Riesz Potentials and the Hardy–Littlewood–Sobolev Inequality: Examples

The bump. By [F3] choose $0<r<R$ and a smooth $\rho :{\mathbb{R}}^n\to [0,1]$ with $\rho =1$ on $\bar{B}(0,r)$ and support in $B(0,R)$; then $\rho \in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$ is real, nonnegative and nonzero, and it is Lebesgue measurable by [F8].

The norm of the bump is finite and positive. Since $0\le \rho \le 1$ and $\rho$ vanishes off the measurable ball $B(0,R)$, monotonicity and the scalar rule of [F6] together with ${\int }_{B(0,R)}1d\lambda =\lambda (B(0,R))<\mathrm{\infty }$ give $\mathrm{\parallel }\rho {\mathrm{\parallel }}_p^p=\int {\rho }^p\le \lambda (B(0,R))<\mathrm{\infty }$. For the lower bound, $\rho =1$ on $B(0,r)$, so again by [F6] $\mathrm{\parallel }\rho {\mathrm{\parallel }}_p^p\ge {\int }_{B(0,r)}{\rho }^p=\lambda (B(0,r))>0$. Hence $0<\mathrm{\parallel }\rho {\mathrm{\parallel }}_p<\mathrm{\infty }$.

The potential of the bump is finite and strictly positive everywhere. Fix $x$ and put $S:=\mathrm{\mid }x\mathrm{\mid }+R+1$, so that $B(0,R)\subseteq B(x,S)$. As $\rho$ vanishes off $B(0,R)$ and $0\le \rho \le 1$, monotonicity in [F6], the change-of-variables formula [F4] applied to the substitution $y\mapsto x-y$ (determinant $1$) and the polar formula [F5] give $\int K_{\alpha }(x-y)\rho (y)dy\le {\int }_{B(0,R)}{K}_{\alpha }(x-y)dy\le {\int }_{B(0,S)}{K}_{\alpha }(z)dz=\sigma (S^{n-1})\frac{S^{\alpha }}{\alpha }<\mathrm{\infty }\mathrm{.}$ In particular $I_{\alpha }\rho (x)$ is defined by [F1]. On the other hand, for every $y\in B(0,r)\setminus \{x\}$ one has $0<\mathrm{\mid }x-y\mathrm{\mid }\le \mathrm{\mid }x\mathrm{\mid }+r$, so $K_{\alpha }(x-y)\ge (\mathrm{\mid }x\mathrm{\mid }+r{)}^{\alpha -n}$. The omitted singleton has Lebesgue measure zero, and changing the assigned diagonal value does not affect the integral by [F1]. Since $\rho =1$ on $B(0,r)$, [F6] and [F5] give $I_{\alpha }\rho (x)\ge {\int }_{B(0,r)}(\mathrm{\mid }x\mathrm{\mid }+r{)}^{\alpha -n}dy=(\mathrm{\mid }x\mathrm{\mid }+r{)}^{\alpha -n}\lambda (B(0,r))>0,$ where positivity of the ball measure is [F5]. Thus $0\le I_{\alpha }\rho (x)<\mathrm{\infty }$ and $I_{\alpha }\rho (x)>0$ for every $x$.

Positivity and finiteness of the target norm. The hypothesis applied to $\rho$ gives $\mathrm{\parallel }{I}_{\alpha }\rho {\mathrm{\parallel }}_q\le C\mathrm{\parallel }\rho {\mathrm{\parallel }}_p<+\mathrm{\infty }$, and since $I_{\alpha }\rho >0$ everywhere by step 2.2 the function $(I_{\alpha }\rho {)}^q$ is nonnegative and strictly positive on the ball $B(0,r)$ of positive measure; if $\int (I_{\alpha }\rho {)}^q$ were zero then [F6] would make $(I_{\alpha }\rho {)}^q$ vanish almost everywhere, contradicting strict positivity on a set of positive measure. Hence $0<\mathrm{\parallel }{I}_{\alpha }\rho {\mathrm{\parallel }}_q<\mathrm{\infty }$.

