omega)}. \end{align}\] ◻

Proof. Let us recall that the digamma function given by \(\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}= \frac{d}{d x}\log \Gamma (x)\) is strictly increasing on \((0,\infty)\) where it derivative \(\psi'(x)=\sum_{n=0}^\infty \frac{1}{(x+n)^2}>0\) is also called the trigamma function. Moreover \(\psi\) has only one positive root \(x_0 \approx 1.461632 \in (1,\frac{3}{2})\). Hence the function \(x\mapsto \Gamma(x)\) is strictly increasing on \((x_0,\infty)\) and strictly decreasing on \((0, x_0)\), in particular \(\Gamma(x_0)=\min_{x\in (0,\infty)}\Gamma(x)\). Furthermore, let \(\gamma \approx 0.5772\) be Euler-Mascheroni constant then some special values of \(\psi\) are given by \[\begin{align} \psi(1)&= -\gamma \approx -0.5772,\qquad && \psi(1/2)= -\gamma -2\log2 \approx -1.9635, \\ \psi(0^+)&= \lim_{x\to 0^+} \psi(x) = -\infty,&& \psi(x+1)=\psi(x)+\frac{1}{x},\quad x>0. \end{align}\] Define the function \(G: (0,1)\to (0,\infty)\) with \[\begin{align} G(s) :=\frac{s(1-s)}{C_{d,s}}=\frac{\pi^{d/2} \, \Gamma(2 - s)}{2^{2s} \, \Gamma\left( \frac{d + 2s}{2} \right)}. \end{align}\] For \(d\geq 2\) or \(\frac{1}{2}\leq s<1\), since \(\psi\) is increasing we have \(\psi(\frac{d+2s}{2})\geq \psi(1)\) and \(\psi(2-s)\geq \psi(1)\). Thus, the logarithmic derivative of \(G\) satisfies \[\begin{align} \frac{d}{ds}\log G(s)= \frac{G'(s)}{G(s)} &= -\psi(2-s) - 2 \log 2 - \psi\Big(\frac{d+2s}{2}\Big) \\ &\leq -2\psi(1) - 2 \log 2 =2\gamma- 2 \log 2<0 . \end{align}\] Therefore, for \(d\geq 2,\) \(G\) is strictly decreasing on \((0,1)\) yielding that \[\begin{align} \frac{\pi^{d/2}}{2d\,\Gamma(d/2)} <G(s)< \frac{\pi^{d/2}}{\Gamma(d/2)}, \qquad \text{for } s \in (0,1), \; d \geq 2. \end{align}\] For \(d=1\) and \(\frac{1}{2}\leq s\leq 1\) we have \(\frac{G'(s)}{G(s)}<0\). Next, since the trigamma function \(\psi'(x)= \frac{d^2}{dx^2}\log\Gamma(x)\) is decreasing, the function \(s\mapsto H(s)\) with \(H(s):=\big(\tfrac{G'(s)}{G(s)} \big)'= \psi'(2-s) -\psi'(\frac{1+2s}{2})\) is increasing. Thus, using \(\psi'(1)= \frac{\pi^2}{6}\), \(\psi'(\frac{1}{2})= \frac{\pi^2}{2}\) and the formula \(\psi'(x+1)=\psi'(x)-\frac{1}{x^2}\), we have \[\begin{align} H(s)\leq H(\tfrac12)=\psi'(\frac{3}{2})- \psi'(1)= \frac{\pi^2}{3} - 4<0&& \text{for 0<s<\frac{1}{2}}. \end{align}\] This implies that \(s\mapsto \frac{G'(s)}{G(s)}\) is decreasing on \((0,\frac{1}{2})\), in particular \[\begin{align} \frac{G'(s)}{G(s)}\leq \frac{G'(1/16)}{G(1/16)}<0,\qquad \frac{1}{16}<s<\frac{1}{2}. \end{align}\] Thus we have \(\frac{G'(s)}{G(s)}<0\) for \(\frac{1}{16}<s<1\). Altogether we find that \(G\) is strictly decreasing on \((\frac{1}{16},1)\). Recalling \(G(s)= \frac{\sqrt{\pi}\,\Gamma(2-s)}{2^{2s}\,\Gamma\!\left(\frac{1+2s}{2}\right)}\) and \(\Gamma\left(\tfrac{31}{16}\right)= \tfrac{15}{16}\Gamma\left(\tfrac{15}{16}\right)\approx 0.97<1\) we get \[\begin{align} \frac{1}{2}=G(1)<G(s)\leq G(\frac{1}{16})= \frac{\sqrt{\pi}}{2^{1/8}\,\Gamma\big(\frac{9}{16}\big)} \Gamma\big(\frac{31}{16}\big) <\frac{\sqrt{\pi}}{2^{1/8}\,\Gamma\big(\frac{9}{16}\big)}, \quad\text{\frac{1}{16}\leq s<1}. \end{align}\] It remains the case \(d=1\) and \(0<s<\tfrac{1}{16}\) we have \(2-s \in (\tfrac{31}{16},2) \subset (x_0,\infty)\) and \(\tfrac{1+2s}{2} \in (\tfrac12,\tfrac{9}{16}) \subset (0,x_0).\) Since the digamma function is strictly increasing we find that \[\begin{align} \frac{d}{d s} \log \Big( \frac{1}{2^{2s}\Gamma\big(\frac{1+2s}{2}\big)} \Big) = -2\log 2 -\psi\big(\tfrac{1+2s}{2}\big) \geq -2\log 2 -\psi\big(\frac{9}{16}\big)\approx 0.26>0. \end{align}\] Whence, \(s \mapsto \frac{1}{2^{2s}\,\Gamma\!\left(\frac{1+2s}{2}\right)}\) is strictly increasing on \((0,\tfrac{1}{16}).\) On the other hand, since \(\Gamma\) is increasing on \((x_0,\infty)\) we deduce that \[\begin{align} \Gamma\Big(\frac{31}{16}\Big) &\leq\Gamma(2-s) \leq 1\quad \text{and}\quad 1 \leq \frac{\sqrt{\pi}}{2^{2s}\Gamma\Big(\frac{1+2s}{2}\Big)} \leq \frac{\sqrt{\pi}}{2^{1/8}\Gamma\Big(\frac{9}{16}\Big)}. \end{align}\] It follows that, for \(0<s<\frac{1}{16}\), \[\begin{align} \frac{24}{25}<0.97\approx \Gamma\Big(\frac{31}{16}\Big) \leq G(s) = \frac{\sqrt{\pi}\,\Gamma(2-s)}{2^{2s}\,\Gamma\big(\tfrac{1+2s}{2}\big)} \leq \frac{\sqrt{\pi}}{ 2^{1/8}\Gamma\big(\tfrac{9}{16}\big)}< 1.04 =\frac{26}{25}. \end{align}\] Hence we have \[\begin{align} \frac{1}{2}<G(s)\leq \frac{\sqrt{\pi}}{2^{1/8}\,\Gamma\left(\frac{9}{16}\right)}<\frac{26}{25} \qquad\text{for \,\, 0<s<1}. \end{align}\] ◻

