Near and far bounds for a Riesz potential

Local integrability. Let $f$ be a measurable representative of the class with $N_p(f)=\mathrm{\parallel }f{\mathrm{\parallel }}_p<\mathrm{\infty }$, and let $B(x,r)$ be any Euclidean ball. By [F4] with the conjugate pair $p,p^{\mathrm{\prime }}$, whose second exponent is finite because $p>1$, ${\int }_{B(x,r)}\mathrm{\mid }f\mathrm{\mid }\le \mathrm{\parallel }f\cdot 1_{B(x,r)}{\mathrm{\parallel }}_p\mathrm{\parallel }{1}_{B(x,r)}{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\le \mathrm{\parallel }f{\mathrm{\parallel }}_p\lambda (B(x,r){)}^{1\mathrm{/}{p}^{\mathrm{\prime }}}<\mathrm{\infty },$ the last inequality using $\lambda (B(x,r))<\mathrm{\infty }$ from [F5]. Hence every measurable representative is locally integrable, and [F3] defines $Mf(x)\in [0,\mathrm{\infty }]$ at every $x$.

Ball bounds. For every $x$ and $r>0$, the inclusion $B(x,r)\subseteq x+(-r,r{)}^n$ holds by [F12], and the open cube is Lebesgue measurable with measure $(2r{)}^n$ by [F5]; monotonicity in [F5] gives $\lambda (B(x,r))\le (2r{)}^n\mathrm{.}$ Combining with the averaging inequality of [F3], for every $r>0$, ${\int }_{B(x,r)}\mathrm{\mid }f\mathrm{\mid }\le \lambda (B(x,r))Mf(x),\lambda (B(x,r){)}^{-1}{\int }_{B(x,r)}\mathrm{\mid }f\mathrm{\mid }\le Mf(x)\mathrm{.}$

Measurability of the integrands. Fix $x$. The map $y\mapsto x-y$ is continuous, hence Borel measurable, and $K_{\alpha }$ is Borel measurable because it is continuous off the origin and takes the finite value $0$ there; by the composition clause of [F8] the map $y\mapsto K_{\alpha }(x-y)$ is Borel measurable, hence Lebesgue measurable by the Borel-containment clause of [F8]. The modulus $\mathrm{\mid }f\mathrm{\mid }$ and the components of $f$ are measurable by [F2], so the product clause of [F8] makes $y\mapsto K_{\alpha }(x-y)\mathrm{\mid }f(y)\mathrm{\mid }$ and each component of $y\mapsto K_{\alpha }(x-y)f(y)$ Lebesgue measurable.

Far kernel integral. Fix $x$ and $R>0$, and define $h_R(z):=\mathrm{\mid }z{\mathrm{\mid }}^{(\alpha -n)p^{\mathrm{\prime }}}{1}_{\{\mathrm{\mid }z\mathrm{\mid }\ge R\}}$. The function $z\mapsto \mathrm{\mid }z\mathrm{\mid }$ is continuous and $t\mapsto t^{(\alpha -n)p^{\mathrm{\prime }}}{1}_{\{t\ge R\}}$ is Borel, so $h_R$ is Borel, nonnegative and hence Lebesgue measurable by [F8]. The affine map $T(y):=y-x$ has $T(T^{-1})$ equal to the identity with inverse $y\mapsto y+x$, identity derivative and determinant $1$, so it is a $C^1$ diffeomorphism of ${\mathbb{R}}^n$ by [F6]; applying the change-of-variables formula of [F6] to $h_R$ gives ${\int }_{{\mathbb{R}}^n}\mathrm{\mid }y-x{\mathrm{\mid }}^{(\alpha -n)p^{\mathrm{\prime }}}{1}_{\{\mathrm{\mid }y-x\mathrm{\mid }\ge R\}}dy={\int }_{{\mathbb{R}}^n}{h}_R(z)dz\mathrm{.}$ The right-hand integral is radial and $h_R$ is Borel, so the polar formula [F7] computes it as $\sigma (S^{n-1}){\int }_R^{\mathrm{\infty }}{r}^{(\alpha -n)p^{\mathrm{\prime }}+n-1}dr=\sigma (S^{n-1})\frac{R^{(\alpha -n)p^{\mathrm{\prime }}+n}}{(n-\alpha )p^{\mathrm{\prime }}-n},$ where the antiderivative is evaluated at the convergent upper end because the exponent $(\alpha -n)p^{\mathrm{\prime }}+n-1<-1$; that inequality is equivalent to $(n-\alpha )p^{\mathrm{\prime }}>n$, which in turn is equivalent to $p<n\mathrm{/}\alpha$, the hypothesis.

