On the sharp Hardy inequality in Sobolev–Slobodeckiĭ spaces

Abstract

We study the sharp constant in the Hardy inequality for fractional Sobolev spaces defined on open subsets of the Euclidean space. We first list some properties of such a constant, as well as of the associated variational problem. We then restrict the discussion to open convex sets and compute such a sharp constant, by constructing suitable supersolutions by means of the distance function. Such a method of proof works only for $s\,p\ge 1$ or for $\Omega$ being a half-space. We exhibit a simple example suggesting that this method can not work for $s\,p<1$ and $\Omega$ different from a half-space. The case $s\,p<1$ for a generic convex set is left as an interesting open problem, except in the Hilbertian setting (i.e. for $p=2$): in this case we can compute the sharp constant in the whole range $0<s<1$. This completes a result which was left open in the literature.

Appendices

Appendix A: Negative powers of the distance in the borderline case $s=1/2$

We consider for $-1<\beta <0$

$$\begin{aligned} U_\beta (t)=d_I(t)^\beta =\left( \min \{t,1-t\}\right) ^\beta ,\qquad \text{ for } t\in I=(0,1). \end{aligned}$$

We extend this function by 0 outside the interval I. We want to estimate its fractional Laplacian of order 1/2.

Lemma A.1

Under the assumptions above, we have

$$\begin{aligned} (-\Delta )^\frac{1}{2} U_\beta \le \beta \,H(t)\,U_\beta (t)+\frac{2}{t\,(1-t)}\,U_\beta (t),\qquad \text{ in } I, \end{aligned}$$

(A.1)

in weak sense, where H is the continuous function on $I{\setminus }\{1/2\}$ defined by

$$\begin{aligned} H(t)=-\frac{2}{t\,(1-t)}+\frac{2}{d_I(t)}\,\log \left( \frac{4\,t\,(1-t)}{(1-2\,t)^2}\right) . \end{aligned}$$

This has the following properties:

• it is symmetric with respect to 1/2, i.e.

  $$\begin{aligned} H(t)=H(1-t),\qquad \text{ for } t\in I{\setminus }\left\{ \frac{1}{2}\right\} ; \end{aligned}$$

• it belongs to $L^q_{\textrm{loc}}(I)$ for every $1\le q<\infty$;

• it satisfies

  $$\begin{aligned} H(t)\sim - 4\,\log \left( \frac{1}{2}-t\right) ^2,\qquad \text{ for } t\rightarrow \frac{1}{2}, \end{aligned}$$

  thus the right-hand side of (A.1) diverges to $-\infty$ as t approaches 1/2.

In particular, in this case the function $U_\beta$ is not even locally weakly (1/2)-superharmonic on I.

Proof

Observe that $U_\beta \in W^{s,2}_{\textrm{loc}}(I)\cap L^1_{1}({\mathbb {R}})$, under the assumption $-1<\beta <0$. We first show that $U_\beta$ satisfies (A.1) in $I{\setminus }\{1/2\}$. Let us take $\varphi \in C^\infty _0(I{\setminus }\{1/2\})$ non-negative, then there exists $0<\delta _0<1/4$ such that its support is contained in the set

$$\begin{aligned} {\mathcal {I}}_{\delta _0}=\left[ \delta _0,\frac{1}{2}-\delta _0\right] \cup \left[ \frac{1}{2}+\delta _0,1-\delta _0\right] . \end{aligned}$$

For every $\varepsilon >0$ and $t\in I$, we set

$$\begin{aligned} {\mathfrak {I}}_\varepsilon (t)=(t-\varepsilon ,t+\varepsilon )\qquad \text{ and } \qquad {\mathcal {D}}_\varepsilon =\Big \{(t,y)\in {\mathbb {R}}\times {\mathbb {R}}\, :\, t-\varepsilon \le y\le t+\varepsilon \Big \}. \end{aligned}$$