Proof. Since \(\eta= \theta\sigma +(1-\theta) s\) then Hölder inequality implies \[\begin{align} \int_{\mathop{\mathrm{\mathbb{R}}}^d}|\hat{u}(\xi)|^2 |\xi|^{2\theta \sigma} |\xi|^{2(1-\theta)s}\,\mathrm{d}\xi &\leq \Big( \int_{\mathop{\mathrm{\mathbb{R}}}^d}|\hat{u}(\xi)|^2 |\xi|^{2\sigma}\,\mathrm{d}\xi\Big)^{\theta}\Big( \int_{\mathop{\mathrm{\mathbb{R}}}^d}|\hat{u}(\xi)|^2 |\xi|^{2s}\,\mathrm{d}\xi\Big)^{1-\theta}. \end{align}\] In other words, by the identity 12 we get \([u]^*_{H^\eta (\mathop{\mathrm{\mathbb{R}}}^d)}\leq[u]^{*\,\theta}_{H^s (\mathop{\mathrm{\mathbb{R}}}^d)} [u]^{*\,(1-\theta)}_{H^\sigma(\mathop{\mathrm{\mathbb{R}}}^d)}.\) Given that \([u]^{*\,2}_{H^{t}(\mathop{\mathrm{\mathbb{R}}}^d)}= B_{d,2,t} [u]^2_{H^{t}(\mathop{\mathrm{\mathbb{R}}}^d)}\), we conclude using Lemma 7 that \[\begin{align} [u]_{H^\eta (\mathop{\mathrm{\mathbb{R}}}^d)}\leq \big(\tfrac{B_{d,2,\sigma}^{\theta}B_{d,2,s}^{1-\theta}}{B_{d,2,\eta}} \big)^{1/2} [u]_{H^\sigma(\mathop{\mathrm{\mathbb{R}}}^d)}^{\theta}\cdot [u]_{H^s (\mathop{\mathrm{\mathbb{R}}}^d)}^{(1-\theta)} \leq \sqrt{2d} [u]_{H^\sigma(\mathop{\mathrm{\mathbb{R}}}^d)}^{\theta}\cdot [u]_{H^s (\mathop{\mathrm{\mathbb{R}}}^d)}^{(1-\theta)}. \end{align}\] ◻

Proof. The existence of subsequence \(s_n\to 1^-\) such that \(\|u_{s_n}-u\|_{L^p(\Omega)}\xrightarrow{s_n\to 1^-}0\) for some \(u\in BV_p(\Omega)\), which additionally satisfies the second claim, can be found in [20]; see also [4]. Next, note that by [6] we have \[\begin{align} \label{eq:xxuniform-bound} s(1-s) \iint_{\Omega \times \Omega} \frac{|u(x)-u(y)|^p}{|x-y|^{d+sp}} \,\mathrm{d}y\,\mathrm{d}x\leq C(d,p,\Omega) \begin{cases} \|u\|^p_{W^{1,p}(\Omega)}&1<p<\infty\\ \|u\|_{BV(\Omega)} &p=1. \end{cases} \end{align}\tag{15}\] Thus by assumption we find that \(\|u_{s_n}\|_{W^{s_n,p}(\Omega)}+\|u\|_{W^{s_n,p}(\Omega)}\leq C(d,p,\Omega)\). Let us consider \(0\leq \eta < s_n <1\), for \(n\) large enough, so that \(u_{s_n}, u \in W^{\eta,p}(\Omega)\). The robust interpolation, Theorem 28, with \(\theta=\frac{\eta}{s_n}\) implies \[\begin{align} \|u_{s_n}-u\|_{W^{\eta,p}(\Omega)} &\leq C \|u_{s_n}-u\|^{\frac{\eta}{s_n}}_{W^{s_n,p}(\Omega)} \|u_{s_n}-u\|^{1-\frac{\eta}{s_n}}_{L^{p}(\Omega)} \leq C \|u_{s_n}-u\|^{1-\frac{\eta}{s_n}}_{L^{p}(\Omega)} \xrightarrow{n\to\infty} 0. \end{align}\] Note that \(u_{s_n}, u \in W^{\tau,p}(\Omega)\) for \(\frac{1}{p} < \tau \le s_n < 1\). By the continuity of the trace operator and the previous claims, we have \[\begin{align} \|\gamma^{s_n}_0(u_{s_n}) - \gamma^1_0(u)\|_{L^p(\partial \Omega)} &+ \|\gamma^{s_n}_0(u_{s_n}) - \gamma^1_0(u)\|_{W^{\tau-\frac{1}{p},p}(\partial \Omega)} \\ &\leq C_\tau \|u_{s_n} - u\|_{W^{\tau,p}(\Omega)} \xrightarrow{n \to \infty} 0, \end{align}\] where \(C_\tau = C(d,p,\Omega,\tau) > 0\) is a constant. This proves the third claim. ◻

Proof. Assume there exist no such \(s_0\) and \(B\) exist. Taking \(s_0 = 1-\frac{1}{2^n}>\frac{1}{p}\) with \(n \geq 1\) large and \(B = 2^n\) there exist \(s_n \in (1-\frac{1}{2^n},1)\) and \(u_{s_n} \in W^{s_n, p}_0(\Omega)\) such that \[\begin{align} \|u_{s_n}\|_{L^p(\Omega)} = 1\quad \text{and}\quad s_n(1-s_n)\iint_{ \Omega \times \Omega}\frac{|u_{s_n}(x)-u_{s_n}(y)|^p}{|x-y|^{d+s_sp}}\,\mathrm{d}y \,\mathrm{d}x<\frac{1}{2^n}. \end{align}\] In particular, \(u_{s_n}\in W^{s_n, p}_0(\Omega)\), \(\gamma^{s_n}_0 u_{s_n}=0\) and \[\begin{align} \sup_{n\geq 1}\Big( \int_\Omega |u_{s_n}(x)|^p\,\mathrm{d}x+ s_n(1-s_n)\iint_{ \Omega \times \Omega}\frac{|u_{s_n}(x)-u_{s_n}(y)|^p}{|x-y|^{d+s_sp}}\,\mathrm{d}y \,\mathrm{d}x\Big)<2. \end{align}\] By the asymptotic compactness Theorem 30 there is \(u \in W^{1,p}(\Omega)\) and a subsequence still denoted \((u_{s_n})_n\) such that \(\|u_{s_n}-u\|_{W^{\eta,p}(\Omega)}\to0\), \(0\leq \eta<1\) as \(n\to \infty\) and \[\begin{align} K_{d,p} \frac{|\mathbb{S}^{d-1}|}{p}\|\nabla u\|_{L^p(\Omega)}^p \leq \liminf_{n\to\infty} s_n(1-s_n)\iint_{ \Omega \times \Omega}\frac{|u_{s_n}(x)-u_{s_n}(y)|^p}{|x-y|^{d+s_sp}}\,\mathrm{d}y \,\mathrm{d}x= 0. \end{align}\] In particular \(\|u\|_{L^p(\Omega)}=1\) and \(\nabla u\)=0 a.e on \(\Omega\). For fixed \(\eta\in (\frac{1}{p}, 1)\) and \(N\geq 1\) sufficiently large such that \(\eta<s_n<1\) for all \(n\geq N\) we have \(W^{s_n,p}(\Omega)\subset W^{\eta,p}(\Omega)\). This implies \(\gamma^{s_n}_0 v= \gamma^\eta_0v\) for all \(v\in W^{s_n,p}(\Omega)\cap C^1(\overline{\Omega})\). By density argument, it is not difficult to conclude that \(\gamma^\eta_0\mid_{W^{s_n,p}(\Omega)}= \gamma^{s_n}_0\). Since, \(u_{s_n}\in W^{s_n,p}_0(\Omega)\), it follows that \[\begin{align} \gamma^\eta_0 (u_{s_n})= \gamma^{s_n}_0 (u_{s_n})=0\qquad\text{for all n\geq N}. \end{align}\] Therefore, by