The critical Riesz potential can diverge and be essentially unbounded

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 $I_{\alpha }f(x)\ge (\frac{3}{2}{)}^{\alpha -n}{\int }_{A_x}\mathrm{\mid }y{\mathrm{\mid }}^{\alpha -n}\mathrm{\mid }f(y)\mathrm{\mid }dy=(\frac{3}{2}{)}^{\alpha -n}{\int }_{A_x}\mathrm{\mid }y{\mathrm{\mid }}^{-n}(\log (e\mathrm{/}\mathrm{\mid }y\mathrm{\mid }){)}^{-1}dy\mathrm{.}$

Evaluation of the annular integral. The integrand in step 2.2 is a nonnegative Borel radial function equal to $r^{-1}(\log (e\mathrm{/}r){)}^{-1}$ in the radial variable, so [F3] and step 1.2 give ${\int }_{A_x}\mathrm{\mid }y{\mathrm{\mid }}^{-n}(\log (e\mathrm{/}\mathrm{\mid }y\mathrm{\mid }){)}^{-1}dy=\sigma (S^{n-1}){\int }_{2\mathrm{\mid }x\mathrm{\mid }}^{e^{-1}}{r}^{-1}(\log (e\mathrm{/}r){)}^{-1}dr=\sigma (S^{n-1}){\int }_2^{\log (e\mathrm{/}(2\mathrm{\mid }x\mathrm{\mid }))}{u}^{-1}du\mathrm{.}$ The upper limit exceeds $2$ because $\mathrm{\mid }x\mathrm{\mid }<e^{-1}\mathrm{/}2$, so [F5] evaluates the last integral as $\log (\log (e\mathrm{/}(2\mathrm{\mid }x\mathrm{\mid })))-\log 2$, and the lower bound of step 2.2 reads $I_{\alpha }f(x)\ge (\frac{3}{2}{)}^{\alpha -n}\sigma (S^{n-1})[\log (\log \frac{e}{2\mathrm{\mid }x\mathrm{\mid }})-\log 2]\mathrm{.}$

Essential unboundedness. Fix $T>0$. Since $\log \log (e\mathrm{/}(2r))\to +\mathrm{\infty }$ as $r\to 0^{+}$ and ${\left(\frac{3}{2}\right)}^{\alpha -n}\sigma (S^{n-1})>0$ by [F3], there is $\delta \in (0,e^{-1}\mathrm{/}2)$ with ${\left(\frac{3}{2}\right)}^{\alpha -n}\sigma (S^{n-1})[\log \log (e\mathrm{/}(2\delta ))-\log 2]>T$. Step 3.1 then gives $I_{\alpha }f(x)>T$ for every $x$ with $0<\mathrm{\mid }x\mathrm{\mid }<\delta$, so the superlevel set $\{\mathrm{\mid }{I}_{\alpha }f\mathrm{\mid }>T\}$ contains the punctured ball $B(0,\delta )\setminus \{0\}$. That punctured ball contains the annulus $\{\delta \mathrm{/}2<\mathrm{\mid }x\mathrm{\mid }<\delta \}$, a Borel set whose measure $\sigma (S^{n-1}){\int }_{\delta \mathrm{/}2}^{\delta }{s}^{n-1}ds>0$ is positive by [F3]; at $x=0$ the defining absolute integral is $+\mathrm{\infty }$ by step 1.4, so $I_{\alpha }f(0)$ is not assigned a finite value. Hence $\{\mathrm{\mid }{I}_{\alpha }f\mathrm{\mid }>T\}$ is not a null set for any $T$, and no constant $T$ can bound $\mathrm{\mid }{I}_{\alpha }f\mathrm{\mid }$ almost everywhere.

Conclusion. Steps 2.1, 1.4 and 4.1 exhibit a function $f\in L^{p_0}({\mathbb{R}}^n;\mathbb{C})$ whose Riesz integral diverges at the origin and whose finite values are essentially unbounded on every neighbourhood of the origin, so the critical case $p_0=n\mathrm{/}\alpha$ admits neither a finite raw potential at every point nor a bounded $L^{p_0}\to L^{\mathrm{\infty }}$ estimate. This shows the necessity of the strict range $1<p<n\mathrm{/}\alpha$ in the strong theorem of this pair. Countable Choice is used only through the polar, measure, and Lebesgue change-of-variables interfaces [F3]-[F4] and [F8]-[F10]; no other choice principle is invoked.