By using Fubini’s Theorem and a change of variable, we can write as usual

$$\begin{aligned}&\iint _{{\mathbb {R}}\times {\mathbb {R}}} \frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\big (\varphi (t)-\varphi (y)\big )}{|t-y|^{2}}\,dt\,dy\nonumber \\&\quad =\lim _{\varepsilon \rightarrow 0} \iint _{({\mathbb {R}}\times {\mathbb {R}}){\setminus }{\mathcal {D}}_\varepsilon }\frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\big (\varphi (t)-\varphi (y)\big )}{|t-y|^{2}}\,dt\,dy\nonumber \\&\quad =2\,\lim _{\varepsilon \rightarrow 0} \int _{{\mathbb {R}}} \left( \int _{{\mathbb {R}}{\setminus } {\mathfrak {I}}_\varepsilon (t)} \frac{U_\beta (t)-U_\beta (y)}{|t-y|^{2}}\,dy\right) \,\varphi (t)\,dt\nonumber \\&\quad =2\,\lim _{\varepsilon \rightarrow 0} \int _{{\mathcal {I}}_{\delta _0}} \left( \int _{{\mathbb {R}}{\setminus } {\mathfrak {I}}_\varepsilon (t)} \frac{U_\beta (t)-U_\beta (y)}{|t-y|^{2}}\,dy\right) \,\varphi (t)\,dt. \end{aligned}$$

(A.2)

We first observe that, by using that $U_\beta (t)=U_\beta \left( 1-t\right)$, we have for $t\in {\mathcal {I}}_{\delta _0}$

$$\begin{aligned} \begin{aligned} \int _{{\mathbb {R}}{\setminus } {\mathfrak {I}}_\varepsilon (t)} \frac{U_\beta (t)-U_\beta (y)}{|t-y|^{2}}\,dy&=\int _{{\mathbb {R}}{\setminus } {\mathfrak {I}}_\varepsilon (t)} \frac{U_\beta (1-t)-U_\beta (1-y)}{|t-y|^{2}}\,dy\\&=\int _{{\mathbb {R}}{\setminus } {\mathfrak {I}}_\varepsilon (t)}\frac{U_\beta (1-t)-U_\beta (1-y)}{|(1-t)-(1-y)|^{2}}\,dy\\&=\int _{{\mathbb {R}}{\setminus } {\mathfrak {I}}_\varepsilon (1-t)} \frac{U_\beta (1-t)-U_\beta (\tau )}{|(1-t)-\tau |^{2}}\,d\tau . \end{aligned} \end{aligned}$$

This shows that it is sufficient to consider $t\in [\delta _0,1/2-\delta _0]$. For almost every $t\in [\delta _0,1/2-\delta _0]$ and every $0<\varepsilon < \delta _0$, we have⁴

$$\begin{aligned} \begin{aligned} \int _{{\mathbb {R}}{\setminus } {\mathfrak {I}}_\varepsilon (t)} \frac{U_\beta (t)-U_\beta (y)}{|t-y|^{2}}\,dy&=\int _0^{t-\varepsilon }\frac{t^\beta -y^\beta }{|t-y|^{2}}\,dy+\int _{t+\varepsilon }^\frac{1}{2} \frac{t^\beta -y^\beta }{|t-y|^{2}}\,dy\\&\quad +\int _\frac{1}{2}^1 \frac{t^\beta -(1-y)^\beta }{(y-t)^{2}}\,dy\\&\quad +\int _1^{+\infty } \frac{t^\beta }{(y-t)^{2}}\,dy+\int _{-\infty }^0 \frac{t^\beta }{(t-y)^{2}}\,dy\\&=\int _0^{t-\varepsilon }\frac{t^\beta -y^\beta }{|t-y|^{2}}\,dy +\int _{t+\varepsilon }^\frac{1}{2} \frac{t^\beta -y^\beta }{|t-y|^{2}}\,dy\\&\quad +\int _\frac{1}{2}^1 \frac{t^\beta -(1-y)^\beta }{(y-t)^{2}}\,dy\\&\quad +\frac{U_\beta (t)}{(1-t)}+\frac{U_\beta (t)}{t}. \end{aligned} \end{aligned}$$

To estimate the remaining integrals, we use the “above tangent” property for the convex function $\tau \mapsto \tau ^\beta$, to infer that

$$\begin{aligned} t^\beta -y^\beta \le \beta \,t^{\beta -1}\,(t-y),\qquad \text{ for } y\in \left( 0,t-\varepsilon \right) \cup \left( t+\varepsilon ,\frac{1}{2}\right) , \end{aligned}$$

and

$$\begin{aligned} t^\beta -(1-y)^\beta \le \beta \,t^{\beta -1}\,(t+y-1),\qquad \text{ for } y\in \left( \frac{1}{2},1\right) . \end{aligned}$$