definition of the trace norm we have \[\begin{align} &\|\gamma^1_0(u)\|_{W^{\eta-\frac{1}{p},p}(\partial\Omega)} =\|\gamma^{s_n}_0(u_{s_n}) -\gamma^1_0(u)\|_{W^{\eta-\frac{1}{p},p}(\partial\Omega)}\leq \|u_{s_n}-u\|_{W^{\eta,p}(\Omega)} \xrightarrow{s_n\to 1^-}0. \end{align}\] It follows that \(\gamma^1_0 u=0\). Since \(\partial\Omega\) is Lipschitz, we deduce that \(u \in W^{1,p}_0(\Omega)\). In conclusion, \(u \in W^{1,p}_0(\Omega)\), \(\nabla u = 0\) a.e. in \(\Omega\) and \(\|u\|_{L^p(\Omega)}=1\). Consequently, the zero extension \(\widetilde{u}\) of \(u\) belongs to \(W^{1,p}(\mathop{\mathrm{\mathbb{R}}}^d)\) and satisfies \(\nabla \widetilde{u}= \widetilde{\nabla u}= 0\) a.e. in \(\mathop{\mathrm{\mathbb{R}}}^d\). Hence \(\widetilde{u}\) is constant on \(\mathop{\mathrm{\mathbb{R}}}^d\) and we have \(\widetilde{u}\equiv0\) on \(\mathop{\mathrm{\mathbb{R}}}^d\), since \(\widetilde{u}= 0\) on \(\mathop{\mathrm{\mathbb{R}}}^d \setminus \Omega\). This clearly implies that \(u \equiv 0\) on \(\Omega\). Thereby contradicting the fact that \(\|u\|_{L^p(\Omega)}=1\) and thus our initial assumption was wrong. ◻

Proof of Theorem 3. Throughout this proof, we assume \(sp>1\), let \(C>0\) denote a generic constant independent of \(s\), while \(s_* \in (\tfrac{1}{p}, 1)\) is sufficiently close to \(1\). Given that \((f_s)_s\) in \(L^{p'}(\Omega)\) converges weakly as \(s\to 1^-\) we may assume thanks to the uniform boundedness principle that \((\|f_s\|_{L^{p'}(\Omega)})_{s\in (s_*,1)}\) is bounded; namely, \(\sup_{s_*<s<1} \| f_s\|_{L^{p'}(\Omega)} < \infty.\) Now let \(\overline{g}_1, h_s \in W^{1,p}(\Omega)\) such that \(\gamma^1_0(\overline{g}_1)= g_1\), \(\gamma^1_0 (h_s)= g_s-g_1\) and \[\begin{align} \|h_s\|_{W^{1,p}(\Omega)}\leq 1-s+ \|g_s-g_1\|_{W^{1-\frac{1}{p},p}(\partial \Omega)}. \end{align}\] The function \(\overline{g}_s= h_s +\overline{g}_1\in W^{1,p}(\Omega)\) satisfies \(\gamma^1_0(\overline{g}_s) = \gamma^1_0(h_s+\overline{g}_1) = g_s\) and \[\begin{align} \label{eq:gs-frac-extens} \|\overline{g}_s-\overline{g}\|_{W^{1,p}(\Omega)}\leq 1-s+ \|g_s-g\|_{W^{1-\frac{1}{p},p}(\partial \Omega)}\xrightarrow{s\to1^-}0. \end{align}\tag{16}\] Hence, replacing \(g_s\) by \(\overline{g}_s\) if necessary, we may assume without loss of generality that \(g_s\in W^{1,p}(\Omega)\) such that \(\|g_s-g_1 \|_{W^{1,p}(\Omega)}\xrightarrow{s\to1^-}0.\) In particular, taking into account Theorem 25 we get \[\begin{align} \label{eq:xgs-conv-uni} \|g_s-g_1 \|_{W^{s,p}(\Omega)}\leq C\|g_s-g_1 \|_{W^{1,p}(\Omega)}\xrightarrow{s\to1^-}0. \end{align}\tag{17}\] Thus, up to relabeling \(s_*\), we can assume \(\|g_s-g \|_{W^{s,p}(\Omega)}<1\) for \(s\in (s_*, 1)\), so that taking into account Theorem 25 we get \[\begin{align} \sup_{s_*<s<1}\|g_s \|_{W^{s,p}(\Omega)}\leq 1+ \sup_{s_*<s<1} \|g \|_{W^{s,p}(\Omega)}\leq 1+ C \|g\|^p_{W^{1,p}(\Omega)}<\infty. \end{align}\] Accordingly, for \(s_*\) close enough to \(1\), we obtain the uniform estimate \[\begin{align} \label{eq:uniform-bdd-f-g-D-regio} M:= \sup_{s_*<s<1}\big(\|f_s \|_{L^{p'}\Omega)}+\|g_s \|_{W^{s,p}(\Omega)}\big)<\infty. \end{align}\tag{18}\] By the robust Poincaré-Friedrichs inequality, Theorem 31, we can assume that \(s_* \in (\frac{1}{p},1)\) and \(C>0\) are such that, for all \(s\in ( s_*,1)\) and \(v \in W^{s,p}_0(\Omega)\) we have \[\begin{align} \label{eq:uniform-coercivity-D-regio} \|v\|^p_{W^{s,p}(\Omega)}\leq C\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(v,v). \end{align}\tag{19}\] Each \(u_s\in W^{s,p}(\Omega)\), \(s_*<s\leq 1\), uniquely satisfies the minimization problem \[\begin{align} \mathop{\mathrm{\mathcal{J}}}^{s,p}_0(u_s) &= \min_{v-g_s\in W^{s,p}_0(\Omega)} \mathop{\mathrm{\mathcal{J}}}^{s,p}_0(v),\\ \mathop{\mathrm{\mathcal{J}}}^{s,p}_0(v) &= \tfrac{1}{p} \mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(v,v) -\int_\Omega f_s(v-g_s)\,\mathrm{d}x. \end{align}\] In particular, upon deriving the Euler Lagrange equation \(u_s\) verifies \[\begin{align} \mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s,u_s-g_s) = \int_\Omega f_s ( u_s-g_s)\,\mathrm{d}x \quad \text{and}\quad u_s-g_s\in W^{s,p}_0(\Omega). \end{align}\] Since \(u_s-g_s\in W^{s,p}_0(\Omega),\) the estimate 19 implies that \[\begin{align} \begin{aligned} \| u_s-g_s \|_{W^{s,p}(\Omega)} \leq C\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s,u_s)^{\frac{1}{p}}+ C\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(g_s,g_s)^{\frac{1}{p}}. \end{aligned} \end{align}\] Using this estimate we also find that \[\begin{align} \begin{aligned} &\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s,u_s) =\int_\Omega