The scaling identities. For $\lambda >0$ define ${\rho }_{\lambda }(x):=\rho (\lambda x)$; it is again a complex smooth compactly supported function, and $I_{\alpha }{\rho }_{\lambda }$ is finite everywhere by the computation of step 2.2 applied to the support of ${\rho }_{\lambda }$. The change-of-variables formula [F4] applied to the linear map $x\mapsto \lambda x$, whose determinant is ${\lambda }^n$ and whose inverse is $x\mapsto {\lambda }^{-1}x$, gives $\mathrm{\parallel }{\rho }_{\lambda }{\mathrm{\parallel }}_p^p=\int \rho (\lambda x{)}^{p}dx={\lambda }^{-n}\int \rho (z{)}^{p}dz={\lambda }^{-n}\mathrm{\parallel }\rho {\mathrm{\parallel }}_p^p,$ that is $\mathrm{\parallel }{\rho }_{\lambda }{\mathrm{\parallel }}_p={\lambda }^{-n\mathrm{/}p}\mathrm{\parallel }\rho {\mathrm{\parallel }}_p$. For the potential, the same substitution $z=\lambda y$ in the defining integral and the homogeneity $K_{\alpha }(x-{\lambda }^{-1}z)={\lambda }^{n-\alpha }{K}_{\alpha }(\lambda x-z)$ give $I_{\alpha }{\rho }_{\lambda }(x)=\int K_{\alpha }(x-y)\rho (\lambda y)dy={\lambda }^{-n}\int K_{\alpha }(x-{\lambda }^{-1}z)\rho (z)dz={\lambda }^{-\alpha }\int K_{\alpha }(\lambda x-z)\rho (z)dz={\lambda }^{-\alpha }(I_{\alpha }\rho )(\lambda x),$ so applying [F4] once more yields $\mathrm{\parallel }{I}_{\alpha }{\rho }_{\lambda }{\mathrm{\parallel }}_q={\lambda }^{-\alpha -n\mathrm{/}q}\mathrm{\parallel }{I}_{\alpha }\rho {\mathrm{\parallel }}_q$.

The scale inequality. The hypothesis applied to the legitimate test function ${\rho }_{\lambda }$ gives $\mathrm{\parallel }{I}_{\alpha }{\rho }_{\lambda }{\mathrm{\parallel }}_q\le C\mathrm{\parallel }{\rho }_{\lambda }{\mathrm{\parallel }}_p$; substituting step 4.1, ${\lambda }^{-\alpha -n\mathrm{/}q}\mathrm{\parallel }{I}_{\alpha }\rho {\mathrm{\parallel }}_q\le C{\lambda }^{-n\mathrm{/}p}\mathrm{\parallel }\rho {\mathrm{\parallel }}_p(\lambda >0)\mathrm{.}$ Dividing the positive quantities by $\mathrm{\parallel }{I}_{\alpha }\rho {\mathrm{\parallel }}_q$, which is finite and nonzero by step 3.1, and multiplying by ${\lambda }^{\alpha +n\mathrm{/}q}$ gives ${\lambda }^{\beta }\le C^{\mathrm{\prime }}:=\frac{C\mathrm{\parallel }\rho {\mathrm{\parallel }}_p}{\mathrm{\parallel }{I}_{\alpha }\rho {\mathrm{\parallel }}_q}<\mathrm{\infty }(\lambda >0),\beta :=\frac{n}{p}-\alpha -\frac{n}{q}\mathrm{.}$