Near shell estimate. Fix $x$ with $Mf(x)<\mathrm{\infty }$ and $R>0$. For $j\ge 0$ put $S_j:=\{y:2^{-j-1}R\le \mathrm{\mid }x-y\mathrm{\mid }<2^{-j}R\}\mathrm{.}$ The sets $S_j$ are pairwise disjoint Lebesgue measurable sets with union $B(x,R)\setminus \{x\}$: a point $y$ lies in exactly one shell according to the dyadic size of $\mathrm{\mid }x-y\mathrm{\mid }\in (0,R)$, and the excluded point is exactly $x$. On $S_j$ one has $\mathrm{\mid }x-y\mathrm{\mid }<2^{-j}R$, so the ball $B(x,2^{-j}R)$ contains $S_j$, and since $\alpha -n<0$ reverses the inequality at the positive lower endpoint $2^{-j-1}R\le \mathrm{\mid }x-y\mathrm{\mid }$, $K_{\alpha }(x-y)=\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha -n}\le (2^{-j-1}R{)}^{\alpha -n}(y\in S_j)\mathrm{.}$ Therefore, by monotonicity and the averaging bound of step 1.2, ${\int }_{S_j}{K}_{\alpha }(x-y)\mathrm{\mid }f(y)\mathrm{\mid }dy\le (2^{-j-1}R{)}^{\alpha -n}{\int }_{B(x,2^{-j}R)}\mathrm{\mid }f\mathrm{\mid }\le (2^{-j-1}R{)}^{\alpha -n}(2\cdot 2^{-j}R{)}^{n}Mf(x)=2^{2n-\alpha }{R}^{\alpha }{2}^{-j\alpha }Mf(x)\mathrm{.}$

Far bound. The $p^{\mathrm{\prime }}$-th root of the value in step 1.4 is $C^{\mathrm{\prime }}{R}^{\alpha -n\mathrm{/}p},C^{\mathrm{\prime }}:=\sigma (S^{n-1}{)}^{1\mathrm{/}{p}^{\mathrm{\prime }}}((n-\alpha )p^{\mathrm{\prime }}-n{)}^{-1\mathrm{/}{p}^{\mathrm{\prime }}},$ using $(\alpha -n)p^{\mathrm{\prime }}+n=(\alpha -n+n\mathrm{/}{p}^{\mathrm{\prime }})p^{\mathrm{\prime }}$ and $\alpha -n+n\mathrm{/}{p}^{\mathrm{\prime }}=\alpha -n\mathrm{/}p$; the constant $C^{\mathrm{\prime }}$ is finite and positive because $\sigma (S^{n-1})$ is finite by [F7] and $(n-\alpha )p^{\mathrm{\prime }}-n>0$. Hence $F_R(x)={\int }_{{\mathbb{R}}^n}\mathrm{\mid }y-x{\mathrm{\mid }}^{\alpha -n}{1}_{\{\mathrm{\mid }y-x\mathrm{\mid }\ge R\}}\mathrm{\mid }f(y)\mathrm{\mid }dy\le \left({\int }_{{\mathbb{R}}^n}\right(\mathrm{\mid }y-x{\mathrm{\mid }}^{\alpha -n}{1}_{\{\mathrm{\mid }y-x\mathrm{\mid }\ge R\}}{)}^{p^{\mathrm{\prime }}}dy{)}^{1\mathrm{/}{p}^{\mathrm{\prime }}}\mathrm{\parallel }f{\mathrm{\parallel }}_p\le C^{\mathrm{\prime }}{R}^{\alpha -n\mathrm{/}p}\mathrm{\parallel }f{\mathrm{\parallel }}_p<\mathrm{\infty },$ where the first inequality is Hölder [F4] applied to the pair $\mathrm{\mid }f\mathrm{\mid }$ and the radial weight, and the weight's exact $L^{p^{\mathrm{\prime }}}$ norm is the quantity computed in step 1.4. The estimate uses no hypothesis on $Mf(x)$, so it holds at every $x$ and every $R>0$, and in particular proves finiteness of $F_R(x)$ everywhere.