f_s ( u_s-g_s)\,\mathrm{d}x +\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s,g_s)\\ &\leq \| u_s-g_s \|_{W^{s,p}(\Omega)} \|f_s\|_{L^{p'}\Omega)}+\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s,u_s)^{\frac{1}{p'}}\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(g_s,g_s)^{\frac{1}{p}}\\ &\leq C \|f_s\|_{L^{p'}\Omega)}\big( \mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s,u_s)^{\frac{1}{p}}+\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(g_s,g_s)^{\frac{1}{p}}\big) +\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s,u_s)^{\frac{1}{p'}}\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(g_s,g_s)^{\frac{1}{p}}. \end{aligned} \end{align}\] In view of the Young inequality \(|ab|\leq \frac{\delta^p |a|^p}{p}+ \frac{|b|^{p'}}{p'\delta^{p'}}\), \(a,b\in \mathop{\mathrm{\mathbb{R}}}\) and \(\delta>0\), we get

\[\begin{align} \mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s,u_s)^{\frac{1}{p'}}\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(g_s,g_s)^{\frac{1}{p}} &\leq \frac{\delta^{p}\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s,u_s)}{p'}+ \frac{\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(g_s,g_s)}{p\delta^{\frac{p^2}{p'}}},\\ C\|f_s\|_{L^{p'}\Omega)} \mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s,u_s)^{\frac{1}{p}} &\leq \frac{\delta^p\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s,u_s)}{p}+ \frac{C^{p'}\|f_s\|_{L^{p'}\Omega)} ^{p'}}{p'\delta^{p'}},\\ C\|f_s\|_{L^{p'}\Omega)}\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(g_s,g_s)^{\frac{1}{p}} &\leq \frac{\delta^{p}\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(g_s,g_s)}{p} + \frac{C^{p'}\|f_s\|_{L^{p'}\Omega)} ^{p'}}{p'\delta^{p'}}. \end{align}\] Inserting altogether in the previous estimate gives \[\begin{align} \mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s,u_s) &\leq \delta^{p}\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s,u_s) + ( \frac{1}{p\delta^{\frac{p^2}{p'}}}+ \frac{\delta^{p}}{p}) \mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(g_s,g_s) + \frac{2C^{p'}}{p'\delta^{p'}} \|f_s\|_{L^{p'}\Omega)} ^{p'}. \end{align}\] Taking in particular \(\delta^p=\frac{1}{2}\) we arrive at \[\begin{align} \mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s,u_s)&\leq C(\|f_s\|_{L^{p'}\Omega)} ^{p'} + \mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(g_s,g_s)) . \end{align}\] On the other hand, the coercivity estimate 19 implies \[\begin{align} \label{eq:xlp-bound-dirich} \begin{aligned} \|u_s\|_{L^p(\Omega)} &\leq \|g_s\|_{L^p(\Omega)}+ C\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s-g_s,u_s-g_s)^{\frac{1}{p}}\\ &\leq C\|g_s\|_{W^{s,p}(\Omega) }+ C\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(u_s,u_s)^{\frac{1}{p}}. \end{aligned} \end{align}\tag{20}\] By combining everything together and accounting 18 leads to the uniform estimate \[\begin{align} \label{eq:unifo-bdedness-D} \|u_s\|_{W^{s,p}(\Omega)} &\leq C\big(\|f_s \|^{\frac{1}{p-1}}_{L^{p'}\Omega)}+\|g_s \|_{W^{s,p}(\Omega)}\big) \leq C \qquad s\in (s_*,1) \end{align}\tag{21}\] with a generic constant \(C>0\) independent of \(s\). Accordingly, by the asymptotic compactness Theorem 30, there is \(u\in W^{1,p}(\Omega)\) and subsequence \(s_n\to 1^-\) such that \[\begin{align} &\|u_{s_n} -u\|_{W^{\eta,p}(\Omega)}\xrightarrow{s_n\to 1^-}0 \qquad\text{for all 0\leq \eta<1}, \\ &\|\gamma^{s_n}_0(u_{s_n}) -\gamma^1_0(u)\|_{W^{\tau-\frac{1}{p},p}(\partial\Omega)}\xrightarrow{s_n\to 1^-}0\qquad\text{for all \frac{1}{p} <\tau<1},\\ &\mathop{\mathrm{\mathcal{E}}}^{1,p}_\Omega(u,u) \leq \liminf_{n\to\infty}\mathop{\mathrm{\mathcal{E}}}^{s_n,p}_\Omega(u_{s_n}, u_{s_n}). \end{align}\] Since \(\gamma^{s_n}_0 (u_{s_n})= \gamma^{s_n}_0(g_{s_n})\), \(\gamma^{\tau}_0|_{W^{s_n,p}(\Omega)}=\gamma^{s_n}_0\) and \(\gamma^{s_n}_0|_{W^{1,p}(\Omega)}= \gamma^{1}_0\) it follows that \[\begin{align} \|\gamma^1_0(u)-\gamma^1_0(g_1)\|_{L^{p}(\partial\Omega)} &\leq \|\gamma^{s_n}_0(u_{s_n}) -\gamma^1_0(u)\|_{L^{p}(\partial\Omega)} + \|\gamma^{s_n}_0(g_{s_n}) -\gamma^1_0( g_1)\|_{L^{p}(\partial\Omega)}\\ &\leq C_\tau \|\gamma^{s_n}_0(u_{s_n}) -\gamma^1_0(u)\|_{W^{\tau-\frac{1}{p},p}(\partial\Omega)} + C\|g_{s_n}- g_1\|_{W^{1,p}(\Omega)} \to0, \end{align}\] as \(s_n\to 1^-\). This implies \(\gamma^1_0(u)= \gamma^1_0(g_1)\), i.e., \(u-g_1\in W^{1,p}_0(\Omega)\). From the fact that \(u_{s_n}-g_{s_n}\to u-g_{1}\) in \(L^p(\Omega)\) and \(f_{s_n}\rightharpoonup f_1\) we find that \[\begin{align} \liminf_{n\to\infty}\mathop{\mathrm{\mathcal{J}}}_0^{s_n,p}(u_{s_n})= \liminf_{n\to\infty}\big(\tfrac{1}{p} \mathop{\mathrm{\mathcal{E}}}_\Omega^{s_n,p}(u_{s_n},u_{s_n} )-\int_\Omega f_{s_n}(u_{s_n}-g_{s_n})\,\mathrm{d}x \big) \geq \mathop{\mathrm{\mathcal{J}}}^{1,p}_0(u). \end{align}\] Now, we want show that \(u=u_1\). Let \(v\in g_1+W^{1,p}_0(\Omega)\subset g_1+W^{s,p}_0(\Omega)\). Define \(v_s=v-g_1+g_s\) so that \(v_s\in g_s+ W^{s,p}_0(\Omega)\). Using \(v_s-v=-g_1+g_s\), we find that \[\begin{align} \big|\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(v_s,v_s)^{\frac{1}{p}}-\mathop{\mathrm{\mathcal{E}}}^{1,p}_\Omega(v,v)^{\frac{1}{p}}\big| &\leq \mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(v_s-v,v_s-v)^{\frac{1}{p}}+ \big|\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(v,v)^{\frac{1}{p}}-\mathop{\mathrm{\mathcal{E}}}^{1,p}_\Omega(v,v)^{\frac{1}{p}}\big|\\ &\leq \|g_s-g\|_{W^{s,p}(\Omega)}+ \big|\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(v,v)^{\frac{1}{p}}-\mathop{\mathrm{\mathcal{E}}}^{1,p}_\Omega(v,v)^{\frac{1}{p}}\big|. \end{align}\] Given that \(\Omega\) is Lipschitz, we have \(\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(v,v)\xrightarrow{s\to1^-}\mathop{\mathrm{\mathcal{E}}}^{1,p}_\Omega(v,v)\) and from 17 we have \(\|g_s-g_1\|_{W^{s,p}(\Omega)}\xrightarrow{s\to1^-}0\), it follows that \(\mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(v_s,v_s) \xrightarrow{s\to 1^-} \mathop{\mathrm{\mathcal{E}}}^{1,p}_\Omega(v,v).\) Noting that \(v_s-g_s= v-g_1\in W^{1,p}_0(\Omega)\), this and \(f_{s}\rightharpoonup f_1\) yields \[\begin{align} \lim_{s\to1^-}\mathop{\mathrm{\mathcal{J}}}^{s,p}_0(v_s) &= \lim_{s\to1^-}\big(\tfrac{1}{p} \mathop{\mathrm{\mathcal{E}}}^{s,p}_\Omega(v_s, v_s)-\int_\Omega f_{s} (u_{s}-g_{s})\,\mathrm{d}x \big)\\ &=\tfrac{1}{p}\mathop{\mathrm{\mathcal{E}}}^{1,p}(v,v) -\int_\Omega f_1(u-g_{1})\,\mathrm{d}x= \mathop{\mathrm{\mathcal{J}}}^{1,p}_0(v). \end{align}\] Since, each \(u_{s_n}\) satisfies so that \(\mathop{\mathrm{\mathcal{J}}}_0^{s_n,p}(u_{s_n})\leq \mathop{\mathrm{\mathcal{J}}}_0^{s_n,p}(v_{s_n})\), we deduce that \[\begin{align} \mathop{\mathrm{\mathcal{J}}}^{1,p}_0(u)\leq \liminf_{n\to\infty}\mathop{\mathrm{\mathcal{J}}}_0^{s_n,p}(u_{s_n})\leq \liminf_{n\to\infty}\mathop{\mathrm{\mathcal{J}}}_0^{s_n,p}(v_{s_n})= \mathop{\mathrm{\mathcal{J}}}^{1,p}_0(v). \end{align}\] In conclusion, \(\|u_{s_n}-u\|_{W^{\eta,p}(\Omega)}\to 0\) as \(s_n\to1^-\) for all \(0\leq \eta<1\) and \[\begin{align} \mathop{\mathrm{\mathcal{J}}}^{1,p}_0(u)&= \min_{v-g_1\in W^{1,p}_0(\Omega)} \mathop{\mathrm{\mathcal{J}}}^{1,p}_0(v). \end{align}\] In other words, \(u= u_1\in W^{1,p}(\Omega)\) is the unique weak solution to the local Dirichlet. The uniqueness of \(u_1\) implies that \(\|u_{s}-u_1\|_{W^{\eta,p}(\Omega)}\to 0\) as \(s\to1^-\) for all \(0\leq \eta<1\). ◻

Proof. The case \(s=1\) albeit standard, is somewhat technical and can be adapted for instance from [41] so that \(\partial_{x_i} \overline{u} = \overline{\partial_{x_i} u}\) for \(i = 1,\dots,d-1\) and \(\partial_{x_d} \overline{u} = -\overline{\partial_{x_d} u}\). For \(1\leq p\leq \infty\), the change of variables \(z=(z',z_d)= (x',-x_d)\) yields \[\begin{align} \|\overline{u}\|_{L^p(\Omega)} = 2^{1/p} \|u\|_{L^p(\Omega_+)}\quad \text{and}\quad \|\nabla \overline{u}\|_{W^{1,p}(\Omega)} = 2^{1/p}\|\nabla u\|_{W^{1,p}(\Omega_+)}. \end{align}\] Now we turn our attention to the case \(0\leq s<1\). For \(x\in \mathop{\mathrm{\mathbb{R}}}^d_+\) and \(y\in\mathop{\mathrm{\mathbb{R}}}^d_-\) we have \((x_d-y_d)^2\geq (x_d+y_d)^2\) and therefore \[\begin{align} \int_{\Omega}\int_{\Omega} \frac{|\overline{u}(x)-\overline{u}(y)|^p}{|x-y|^{d+sp}} &\,\mathrm{d}y\,\mathrm{d}x = \int_{ \Omega_-}\int_{ \Omega_-} \frac{|u(x',-x_d)-u(y',-y_d)|^p}{|x-y|^{d+sp}} \,\mathrm{d}y\,\mathrm{d}x\\ &+\int_{\Omega_+}\int_{\Omega_+} \frac{|u(x)-u(y)|^p}{|x-y|^{d+sp}} \,\mathrm{d}y \,\mathrm{d}x+ 2\int_{\Omega_+}\int_{ \Omega_-} \frac{|u(x)-u(y',-y_d)|^p}{|x-y|^{d+sp}} \,\mathrm{d}y\,\mathrm{d}x \\ & \leq 4\int_{\Omega_+}\int_{\Omega_+} \frac{|u(x)-u(y)|^p}{|x-y|^{d+sp}} \,\mathrm{d}y \,\mathrm{d}x. \end{align}\] By the same token, it readily follows that \[\begin{align} [\overline{u}] _{W^{s,\infty}(\Omega)} \leq [u]_{W^{s,\infty}(\Omega_+)}\qquad s\in (0,1). \end{align}\] Therefore, \(\mathop{\mathrm{\mathbb{R}}}^d_+\) is a \(W^{s,p}\)-extension domain since if \(\Omega_+=\mathop{\mathrm{\mathbb{R}}}^d_+\) then \(\Omega=\mathop{\mathrm{\mathbb{R}}}^d\). ◻