The exponent vanishes. Suppose $\beta \mathrm{\ne }0$. Then $1\mathrm{/}\beta \in \mathbb{R}$ and $C^{\mathrm{\prime }}+1>0$, so $\lambda :=(C^{\mathrm{\prime }}+1{)}^{1\mathrm{/}\beta }$ is a positive real number; by the real-power laws [F7] applied with $a=C^{\mathrm{\prime }}+1$, $r=1\mathrm{/}\beta$ and $s=\beta$, ${\lambda }^{\beta }=((C^{\mathrm{\prime }}+1{)}^{1\mathrm{/}\beta }{)}^{\beta }=(C^{\mathrm{\prime }}+1{)}^{(1\mathrm{/}\beta )\beta }=C^{\mathrm{\prime }}+1>C^{\mathrm{\prime }},$ contradicting ${\lambda }^{\beta }\le C^{\mathrm{\prime }}$ for every $\lambda >0$ as established in step 5.1. Therefore $\beta =0$, which is precisely $n\mathrm{/}p-\alpha -n\mathrm{/}q=0$, equivalently $1\mathrm{/}q=1\mathrm{/}p-\alpha \mathrm{/}n$.

Conclusion. A uniform bound $\mathrm{\parallel }{I}_{\alpha }f{\mathrm{\parallel }}_q\le C\mathrm{\parallel }f{\mathrm{\parallel }}_p$ over the complex smooth compactly supported functions forces $1\mathrm{/}q=1\mathrm{/}p-\alpha \mathrm{/}n$; the argument uses only the homogeneity of the kernel, a nonzero nonnegative bump, and the exact dilation identities, so the strict-range Hardy-Littlewood-Sobolev theorem is not a premise of this necessity statement. Countable Choice enters only through the change-of-variables and polar interfaces [F4], [F5] and [F9].

The density and its norm. By [F3] the ball $B(0,ϵ)$ is measurable with $0<\lambda (B(0,ϵ))<\mathrm{\infty }$, so $f_{ϵ}=\lambda (B(0,ϵ){)}^{-1}{1}_{B(0,ϵ)}$ is a well-defined nonnegative measurable function with $\mathrm{\mid }{f}_{ϵ}{\mathrm{\mid }}^1$ integrable and ${\int }_{{\mathbb{R}}^n}{f}_{ϵ}d\lambda =\frac{\lambda (B(0,ϵ))}{\lambda (B(0,ϵ))}=1;$ in particular $\mathrm{\parallel }{f}_{ϵ}{\mathrm{\parallel }}_1=1$ and $f_{ϵ}\in L^1({\mathbb{R}}^n;\mathbb{C})$.

The pointwise lower bound. Fix $x$ with $\mathrm{\mid }x\mathrm{\mid }>2ϵ$. For every $y\in B(0,ϵ)$ the triangle inequality for the Euclidean norm gives $\mathrm{\mid }x-y\mathrm{\mid }\le \mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid }<\mathrm{\mid }x\mathrm{\mid }+ϵ<3\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2$; since $\alpha -n<0$, raising the positive numbers to the negative power reverses the inequality and $K_{\alpha }(x-y)=\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha -n}\ge {\left(\frac{3\mathrm{\mid }x\mathrm{\mid }}{2}\right)}^{\alpha -n}={\left(\frac{3}{2}\right)}^{\alpha -n}\mathrm{\mid }x{\mathrm{\mid }}^{\alpha -n}\mathrm{.}$ As $f_{ϵ}\ge 0$ and $\int f_{ϵ}=1$, monotonicity and the scalar rule of [F6] applied to the definition [F1] give $I_{\alpha }{f}_{ϵ}(x)=\int K_{\alpha }(x-y)f_{ϵ}(y)dy\ge {\left(\frac{3}{2}\right)}^{\alpha -n}\mathrm{\mid }x{\mathrm{\mid }}^{\alpha -n}\int f_{ϵ}={\left(\frac{3}{2}\right)}^{\alpha -n}\mathrm{\mid }x{\mathrm{\mid }}^{\alpha -n},$ the pointwise absolute convergence being a consequence of the same finite upper bound since $K_{\alpha }(x-y)\le (\mathrm{\mid }x\mathrm{\mid }-ϵ{)}^{\alpha -n}$ on the support for the upper estimate.