These yield

$$\begin{aligned} \begin{aligned}&\int _0^{t-\varepsilon } \frac{t^\beta -y^\beta }{|t-y|^{2}}\,dy+\int _{t+\varepsilon }^\frac{1}{2} \frac{t^\beta -y^\beta }{|t-y|^{2}}\,dy+\int _\frac{1}{2}^1 \frac{t^\beta -(1-y)^\beta }{(y-t)^{2}}\,dy\\&\quad \le \beta \,t^{\beta -1}\,\int _0^{t-\varepsilon } \frac{t-y}{|t-y|^{2}}\,dy+\beta \,t^{\beta -1}\,\int _{t+\varepsilon }^\frac{1}{2} \frac{t-y}{|t-y|^{2}}\,dy+\beta \,t^{\beta -1}\\&\quad \int _\frac{1}{2}^1 \frac{(t+y-1)}{(y-t)^{2}}\,dy. \end{aligned} \end{aligned}$$

The last integrals can be explicitly computed. We have

$$\begin{aligned} \begin{aligned} \int _0^{t-\varepsilon } \frac{t-y}{|t-y|^{2}}\,dy&+\int _{t+\varepsilon }^{\frac{1}{2}} \frac{t-y}{|t-y|^{2}}\,dy =\log t-\log \left( \frac{1}{2}-t\right) , \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \int _{\frac{1}{2}}^1 \frac{(t+y-1)}{(y-t)^2}\,dy&=t\,\int _\frac{1}{2}^1 \frac{1}{(y-t)^2}\,dy+\int _{\frac{1}{2}}^1 \frac{y-1}{(y-t)^{2}}\,dy\\&=t\,\left[ \left( \frac{1}{2}-t\right) ^{-1}-(1-t)^{-1}\right] \\&\quad -\frac{1}{2}\,\left( \frac{1}{2}-t\right) ^{-1}+\left[ \log (1-t)-\log \left( \frac{1}{2}-t\right) \right] . \end{aligned} \end{aligned}$$

This finally gives that for $t\in [\delta _0,1/2-\delta _0]$, we have

$$\begin{aligned} \int _{{\mathbb {R}}{\setminus } {\mathfrak {I}}_\varepsilon (t)} \frac{U_\beta (t)-U_\beta (y)}{|t-y|^{1+2\,s}}\,dy\le \beta \,t^{\beta -1}\,G(t)+\frac{U_\beta (t)}{t\,(1-t)}, \end{aligned}$$

(A.3)

where we set for simplicity

$$\begin{aligned} \begin{aligned} G(t)&=\left[ \log t-\log \left( \frac{1}{2}-t\right) \right] +t\,\left[ \left( \frac{1}{2}-t\right) ^{-1}-(1-t)^{-1}\right] \\&\quad -\frac{1}{2}\,\left( \frac{1}{2}-t\right) ^{-1}+\left[ \log (1-t)-\log \left( \frac{1}{2}-t\right) \right] . \end{aligned} \end{aligned}$$

With simple manipulations, we see that this can be also written as

$$\begin{aligned} \begin{aligned} G(t)&=-\frac{1}{1-t}+\log \left( \frac{4\,t\,(1-t)}{\left( 1-2\,t\right) ^2}\right) , \end{aligned} \end{aligned}$$

and thus this is a continuous function on (0, 1/2) such that

$$\begin{aligned} G\in L^q_{\textrm{loc}}((0,1/2]),\ \text{ for } \text{ every } 1\le q<\infty \qquad \text{ and } \qquad \lim _{t\rightarrow \left( \frac{1}{2}\right) ^-} G(t)=+\infty , \end{aligned}$$

because of the logarithmic term. If we now define

$$\begin{aligned} H(t):=2\,\frac{G(t)}{t},\; \text{ for } t\in \left( 0,\frac{1}{2}\right) \; \text{ and } \; H(t):=H(1-t),\quad \text{ for } t\in \left( \frac{1}{2},1\right) , \end{aligned}$$

from (A.2) and (A.3), by recalling that $\varphi$ is non-negative we finally get

$$\begin{aligned} \begin{aligned} \iint _{{\mathbb {R}}\times {\mathbb {R}}}&\frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\big (\varphi (t)-\varphi (y)\big )}{|t-y|^{2}}\,dt\,dy\\&\le \int _I \left[ \beta \,H(t)+ \frac{2}{t\,(1-t)}\right] \,U_\beta (t)\,\varphi (t)\,dt. \end{aligned} \end{aligned}$$

This shows that $U_\beta$ is a local weak subsolution of the the claimed Eq. (A.1), at least in the open set $I{\setminus }\{1/2\}$.