Proof. We only prove for \(0\leq s <1\) as the case \(s=1\) is easily yielded by the same token. Clearly for \(1\leq p< \infty\) we have \[\begin{align} |\varphi(x)u(x)-\varphi(y)u(y)|^p &\leq 2^{p-1}\Big( \|\varphi\|^p_{L^\infty(\Omega)}|u(x)-u(y)|^p+ |u(x)|^p|\varphi(x)-\varphi(y)|^p \Big) \end{align}\] for all \(x,y\in \Omega\). This immediately implies \[\begin{align} [u\varphi ]^p_{W^{s,p}(\Omega)} \leq 2^{p-1}\big(\|\varphi\|^p_{L^\infty(\Omega)} [u]^p_{W^{s,p}(\Omega)} + \|u\|^p_{L^\infty(\Omega)} [u]^p_{W^{s,p}(\Omega)}\big). \end{align}\] Now if \(\varphi\in C^{0,1}(\Omega)\) then \(|\varphi(x)-\varphi(y)|\leq \|\varphi\|_{W^{1,\infty}(\Omega)}(1\land |x-y|^p)\), so that \[\begin{align} |\varphi(x)u(x)-\varphi(y)u(y)|^p &\leq 2^{p-1} \|\varphi\|^p_{W^{1,\infty}(\Omega)} \Big( |u(x)-u(y)|^p+ |u(x)|^p(1\land |x-y|^p)\Big). \end{align}\] Moreover, we note that \[\begin{align} \int_{\Omega}\int_{\Omega} \frac{|u(x)|^p(1\land |x-y|^p)}{|x-y|^{d+sp}} \,\mathrm{d}y \,\mathrm{d}x &\leq \|u\|^p_{L^p(\Omega)} \int_{\mathop{\mathrm{\mathbb{R}}}^d} \frac{1\land |h|^p}{|h|^{d+sp}}\,\mathrm{d}h = \frac{|\mathbb{S}^{d-1}|}{ps(1-s)} \|u\|^p_{L^p(\Omega)}. \end{align}\] Therefore, integrating both sides yields \[\begin{align} s(1-s)&\int_{\Omega}\int_{\Omega} \frac{|\varphi(x)u(x)-\varphi(y)u(y)|^p}{|x-y|^{d+sp}} \,\mathrm{d}y \,\mathrm{d}x \\ &\leq 2^{p-1} \|\varphi\|^p_{W^{1,\infty}(\Omega)} s(1-s)\int_{\Omega}\int_{\Omega} \frac{|u(x)|^p(1\land |x-y|^p)}{|x-y|^{d+sp}} \,\mathrm{d}x\,\mathrm{d}y\\ & \qquad\qquad\qquad + 2^{p-1} \|\varphi\|^p_{W^{1,\infty}(\Omega)} s(1-s)\int_{\Omega}\int_{\Omega} \frac{|u(x)-u(y)|^p}{|x-y|^{d+sp}} \,\mathrm{d}y \,\mathrm{d}x\\ & \leq 2^{p-1} \|\varphi\|^p_{W^{1,\infty}(\Omega)} \bigg( |\mathbb{S}^{d-1}|\|u\|^p_{L^p(\Omega)}+ s(1-s) \int_{\Omega}\int_{\Omega} \frac{|u(x)-u(y)|^p}{|x-y|^{d+sp}} \,\mathrm{d}y \,\mathrm{d}x\bigg). \end{align}\] Moreover, we have \(\|u\varphi\|_{L^p(\Omega)} \leq \|\varphi \|_{L^\infty(\Omega)} \|u\|_{L^p(\Omega)}\). Finally, while the case \(p=\infty\) is obvious, we deduce that \[\begin{align} \|u\varphi \|_{W^{s,p}(\Omega)} &\leq 2^{1-1/p}(1+|\mathbb{S}^{d-1}|)^{1/p} \|\varphi\|_{W^{1,\infty}(\Omega)} \|u\|_{W^{s,p}(\Omega)}. \end{align}\] Now assume \(\varphi\in C^{0,1}_c(\Omega)\) with \(\mathop{\mathrm{supp}}\varphi\subset K\). Let \(\zeta_\varepsilon(x)=\varepsilon^{-d} \zeta(\varepsilon^{-1}x)\), \(\varepsilon>0\) with \(\zeta\) satisfying \(\zeta\in C^\infty(B_1(0))\), \(\int\zeta =1\) and \(0\leq \zeta\leq 1\). There is bounded set \(\Omega'\subset \Omega\) such that \(\delta:= \mathop{\mathrm{dist}}(\Omega', \partial \Omega)>0\) and for \(\varepsilon>0\) sufficiently small we have \(\mathop{\mathrm{supp}}(u\varphi)*\eta_\varepsilon\subset \mathop{\mathrm{supp}}\varphi+ B_\varepsilon(0)\subset \overline{\Omega'}\subset \Omega\) and \(\Omega'+ B_\varepsilon(0)\subset\Omega\). It follows in particular that \([u\varphi]*\zeta_\varepsilon\in C_c^\infty(\Omega)\) and \(\| [u\varphi]*\zeta_\varepsilon-[u\varphi]\|_{ L^p(\Omega)}\xrightarrow{\varepsilon\to 0}0\). Moreover, the function \((x,y)\mapsto U_\varphi (x, y)=([u\varphi](x)-[u\varphi](y))|x-y|^{-\frac{d }{p}-s}\) satisfies, since \(\Omega'+ B_\varepsilon(0)\subset \Omega\), \(\|U_\varphi (\cdot-\varepsilon z, \cdot-\varepsilon z)-U_\varphi \|_{L^p(\Omega'\times\Omega')}\leq 2^p \|U_\varphi \|_{L^p(\Omega\times\Omega)}=2^p |u|^p_{W^{s,p}(\Omega)}\) and, by continuity of the shift in \(L^p(\Omega'\times \Omega')\), \(\|U_\varphi (\cdot-\varepsilon z, \cdot-\varepsilon z)-U_\varphi \|_{L^p(\Omega'\times\Omega')}\xrightarrow{\varepsilon\to1^-}0\), for each \(z\in B_1(0)\). This in conjunction with dominated convergence theorem yield \[\begin{align} \big|[u\varphi]*\zeta_\varepsilon- &[u\varphi]\big|^p_{W^{s,p}(\Omega)}= \int_{\Omega'}\int_{\Omega'}\frac{|([u\varphi]*\zeta_\varepsilon- [u\varphi])(x)- ([u\varphi]*\zeta_\varepsilon- [u\varphi])(y)|^p}{|x-y|^{d+sp}}\,\mathrm{d}y\,\mathrm{d}x\\ &+2 \int_{\Omega'}|[u\varphi]*\zeta_\varepsilon(x)- [u\varphi](x)|^p\int_{\Omega\setminus \Omega'}\frac{\,\mathrm{d}y}{|x-y|^{d+sp}}\,\mathrm{d}x\\ &\leq \int_{B_1(0)}\zeta(z)\int_{\Omega'}\int_{\Omega'}|U_\varphi (x-\varepsilon z, y-\varepsilon z)-U_\varphi (x, y)|^p\,\mathrm{d}y\,\mathrm{d}x\,\mathrm{d}z \\ & +2 \|[u\varphi]*\eta_\varepsilon-[u\varphi]\|^p_{L^p(\Omega')}\int_{|h|>\delta}\frac{\,\mathrm{d}h}{|h|^{d+sp}}\,\xrightarrow{\varepsilon\to1^-}0. \end{align}\] Thus we obtain \(\| [u\varphi]*\zeta_\varepsilon-[u\varphi]\|_{ W^{s,p}(\Omega)} \to 0\) as \(\varepsilon\to 0\), and hence \(u\varphi \in W^{s,p}_0(\Omega)\). ◻