Computation of the tail. The function $x\mapsto \mathrm{\mid }x{\mathrm{\mid }}^{-n}{\mathbf{1}}_{\{\mathrm{\mid }x\mathrm{\mid }>2ϵ\}}$ is nonnegative and Borel, so polar coordinates [F4] give ${\int }_E\mathrm{\mid }x{\mathrm{\mid }}^{-n}d{\lambda }_n(x)=\sigma (S^{n-1}){\int }_{2ϵ}^{\mathrm{\infty }}{r}^{-1}d{\lambda }_1(r)\mathrm{.}$ To see that the radial Lebesgue integral is infinite, set $a=2ϵ$ and $J_j=[2^{j}a,2^{j+1}a)$ for $j\ge 0$. These disjoint intervals partition $[a,\mathrm{\infty })$; on $J_j$, $r^{-1}\ge (2^{j+1}a{)}^{-1}$ and ${\lambda }_1(J_j)=2^{j}a$ by [F5]. Thus each ${\int }_{J_j}{r}^{-1}d{\lambda }_1\ge 1\mathrm{/}2$, using [F5] and [F6]. Finite additivity and monotonicity imply ${\int }_a^{\mathrm{\infty }}{r}^{-1}d{\lambda }_1\ge N\mathrm{/}2$ for every positive integer $N$, so it is infinite. Finally $0<\sigma (S^{n-1})<\mathrm{\infty }$: applying [F4] to ${\mathbf{1}}_{B(0,1)}$ gives $\lambda (B(0,1))=\sigma (S^{n-1})\mathrm{/}n$, and [F3] makes the ball measure positive and finite. Hence ${\int }_E\mathrm{\mid }x{\mathrm{\mid }}^{-n}d{\lambda }_n(x)=+\mathrm{\infty }$.

The far tail diverges. Since $q_0=n\mathrm{/}(n-\alpha )$ we have $(\alpha -n)q_0=-n$, so on the measurable set $E:=\{\mathrm{\mid }x\mathrm{\mid }>2ϵ\}$ the lower bound of step 1.2 gives $\mathrm{\mid }{I}_{\alpha }{f}_{ϵ}(x){\mathrm{\mid }}^{q_0}\ge {\left(\frac{3}{2}\right)}^{(\alpha -n)q_0}\mathrm{\mid }x{\mathrm{\mid }}^{(\alpha -n)q_0}={\left(\frac{3}{2}\right)}^{-n}\mathrm{\mid }x{\mathrm{\mid }}^{-n}$. If $I_{\alpha }{f}_{ϵ}$ belonged to $L^{q_0}$, then applicability of [F6] to the nonnegative functions $\mathrm{\mid }{I}_{\alpha }{f}_{ϵ}{\mathrm{\mid }}^{q_0}{1}_E$ and $\mathrm{\mid }x{\mathrm{\mid }}^{-n}{1}_E$ would give ${\int }_E\mathrm{\mid }x{\mathrm{\mid }}^{-n}dx\le {\left(\frac{3}{2}\right)}^n{\int }_E\mathrm{\mid }{I}_{\alpha }{f}_{ϵ}{\mathrm{\mid }}^{q_0}\le {\left(\frac{3}{2}\right)}^n\mathrm{\parallel }{I}_{\alpha }{f}_{ϵ}{\mathrm{\parallel }}_{q_0}^{q_0}<\mathrm{\infty }\mathrm{.}$