In order to show that $U_\beta$ is a local weak subsolution on the whole interval I, it is sufficient to use a standard trick to⁵ “fill the hole”. we take $\varphi \in C^\infty _0(I)$ non-negative and for every natural number $n\ge 5$ we take $\psi _n\in C^\infty ({\overline{I}})$ such that

$$\begin{aligned} \psi _n\equiv 1 \text{ on } {\overline{I}}{\setminus }\left[ \frac{1}{2}-\frac{2}{n},\frac{1}{2}+\frac{2}{n}\right] ,\qquad \psi _n\equiv 0 \text{ on } \left[ \frac{1}{2}-\frac{1}{n},\frac{1}{2}+\frac{1}{n}\right] , \end{aligned}$$

and

$$\begin{aligned} 0\le \psi _n\le 1,\qquad |\psi _n'|\le C\,n. \end{aligned}$$

The seminorm of $\psi _n$ can be estimated by using its properties and an interpolation inequality (see [11, Corollary 2.2]), i. e.

$$\begin{aligned}{}[\psi _n]^2_{W^{\frac{1}{2},2}(I)}=[1-\psi _n]_{W^{\frac{1}{2},2}(I)}^2\le & {} C\,\left( \int _I |\psi _n'|^2\,dt\right) ^\frac{1}{2}\,\left( \int _I |1-\psi _n|^2\,dt\right) ^\frac{1}{2}\\= & {} C\,\left( \int _{\frac{1}{2}-\frac{2}{n}}^{\frac{1}{2}+\frac{2}{n}}|\psi _n'|^2\,dt\right) ^\frac{1}{2}\,\left( \int _{\frac{1}{2}-\frac{2}{n}}^{\frac{1}{2}+\frac{2}{n}} |1-\psi _n|^2\,dt\right) ^\frac{1}{2}\le {\widetilde{C}}, \end{aligned}$$

where ${\widetilde{C}}$ does not depend on n. This in particular implies that the sequence $\{\Psi _n\}_{n\ge 5}\subseteq L^2(I\times I)$ defined by

$$\begin{aligned} \Psi _n(t,y):=\frac{\psi _n(t)-\psi _n(y)}{|t-y|},\qquad \text{ for } \text{ a. } \text{ e. } (t,y)\in I\times I, \end{aligned}$$

(A.4)

is bounded in $L^2(I\times I)$, since by construction

$$\begin{aligned} \Vert \Psi _n\Vert _{L^2(I\times I)}=[\psi _n]_{W^{\frac{1}{2},2}(I)}. \end{aligned}$$

Thus, up to a subsequence, it converges weakly in $L^2(I\times I)$. Thanks to the properties of $\psi _n$, such a limit function must coincide with the null one.

The test function $\varphi \,\psi _n$ belongs to $C^\infty _0(I{\setminus }\{1/2\})$ and is non-negative. From the first part we get

$$\begin{aligned}&\iint _{{\mathbb {R}}\times {\mathbb {R}}} \frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\big (\varphi (t)\,\psi _n(t)-\varphi (y)\,\psi _n(y)\big )}{|t-y|^{2}}\,dt\,dy\nonumber \\&\quad \le \int _I {\left[ \beta \,H(t)+ \frac{2}{t\,(1-t)}\right] }\,U_\beta (t)\,\varphi (t)\,\psi _n(t)\,dt. \end{aligned}$$

(A.5)