Proof. Let \(\delta=\mathop{\mathrm{dist}}(K,\partial\Omega)>0\), so that \(|x-y|\geq \delta>0\) for \(x\in K\) and \(y\in\mathop{\mathrm{\mathbb{R}}}^d\setminus \Omega\). In other words \(\mathop{\mathrm{\mathbb{R}}}^d\setminus \Omega\subset \mathop{\mathrm{\mathbb{R}}}^d\setminus B_\delta(x)\) for every \(x\in K\). Hence for \(x\in K\) we find that \[\begin{align} \int_{\mathop{\mathrm{\mathbb{R}}}^d\setminus \Omega} \frac{\,\mathrm{d}y }{|x-y|^{d+sp}} \leq \int_{B^c_\delta (0)}\frac{\,\mathrm{d}h }{|h|^{d+sp}} &\leq \max(1,\delta^{-p}) \int_{\mathop{\mathrm{\mathbb{R}}}^d}\frac{1\land |h|^p }{|h|^{d+sp}} \,\mathrm{d}h= \max(1,\delta^{-p}) \frac{|\mathbb{S}^{d-1}|}{s(1-s)p}. \end{align}\] Since \(\mathop{\mathrm{supp}}u\subset K\subset\Omega\) we can split the seminorm as follows \[\begin{align} s(1-s)\int_{\mathop{\mathrm{\mathbb{R}}}^d}\int_{\mathop{\mathrm{\mathbb{R}}}^d} &\frac{|\widetilde{u}(x)-\widetilde{u}(y)|^p}{|x-y|^{d+sp}} \,\mathrm{d}y\,\mathrm{d}x = s(1-s) \int_{\Omega}\int_{\Omega} \frac{|u(x)-u(y)|^p}{|x-y|^{d+sp}} \,\mathrm{d}y \,\mathrm{d}x \\ &+ 2 s(1-s) \int_{K} |u(x)|^p \left( \int_{\mathop{\mathrm{\mathbb{R}}}^d \setminus \Omega} \frac{\,\mathrm{d}y}{|x-y|^{d+sp}}\right)\,\mathrm{d}x\\ &\leq s(1-s) \int_{\Omega}\int_{\Omega} \frac{|u(x)-u(y)|^p}{|x-y|^{d+sp}} \,\mathrm{d}y \,\mathrm{d}x + 2|\mathbb{S}^{d-1}| \max(1,\delta^{-p})\|u\|^p_{L^p(\Omega )}. \end{align}\] Beside this, it is readily seen that \(\|\widetilde{u}\|_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)} = \|u\|_{L^p(\Omega)}\), while the case \(s=1\) is standard and \(\|\nabla \widetilde{u}\|_{L^p(\mathop{\mathrm{\mathbb{R}}}^d)} = \|\nabla u\|_{L^p(\Omega)}\) if \(u\in W^{1,p}(\Omega)\). Therefore, we find that \[\begin{align} \|\widetilde{u}\|^p_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)} \leq 2(1+ |\mathbb{S}^{d-1}|) \max(1,\delta^{-p}) \|u\|^p_{W^{s,p}(\Omega)}, \end{align}\] for \(s\in [0,1]\) and \(1\leq p< \infty\). The case \(p=\infty\) and \(s\in (0,1)\) is straightforward. ◻

Proof. We only prove the first estimate as the second one is similar. Since \(T : U' \to U\) and \(T^{-1} : U \to U'\) are Lipschitz with constants \(\mathrm{Lip}(T)\) and \(\mathrm{Lip}(T^{-1})\), respectively, it follows from Rademacher’s theorem (see [42]), that both \(T\) and \(T^{-1}\) are differentiable almost everywhere.Moreover , the Jacobian \(DT\) and \(DT^{-1}\) are bounded, with \(\|DT\|_{L^\infty(U')} \leq \mathrm{Lip}(T)\) and \(\|DT^{-1}\|_{L^\infty(U)} \leq \mathrm{Lip}(T^{-1})\). Let \(M>1\) such that \[\begin{align} \|\det D T\|_{L^\infty(U)}+ \|\det D T^{-1}\|_{L^\infty(U')}+ \mathrm{Lip}(T)+\mathrm{Lip}(T^{-1})\leq M. \end{align}\] While the case \(p=\infty\) is obvious, the case \(s=1\) is also standard since (by mollifying \(T\)) one obtains the chain rule \(\nabla (u\circ T)= (\nabla u)\circ T\cdot D T\); see [38]. Now we consider the case \(1\leq p<\infty\) and \(s\in [0,1)\). It is easy to find that \(\|u\circ T\|^p_{L^p(U')} \leq M\|u\|^p_{L^p(U)}\). Next, we have \(|x-y|^{d+sp}\leq M^{d+p}|T^{-1}(x)-T^{-1} (y)|^{d+sp}\) as since \(T\) is Lipschitz. Using the change of variables \(x=T(\hat{x})\) and \(y=T(\hat{y})\) implies \[\begin{align} \int_{U'}\int_{U'}& \frac{|u(T(\hat{x}))-u(T(\hat{y}))|^p}{|\hat{x}-\hat{y}|^{d+sp}} \,\mathrm{d}\hat{y} \,\mathrm{d}\hat{x} \\ & = \int_{U}\int_{U} \frac{|u(x)-u(y)|^p}{|T^{-1}(x)-T^{-1}(y)|^{d+sp}}|\det DT^{-1}(x)| \,\mathrm{d}y |\det DT^{-1}(y)| \,\mathrm{d}x \\ & \leq M^{d+p+2} \int_{U}\int_{U} \frac{|u(x)-u(y)|^p}{|x-y|^{d+sp}}\,\mathrm{d}y \,\mathrm{d}x. \end{align}\] That is \([u\circ T]_{W^{s,p}(U') } \leq C [u]_{W^{s,p}(U)}\) with \(C=M^{\frac{d+2}{p}+1}\). ◻

Proof. Step 1: Local flattening. Since \(\partial\Omega\) is compact Lipschitz and \(\mathop{\mathrm{dist}}(\Omega, \partial \Omega')>0\), there exists a finite number of open sets \(U_1,U_2,\cdots, U_k\) and a finite number of elements \(x_1, x_2, \cdots , x_k\in \partial\Omega\) where \(\overline{U}_j\subset \Omega'\) and \(x_j\in \partial\Omega\cap U_j\), such that \(\partial{\Omega} \subset \bigcup_{j=1}^k U_j\). Moreover, for each \(j\in\{1,\cdots,k\}\) there is a bi-Lipschitz isomorphism \(T_j:Q\to U_j\) with inverse \(T^{-1}_j: U_j\to Q\) such that \[\begin{align} T_j(Q_+)=U_j\cap\Omega, \qquad T_j(Q_0)=U_j\cap\partial\Omega \end{align}\] where \(Q_0:=\{ x\in Q : x_d=0\}\) and \(Q_+=Q\cap \mathop{\mathrm{\mathbb{R}}}^d_+\) with \[\begin{align} Q&:=\{ x=(x',x_d)\in\mathop{\mathrm{\mathbb{R}}}^d : |x'|<1\;and\;|x_d|<1\}. \end{align}\] Step 2: Partition of unity. Clearly, we can write \(\Omega'= \bigcup_{j=1}^k U_j \cup \big(\Omega' \setminus \partial{\Omega} \big)\). Accordingly, by the partition of unity, there exist \(k+1\) smooth functions \(\psi_0\), \(\psi_1\),..., \(\psi_k\in C_c^\infty(\mathop{\mathrm{\mathbb{R}}}^d)\) such that for each \(j\in \{0,... ,k\}\) we have \(\mathop{\mathrm{supp}}\psi_0\subset \mathop{\mathrm{\mathbb{R}}}^d\setminus \partial{\Omega}\), \(\mathop{\mathrm{supp}}\psi_j\subset U_j\), \(0\leq \psi_j \leq 1\) and \(\sum_{j=0}^k \psi_j=1\). We obviously have \(u=\sum_{j=0}^k \psi_j u.\)