Approximate point mass. Let $\phi \in C_c({\mathbb{R}}^n;\mathbb{C})$ be continuous and compactly supported, and fix $\eta >0$. Continuity of $\phi$ at the origin gives $\delta >0$ with $\mathrm{\mid }\phi (y)-\phi (0)\mathrm{\mid }<\eta$ whenever $\mathrm{\mid }y\mathrm{\mid }<\delta$. For every $0<ϵ<\delta$ linearity of the integral [F7] together with the normalization $\int f_{ϵ}=1$ gives $\int \phi f_{ϵ}d\lambda -\phi (0)={\int }_{B(0,ϵ)}(\phi (y)-\phi (0))f_{ϵ}(y)dy,$ and the triangle inequality [F7] and monotonicity of the nonnegative integral [F6] bound its modulus by $\eta \int f_{ϵ}=\eta$. Hence $\int \phi f_{ϵ}\to \phi (0)$ as $ϵ\to 0^{+}$: the normalized balls converge to the point mass at the origin against continuous compactly supported tests.

No strong endpoint estimate. Steps 2.1 and 1.3 are contradictory: if $I_{\alpha }{f}_{ϵ}\in L^{q_0}$ then ${\int }_E\mathrm{\mid }x{\mathrm{\mid }}^{-n}dx<\mathrm{\infty }$, but that integral equals $+\mathrm{\infty }$. Hence $I_{\alpha }{f}_{ϵ}\mathrm{\notin }{L}^{q_0}({\mathbb{R}}^n)$ for every $ϵ>0$. Since $\mathrm{\parallel }{f}_{ϵ}{\mathrm{\parallel }}_1=1$ by step 1.1, no constant $C$ satisfies $\mathrm{\parallel }{I}_{\alpha }f{\mathrm{\parallel }}_{q_0}\le C\mathrm{\parallel }f{\mathrm{\parallel }}_1$ for all $f$ in $L^1({\mathbb{R}}^n;\mathbb{C})$: the family $\{f_{ϵ}{\}}_{ϵ>0}$ alone refutes the estimate.

Conclusion. The normalized ball density $f_{ϵ}$ has unit $L^1$ norm, is an approximate point mass, and its potential has the radial lower bound ${\left(\frac{3}{2}\right)}^{\alpha -n}\mathrm{\mid }x{\mathrm{\mid }}^{\alpha -n}$ outside $B(0,2ϵ)$, whose $q_0$-th power is a nonzero multiple of the divergent tail $\mathrm{\mid }x{\mathrm{\mid }}^{-n}$; therefore the strong $L^1\to L^{q_0}$ endpoint fails. The argument exhibits the failure at fixed $ϵ$ without any limit or Fatou step, and no endpoint case is silently substituted into the strict-range theorem. Countable Choice is used only through the polar and measure interfaces [F3]-[F5] and [F8].

Measurability and nonnegativity. On the open set $U=\{0<\mathrm{\mid }y\mathrm{\mid }<e^{-1}\}$ the function $y\mapsto \mathrm{\mid }y{\mathrm{\mid }}^{-\alpha }(\log (e\mathrm{/}\mathrm{\mid }y\mathrm{\mid }){)}^{-1}$ is continuous and positive, since $\mathrm{\mid }y\mathrm{\mid }\mapsto \mathrm{\mid }y\mathrm{\mid }$ and the logarithm are continuous and $\log (e\mathrm{/}\mathrm{\mid }y\mathrm{\mid })>0$ for $\mathrm{\mid }y\mathrm{\mid }<e^{-1}$; on the closed complement ${\mathbb{R}}^n\setminus U=\{0\}\cup \{\mathrm{\mid }y\mathrm{\mid }\ge e^{-1}\}$ the function $f$ is the constant $0$. By [F7] the function $f$ is Borel and Lebesgue measurable, nonnegative, and finite-valued, with $f(0)=0$.

Substitution for the radial profiles. The map $T(r)=\log (e\mathrm{/}r)$ is a decreasing $C^1$ diffeomorphism from $(0,e^{-1})$ onto $(2,\mathrm{\infty })$, with $\mathrm{\mid }{T}^{\mathrm{\prime }}(r)\mathrm{\mid }=1\mathrm{/}r$. Applying [F9] to the nonnegative function $u\mapsto u^{-s}$ gives, for every real $s$, the equality of extended nonnegative Lebesgue integrals ${\int }_0^{e^{-1}}{r}^{-1}(\log (e\mathrm{/}r){)}^{-s}dr={\int }_2^{\mathrm{\infty }}{u}^{-s}du\mathrm{.}$ Here $T(e^{-1})=2$ at the limiting endpoint.

