Hedberg pointwise inequality for Riesz potentials

The degenerate case. Suppose first that $\mathrm{\parallel }f{\mathrm{\parallel }}_p=0$. Then $f=0$ almost everywhere by the definiteness clause of [F2]; for every $x$ the nonnegative integrand $K_{\alpha }(x-\cdot )\mathrm{\mid }f\mathrm{\mid }$ vanishes almost everywhere, so [F6] gives $\int K_{\alpha }(x-y)\mathrm{\mid }f(y)\mathrm{\mid }dy=0<\mathrm{\infty }$, the potential is defined at $x$ by [F1] and $I_{\alpha }f(x)=0$. Also every ball average in [F3] is the integral of a function vanishing almost everywhere, hence is $0$, so $Mf(x)=0$, and the asserted inequality reads $0\le C_{n,\alpha ,p}\cdot 0^{1-\theta }\cdot 0^{\theta }=0$.

The case $Mf(x)=0$. If $Mf(x)=0$, then every ball average of $\mathrm{\mid }f\mathrm{\mid }$ is at most $0$, so ${\int }_{B(x,k)}\mathrm{\mid }f\mathrm{\mid }=0$ for every integer $k\ge 1$; the nonnegative function $\mathrm{\mid }f\mathrm{\mid }$ therefore vanishes almost everywhere on each ball $B(x,k)$ by [F6], and the balls $B(x,k)$ cover ${\mathbb{R}}^n$, so $\mathrm{\mid }f\mathrm{\mid }=0$ almost everywhere by [F7]. Hence $f$ is the zero class, $\mathrm{\parallel }f{\mathrm{\parallel }}_p=0$, and step 1.1 applies. Thus in the remaining case both $M:=Mf(x)$ and $F:=\mathrm{\parallel }f{\mathrm{\parallel }}_p$ are strictly positive, and both are finite by hypothesis and by [F2].

The balanced estimate. Assume $0<M<\mathrm{\infty }$ and $0<F<\mathrm{\infty }$ and put $R:=(F\mathrm{/}M{)}^{p\mathrm{/}n}>0$. Step 2.1 and [F2] give every representative $f$ locally integrable, so $M$ is defined and the splitting lemma [F4] applies at $x$ with this radius: $I_{\alpha }f(x)$ is defined and $\mid I_{\alpha }f(x)\mid \le \int K_{\alpha }(x-y)\mathrm{\mid }f(y)\mathrm{\mid }dy=N_R(x)+F_R(x)\le C_{n,\alpha }{R}^{\alpha }M+D_{n,\alpha ,p}{R}^{\alpha -n\mathrm{/}p}F,$ where $D_{n,\alpha ,p}$ is the far constant of [F4], renamed here to avoid a clash with the constant defined below, the first inequality is [F5], and the equality of the total integral with the sum of the near and far integrals is additivity in [F6] applied on the complementary sets $B(x,R)$ and $\{y:\mathrm{\mid }x-y\mathrm{\mid }\ge R\}$. Since $\theta =\alpha p\mathrm{/}n$ gives $\alpha =\theta n\mathrm{/}p$ and $\alpha -n\mathrm{/}p=(\theta -1)n\mathrm{/}p$, one has $R^{\alpha }=(F\mathrm{/}M{)}^{\alpha p\mathrm{/}n}=(F\mathrm{/}M{)}^{\theta },R^{\alpha -n\mathrm{/}p}=(F\mathrm{/}M{)}^{(\alpha -n\mathrm{/}p)p\mathrm{/}n}=(F\mathrm{/}M{)}^{\theta -1},$ so $R^{\alpha }M=F^{\theta }{M}^{1-\theta }$ and $R^{\alpha -n\mathrm{/}p}F=F^{\theta }{M}^{1-\theta }$; hence $\mathrm{\mid }{I}_{\alpha }f(x)\mathrm{\mid }\le (C_{n,\alpha }+D_{n,\alpha ,p})M^{1-\theta }{F}^{\theta }$.

Conclusion of the estimate. Setting $C_{n,\alpha ,p}:=C_{n,\alpha }+D_{n,\alpha ,p}$, with $D_{n,\alpha ,p}$ the far constant of [F4], step 3.1 gives the asserted bound in the nondegenerate case; together with steps 1.1 and 2.1 every case is covered, the exponential factors are the exact positive powers $\theta \in (0,1)$ and $1-\theta \in (0,1)$ of finite nonnegative quantities, and no expression $0^0$ or ${\mathrm{\infty }}^0$ occurs. Absolute convergence at $x$ is the finiteness of $N_R(x)+F_R(x)$ from step 3.1.

Choice accounting. The argument uses Countable Choice only through the maximal-function interface [F3] and the splitting lemma [F4], both of which are stated under Countable Choice, and through the measure and integral facts [F6]-[F7] of the Euclidean Lebesgue framework; no full Axiom of Choice and no choice over an uncountable family is invoked. The hypothesis $Mf(x)<\mathrm{\infty }$ is used only at the single point $x$, and the conclusion is pointwise at that point.