Step 3: Pull-back. Recall that \(T_j: Q\to U_j\) is the isomorphism of class \(C^{0,1}\) with \(T_j(Q_+)=U_j\cap \Omega\), \(j\in\{1,...,k\}\). Let us consider the pull-back of \(u|_{U_j\cap\Omega }\) defined by \[\begin{align} v_j(y):=u( T_j(y) ) \qquad \text{ for any } y\in Q_+. \end{align}\] According to Lemma 11, \(v_j \in W^{s,p}\left( Q_+ \right)\) and we have \[\begin{align} \label{eq:Tj-push-forward} \begin{aligned} [v_j]^p_{W^{s,p}(Q_+) } &\leq C [u]^p_{W^{s,p}(U_j\cap\Omega)} \qquad \text{ and }\qquad \|v_j\|^p_{L^p(Q_+) } \leq C \|u\|^p_{L^p(U_j\cap\Omega)}. \end{aligned} \end{align}\tag{22}\] Here and in the sequel, it will be clear from the context that \(C = C(d,p,\Omega)>0\) denotes a generic constant independent of \(s\) and possibly different step by step.

Step 4: Even reflection. According to Lemma 8, the even reflection \(\overline{v}_j:Q\to \mathop{\mathrm{\mathbb{R}}}\) with \(\overline{v}_j(x)=v_j(x',|x_d|)\) belongs to \(W^{s,p}(Q)\) and satisfies \[\begin{align} [\overline{v}_j]^p_{W^{s,p}(Q)} &\leq 4 [v_j]^p_{W^{s,p}(Q_+)}\qquad \text{ and }\qquad \|\overline{v}_j\|^p_{L^p(Q) } = 2 \|v_j\|^p_{L^p(Q_+)}. \end{align}\] Step 5: Push-forward. Next, after extending \(v_j\) to \(Q\) as \(\overline{v}_j\) we have to push the extension back to \(\Omega\) by defining \[\begin{align} w_j(x):=\overline{v}_j\big(T_j^{-1}(x)\big) \quad \text{for all } x\in U_j . \end{align}\] It is worth noting that \(w_j\equiv u\). Since \(T_j\) is bi-Lipschitz, by arguing as above using Lemma 11, it follows that \(w_j\in W^{s,p}(U_j)\) and \[\begin{align} \label{eq:Tj-pull-back} \begin{aligned} [w_j]^p_{W^{s,p}(U_j)} &\leq C [\overline{v}_j]^p_{W^{s,p}(Q)}\leq C [v_j]^p_{W^{s,p}(Q_+)}\leq C [u]^p_{W^{s,p}(U_j\cap \Omega)}\\ \|w_j\|^p_{L^p(U_j) } &\leq C \|\overline{v}_j\|^p_{L^p(Q)} \leq C\|v_j\|^p_{L^p(Q_+) } \leq C \|u\|^p_{L^p(U_j\cap\Omega)}. \end{aligned} \end{align}\tag{23}\] Step 6: Zero-extension. By definition \(\psi_0u\) and \(\psi_jw_j\), \(j\in\{1,\cdots,k\}\) , have compact support in \(U_j\subset \Omega'\) so that Lemma 9 implies \(\psi_0u,\psi_jw_j\in W^{s,p}_0(U_j)\). By Lemma 10 their corresponding zero the extensions \(\widetilde{\psi_0 u}\) and \(\widetilde{\psi_jw_j}\) to all \(\mathop{\mathrm{\mathbb{R}}}^d\) belong to \(W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)\). Furthermore, since \(\psi_0u,\psi_jw_j\in W^{s,p}_0(U_j)\) and \(\overline{U}_j\subset \Omega'\) it is not difficult to verify that \(\widetilde{\psi_0 u}, \widetilde{\psi_jw_j}\in W^{s,p}_0(\Omega')\). In sum, \(\widetilde{\psi_0 u}, \widetilde{\psi_jw_j}\in W^{s,p}_0(\Omega')\cap W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)\) and using Lemma 10, Lemma 9 and the estimate 23 we get \[\begin{align} \tag{24} \|\widetilde{\psi_j\,w_j}\|_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)} \leq C \|\psi_j\,w_j\|_{W^{s,p}(U_j)} &\leq C \|w_j\|_{W^{s,p}(U_j)}\leq C \|u\|_{W^{s,p}(\Omega\cap U_j)}, \\ \tag{25} \|\widetilde{\psi_0\,u}\|_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)}&\leq \,C\,\| \psi_0\,u\|_{W^{s,p}(\Omega)}\,\leq\, C\, \|u\|_{W^{s,p}(\Omega)}. \end{align}\] Step 7: Extension. Finally, we define the desired linear extension operator by \[\begin{align} Eu=\widetilde{\psi_0u} + \sum_{j=1}^k \widetilde{\psi_j w_j}. \end{align}\] It readily it follows that \(\mathop{\mathrm{supp}}Eu\subset \Omega'\), since \(\mathop{\mathrm{supp}}\psi_j\subset U_j\subset \Omega'\). Moreover, we have \(Eu \in W^{s,p}_0(\Omega')\cap W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)\) since \(\widetilde{\psi_0 u}, \widetilde{\psi_jw_j}\in W^{s,p}_0(\Omega')\cap W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)\). Next, since \(w_j\equiv u\) (and hence \(\psi_jw_j\equiv \psi_ju\)) on \(U_j\cap \Omega\), we have \[\begin{align} Eu(x)=\widetilde{\psi_0u}(x)+ \sum_{j=1}^k \widetilde{\psi_j w_j}(x)= \psi_0u(x)+ \sum_{j=1}^k \psi_j w_j(x)= u(x)\quad\text{for x\in \Omega}. \end{align}\] That is \(Eu|_{\Omega}= u\). Moreover, combining 24 with 25 , we get \[\begin{align} \|Eu\|_{W^{s,p}(\mathop{\mathrm{\mathbb{R}}}^d)} \leq C \|u\|_{W^{s,p}(\Omega)} \end{align}\] with \(C=C(d,p,\Omega)\) independent of \(s\in [0,1]\). ◻