Absolute convergence at every nonzero point. Fix $x\mathrm{\ne }0$. Since $(\log (e\mathrm{/}\mathrm{\mid }y\mathrm{\mid }){)}^{-1}<1$ for $0<\mathrm{\mid }y\mathrm{\mid }<e^{-1}$, the function satisfies $\mathrm{\mid }f(y)\mathrm{\mid }\le \mathrm{\mid }y{\mathrm{\mid }}^{-\alpha }$ on its support, so $K_{\alpha }(x-y)\mathrm{\mid }f(y)\mathrm{\mid }\le \mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha -n}\mathrm{\mid }y{\mathrm{\mid }}^{-\alpha }$ there. On the measurable set $\{\mathrm{\mid }x-y\mathrm{\mid }<\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2\}$ one has $\mathrm{\mid }y\mathrm{\mid }>\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2$, so $\mathrm{\mid }y{\mathrm{\mid }}^{-\alpha }<(\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2{)}^{-\alpha }$, and [F9] applied to the change of variables $z=x-y$ together with the polar formula [F3] gives ${\int }_{\mathrm{\mid }x-y\mathrm{\mid }<\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2}\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha -n}\mathrm{\mid }y{\mathrm{\mid }}^{-\alpha }dy\le (\frac{\mathrm{\mid }x\mathrm{\mid }}{2}{)}^{-\alpha }{\int }_{\mathrm{\mid }z\mathrm{\mid }<\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2}\mathrm{\mid }z{\mathrm{\mid }}^{\alpha -n}dz=(\frac{\mathrm{\mid }x\mathrm{\mid }}{2}{)}^{-\alpha }\sigma (S^{n-1}){\int }_0^{\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2}{s}^{\alpha -1}ds<\mathrm{\infty },$ because $\alpha >0$. On the complementary set, $\mathrm{\mid }x-y\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }\mathrm{/}2>0$ gives $\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha -n}\le (\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2{)}^{\alpha -n}$, and [F3] gives ${\int }_{\{\mathrm{\mid }y\mathrm{\mid }<e^{-1}\}}\mathrm{\mid }y{\mathrm{\mid }}^{-\alpha }dy=\sigma (S^{n-1}){\int }_0^{e^{-1}}{s}^{n-1-\alpha }ds<\mathrm{\infty },$ because $n-\alpha >0$. The two regions together cover the support of $f$, so the defining absolute integral of $I_{\alpha }f$ at $x$ is finite: $I_{\alpha }f(x)$ is defined by [F1] at every $x\mathrm{\ne }0$.

Divergence at the origin. At $x=0$ the absolute integrand is $\mathrm{\mid }y{\mathrm{\mid }}^{-n}(\log (e\mathrm{/}\mathrm{\mid }y\mathrm{\mid }){)}^{-1}$ on $0<\mathrm{\mid }y\mathrm{\mid }<e^{-1}$, a nonnegative Borel radial function. Polar coordinates [F3] and Lebesgue change of variables [F9] with $u=\log (e\mathrm{/}r)$ give ${\int }_{{\mathbb{R}}^n}{K}_{\alpha }(-y)\mathrm{\mid }f(y)\mathrm{\mid }d{\lambda }_n(y)=\sigma (S^{n-1}){\int }_2^{\mathrm{\infty }}{u}^{-1}d{\lambda }_1(u)\mathrm{.}$ For $j\ge 1$ and $J_j=[2^j,2^{j+1})$, one has $u^{-1}\ge 2^{-j-1}$ on $J_j$ and ${\lambda }_1(J_j)=2^j$ by [F10], so ${\int }_{J_j}{u}^{-1}d{\lambda }_1\ge 1\mathrm{/}2$. Additivity and monotonicity in [F10] show the integral over $[2,\mathrm{\infty })$ is at least $N\mathrm{/}2$ for every positive integer $N$, hence it is infinite. Since $\sigma (S^{n-1})>0$ by [F3], the defining absolute integral diverges and no finite value of $I_{\alpha }f(0)$ is assigned by [F1].