Representative independence of the absolute integrals. Let $f$ and $g$ be measurable representatives of the same class, so that $f=g$ almost everywhere, and fix $x$. The two nonnegative measurable integrands $h_f:=K_{\alpha }(x-\cdot )\mathrm{\mid }f\mathrm{\mid }$ and $h_g:=K_{\alpha }(x-\cdot )\mathrm{\mid }g\mathrm{\mid }$ of step 1.3 are equal off the null set $\{f\mathrm{\ne }g\}$; hence $\mathrm{\mid }{h}_f-h_g\mathrm{\mid }=0$ almost everywhere and [F10] gives $\int \mathrm{\mid }{h}_f-h_g\mathrm{\mid }=0$. Since $h_f\le h_g+\mathrm{\mid }{h}_f-h_g\mathrm{\mid }$ and $h_g\le h_f+\mathrm{\mid }{h}_f-h_g\mathrm{\mid }$ pointwise, additivity and monotonicity in [F9] give $\int h_f\le \int h_g+0,\int h_g\le \int h_f+0,$ so the two extended nonnegative integrals are equal; in particular one is finite if and only if the other is. Applying this to the restrictions $\{\mathrm{\mid }x-y\mathrm{\mid }<R\}$ and $\{\mathrm{\mid }x-y\mathrm{\mid }\ge R\}$ (each restriction has the same form with the additional indicator) shows that $N_R(x)$ and $F_R(x)$ are independent of the representative, as functions of $x$.

Near bound and finiteness. Since $K_{\alpha }(0)=0$ by [F1], the integrand $K_{\alpha }(x-\cdot )\mathrm{\mid }f\mathrm{\mid }\cdot 1_{\{x\}}$ vanishes identically, so the pointwise identity $K_{\alpha }(x-y)\mathrm{\mid }f(y)\mathrm{\mid }{1}_{B(x,R)}(y)={\sum }_{j\ge 0}{K}_{\alpha }(x-y)\mathrm{\mid }f(y)\mathrm{\mid }{1}_{S_j}(y)$ holds; the partial sums increase to the left-hand side, so [F9] (additivity followed by monotone convergence) gives $N_R(x)={\int }_{B(x,R)}{K}_{\alpha }(x-\cdot )\mathrm{\mid }f\mathrm{\mid }={\sum }_{j\ge 0}{\int }_{S_j}{K}_{\alpha }(x-\cdot )\mathrm{\mid }f\mathrm{\mid }\le 2^{2n-\alpha }{R}^{\alpha }Mf(x){\sum }_{j\ge 0}{2}^{-j\alpha }=\frac{2^{2n-\alpha }}{1-2^{-\alpha }}{R}^{\alpha }Mf(x)<\mathrm{\infty },$ because $\alpha >0$ makes the geometric series converge and $Mf(x)<\mathrm{\infty }$. This proves the near bound and the finiteness of $N_R(x)$ with $C_{n,\alpha }=2^{2n-\alpha }\mathrm{/}(1-2^{-\alpha })$, a constant depending only on $n$ and $\alpha$.

Representative independence of the complex integrals and of the potential. Keep the notation of step 2.3 and suppose now that the common absolute integral is finite at $x$. Then the complex functions $K_{\alpha }(x-\cdot )f$ and $K_{\alpha }(x-\cdot )g$ are both integrable, and they agree almost everywhere, so [F10] applied to their real and imaginary parts gives $\int K_{\alpha }(x-\cdot )f=\int K_{\alpha }(x-\cdot )g$. Consequently the convergence set of the defining integral and its value at every convergent point depend only on the class $f$, and if $N_R(x)$ and $F_R(x)$ are both finite, then additivity in [F9] applied on the complementary measurable sets $B(x,R)$ and its complement gives $\int K_{\alpha }(x-\cdot )\mathrm{\mid }f\mathrm{\mid }=N_R(x)+F_R(x)<\mathrm{\infty },$ so $I_{\alpha }f(x)$ is defined by [F1] and is representative-independent.

Conclusion. The local-integrability assertion is step 1.1; the finiteness and the bound for $N_R$ are step 3.1; the finiteness and the bound for $F_R$ at every point are step 2.2; and representative independence of the convergent integrals and of the total potential is steps 2.3 and 3.2. Countable Choice is used exactly through the maximal-function interface [F3], the ball, box, change-of-variables and polar interfaces [F5]-[F7], the measurability interfaces [F8] and the integral-lattice facts [F9]-[F10], all of which are stated under Countable Choice; no full Axiom of Choice is used. No estimate is asserted at a point with $Mf(x)=\mathrm{\infty }$, and no endpoint case $p=1$ or $p=n\mathrm{/}\alpha$ is claimed.