We wish to pass to the limit in (A.5), as n goes to $\infty$: for the right-hand side, it is easily seen that

$$\begin{aligned} \begin{aligned}&\lim _{n\rightarrow \infty }\int _I {\left[ \beta \,H(t)+ \frac{2}{t\,(1-t)}\right] }\,U_\beta (t)\,\varphi (t)\,\psi _n(t)\,dt\\&=\int _I {\left[ \beta \,H(t)+ \frac{2}{t\,(1-t)}\right] }\,U_\beta (t)\,\varphi (t)\,dt, \end{aligned} \end{aligned}$$

by the Dominated Convergence Theorem. As for the left-hand side, we split the integral as follows:

$$\begin{aligned} \begin{aligned}&\iint _{{\mathbb {R}}\times {\mathbb {R}}} \frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\big (\varphi (t)\,\psi _n(t)-\varphi (y)\,\psi _n(y)\big )}{|t-y|^{2}}\,dt\,dy\\&\quad =\iint _{I''\times I''}\frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\big (\varphi (t)\,\psi _n(t)-\varphi (y)\,\psi _n(y)\big )}{|t-y|^{2}}\,dt\,dy\\&\qquad +2\,\iint _{I'\times ({\mathbb {R}}{\setminus } I'')}\frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\varphi (t)\,\psi _n(t)}{|t-y|^{2}}\,dt\,dy, \end{aligned} \end{aligned}$$

where $I'\subseteq I''\subseteq I$ and $I'$ contains the support of $\varphi$. For the last integral we can easily pass to the limit as n goes to $\infty$, for the first one we proceed as follows

$$\begin{aligned} \begin{aligned}&\iint _{I''\times I''}\frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\big (\varphi (t)\,\psi _n(t)-\varphi (y)\,\psi _n(y)\big )}{|t-y|^{2}}\,dt\,dy\\&\quad =\iint _{I''\times I''}\frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\big (\varphi (t)-\varphi (y)\big )}{|t-y|^{2}}\,\frac{\psi _n(t)+\,\psi _n(y)}{2}dt\,dy\\&\qquad +\iint _{I''\times I''}\frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\big (\psi _n(t)-\psi _n(y)\big )}{|t-y|^{2}}\,\frac{\varphi (t)+\,\varphi (y)}{2}dt\,dy. \end{aligned} \end{aligned}$$

By using that

$$\begin{aligned} \frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\big (\varphi (t)-\varphi (y)\big )}{|t-y|^2}\in L^1(I''\times I''), \end{aligned}$$

and the properties of $\psi _n$, we get that

$$\begin{aligned} \begin{aligned}&\lim _{n\rightarrow \infty }\iint _{I''\times I''}\frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\big (\varphi (t)-\varphi (y)\big )}{|t-y|^{2}}\,\frac{\psi _n(t)+\,\psi _n(y)}{2}dt\,dy\\&\quad =\iint _{I''\times I''}\frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\big (\varphi (t)-\varphi (y)\big )}{|t-y|^{2}}\,dt\,dy, \end{aligned} \end{aligned}$$

again thanks to the Dominated Convergence Theorem. Finally, the last integral is the most delicate one: with the notation (A.4), we can write

$$\begin{aligned} \frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\big (\psi _n(t)-\psi _n(y)\big )}{|t-y|^{2}}\,\frac{\varphi (t)+\,\varphi (y)}{2}=\Phi (t,y)\,\Psi _n(t,y), \end{aligned}$$

where

$$\begin{aligned} \Phi (t,y)=\frac{\Big (U_\beta (t)-U_\beta (y)\Big )}{|t-y|}\,\frac{\varphi (t)+\,\varphi (y)}{2}\in L^2(I''\times I''). \end{aligned}$$

The last property follows from the fact that $U_\beta$ is locally Lipschitz on I. Thus, by using the weak convergence of $\{\Psi _n\}_{n\ge 5}$ previously inferred, we get

$$\begin{aligned} \begin{aligned} \lim _{n\rightarrow \infty }\iint _{I''\times I''}\frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\big (\psi _n(t)-\psi _n(y)\big )}{|t-y|^{2}}\,\frac{\varphi (t)+\,\varphi (y)}{2}dt\,dy=0. \end{aligned} \end{aligned}$$