The function lies in $L^{p_0}$. Since $\alpha p_0=n$, the nonnegative Borel function $\mathrm{\mid }f{\mathrm{\mid }}^{p_0}$ is radial with profile $r^{n-1}{r}^{-\alpha p_0}(\log (e\mathrm{/}r){)}^{-p_0}=r^{-1}(\log (e\mathrm{/}r){)}^{-p_0}$ on $(0,e^{-1})$ and $0$ elsewhere, so [F3] and step 1.2 give ${\int }_{{\mathbb{R}}^n}\mathrm{\mid }f{\mathrm{\mid }}^{p_0}d\lambda =\sigma (S^{n-1}){\int }_0^{e^{-1}}{r}^{-1}(\log (e\mathrm{/}r){)}^{-p_0}dr=\sigma (S^{n-1}){\int }_2^{\mathrm{\infty }}{u}^{-p_0}du\mathrm{.}$ Choose a rational $q$ with $1<q<p_0$, possible because $p_0>1$; then $0\le u^{-p_0}\le u^{-q}$ for $u\ge 1$, and ${\int }_1^{\mathrm{\infty }}{u}^{-q}du$ converges by [F4], so the comparison principle of [F4] makes ${\int }_2^{\mathrm{\infty }}{u}^{-p_0}du$ converge. Its value is finite and $\sigma (S^{n-1})$ is finite by [F3], so $f\in L^{p_0}({\mathbb{R}}^n;\mathbb{C})$.

Lower bound on a punctured annulus. Let $0<\mathrm{\mid }x\mathrm{\mid }<e^{-1}\mathrm{/}2$ and suppose $y$ satisfies $2\mathrm{\mid }x\mathrm{\mid }<\mathrm{\mid }y\mathrm{\mid }<e^{-1}$. Then $\mathrm{\mid }x\mathrm{\mid }<\mathrm{\mid }y\mathrm{\mid }\mathrm{/}2$, so the triangle inequality gives $\mathrm{\mid }x-y\mathrm{\mid }\le \mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid }<3\mathrm{\mid }y\mathrm{\mid }\mathrm{/}2$; because $\alpha -n<0$, raising the positive quantities to the power $\alpha -n$ reverses the inequality and $K_{\alpha }(x-y)\ge (3\mathrm{\mid }y\mathrm{\mid }\mathrm{/}2{)}^{\alpha -n}=(\frac{3}{2}{)}^{\alpha -n}\mathrm{\mid }y{\mathrm{\mid }}^{\alpha -n}\mathrm{.}$ Step 1.3 makes the defining integral at $x$ absolutely convergent, so integrating this lower bound against the nonnegative function $f$ on the measurable annulus $A_x:=\{2\mathrm{\mid }x\mathrm{\mid }<\mathrm{\mid }y\mathrm{\mid }<e^{-1}\}$ and using monotonicity and the scalar rule of [F6] is legitimate and gives $I_{\alpha }f(x)\ge (\frac{3}{2}{)}^{\alpha -n}{\int }_{A_x}\mathrm{\mid }y{\mathrm{\mid }}^{\alpha -n}\mathrm{\mid }f(y)\mathrm{\mid }dy=(\frac{3}{2}{)}^{\alpha -n}{\int }_{A_x}\mathrm{\mid }y{\mathrm{\mid }}^{-n}(\log (e\mathrm{/}\mathrm{\mid }y\mathrm{\mid }){)}^{-1}dy\mathrm{.}$