Finally, we obtain that we can pass to the limit in (A.5) as n goes to $\infty$ and obtain

$$\begin{aligned} \iint _{{\mathbb {R}}\times {\mathbb {R}}} \frac{\Big (U_\beta (t)-U_\beta (y)\Big )\,\big (\varphi (t)-\varphi (y)\big )}{|t-y|^{2}}\,dt\,dy\\ \le \int _I {\left[ \beta \,H(t)+ \frac{2}{t\,(1-t)}\right] }\,U_\beta (t)\,\varphi (t)\,dt, \end{aligned}$$

for every $\varphi \in C^\infty _0(I)$ non-negative, as desired. $\square$

Appendix B: The constant $C_{N,s\,p}$

For $N\ge 2$, $0<s<1$ and $1<p<\infty$, we recall the definition

$$\begin{aligned} C_{N,s\,p}=(N-1)\,\omega _{N-1}\,\int _0^{+\infty } t^{N-2}\,(1+t^2)^{-\frac{N+s\,p}{2}}\,dt. \end{aligned}$$

In the next simple result, we write $C_{N,s\,p}$ in an alternative way. This permits to compare the constant obtained in the proof of Theorem 6.6, with that of [38, Theorem 1.1].

Lemma B.1

Let $N\ge 2$. Then we have

$$\begin{aligned} C_{N,s\,p}=\frac{1}{2}\,\int _{{\mathbb {S}}^{N-1}} |\omega _N|^{s\,p}\,d{\mathcal {H}}^{N-1}(\omega ). \end{aligned}$$

Proof

We first observe that by using spherical coordinates, we have

$$\begin{aligned} \int _{B_1(0)} |x_N|^{s\,p}\,dx= & {} \int _{{\mathbb {S}}^{N-1}} |\omega _N|^{s\,p}\,\left( \int _0^1\varrho ^{N-1+s\,p}\,d\varrho \right) \,d{\mathcal {H}}^{N-1}(\omega )\\= & {} \frac{1}{N+s\,p}\,\int _{{\mathbb {S}}^{N-1}} |\omega _N|^{s\,p}\,d{\mathcal {H}}^{N-1}(\omega ). \end{aligned}$$

Thus, in order to conclude, it is sufficient to show that

$$\begin{aligned} \frac{N+s\,p}{2}\,\int _{B_1(0)} |x_N|^{s\,p}\,dx=(N-1)\,\omega _{N-1}\,\int _0^{+\infty } t^{N-2}\,(1+t^2)^{-\frac{N+s\,p}{2}}\,dt. \end{aligned}$$

We compute again the integral on the left-hand side, this time by using cylindrical coordinates, i.e. we write

$$\begin{aligned} \begin{aligned} \int _{B_1(0)} |x_N|^{s\,p}\,dx&=\int _{-1}^1 \left( \int _{\{x\in B_1(0)\, :\, x_N=\tau \}}d{\mathcal {H}}^{N-1}\right) \,|\tau |^{s\,p}\,d\tau \\&=\omega _{N-1}\,\int _{-1}^1 (1-\tau ^2)^\frac{N-1}{2}\,|\tau |^{s\,p}\,d\tau \\&=2\,\omega _{N-1}\,\int _0^1 (1-\tau ^2)^\frac{N-1}{2}\,\tau ^{s\,p}\,d\tau . \end{aligned} \end{aligned}$$

We now use the change of variable $\tau =(1+t^2)^{-1/2}$. This yields

$$\begin{aligned} \begin{aligned} \int _{B_1(0)} |x_N|^{s\,p}\,dx&=2\,\omega _{N-1}\,\int _0^{+\infty } (1+t^2)^{-\frac{s\,p}{2}}\,\frac{t^{N-1}}{(1+t^2)^\frac{N-1}{2}}\,\frac{t}{(1+t^2)^\frac{3}{2}}\,dt\\&=2\,\omega _{N-1}\,\int _0^{+\infty } t^N\,(1+t^2)^{-\frac{N+2+s\,p}{2}}\,dt. \end{aligned} \end{aligned}$$

In turn, by using an integration by parts, we get

$$\begin{aligned} \begin{aligned} \int _0^{+\infty } t^N\,(1+t^2)^{-\frac{N+2+s\,p}{2}}\,dt&=\left[ -\frac{1}{N+s\,p}\,t^{N-1}\,(1+t^2)^{-\frac{N+s\,p}{2}}\right] _{0}^{+\infty }\\&\quad +\frac{1}{N+s\,p}\,(N-1)\,\int _0^{+\infty } t^{N-2}\,(1+t^2)^{-\frac{N+s\,p}{2}}\,dt. \end{aligned} \end{aligned}$$

This concludes the proof. $\square$