Hardy–Littlewood–Sobolev fractional integration inequality

The exponents. Put $\theta :=\alpha p\mathrm{/}n$, which lies in $(0,1)$ because $0<\alpha p<n$. The exponent $q$ of the statement satisfies $1\mathrm{/}q=1\mathrm{/}p-\alpha \mathrm{/}n=(n-\alpha p)\mathrm{/}(pn)=(1-\theta )\mathrm{/}p$, so $q=p\mathrm{/}(1-\theta )$; hence $q<\mathrm{\infty }$, $q>p$, and $(1-\theta )q=p$, while $\theta q=\alpha pq\mathrm{/}n=q-p$.

Joint product measurability. Identify ${\mathbb{R}}^n\times {\mathbb{R}}^n$ with ${\mathbb{R}}^{2n}$; by [F9] its Borel sigma-algebra is $\mathcal{B}({\mathbb{R}}^n)\otimes \mathcal{B}({\mathbb{R}}^n)$, which is contained in $\mathcal{L}({\mathbb{R}}^n)\otimes \mathcal{L}({\mathbb{R}}^n)$. The difference map $(x,y)\mapsto x-y$ is continuous, hence Borel, so its composition with the Borel function $K_{\alpha }$ is Borel by [F9] and therefore product measurable. The second coordinate projection is measurable into the Lebesgue sigma-algebra since the inverse image of every Lebesgue set $A$ is ${\mathbb{R}}^n\times A$; composing it with measurable $f$ makes $(x,y)\mapsto f(y)$ product measurable. The lattice and product clauses of [F9] then make $(x,y)\mapsto \mathrm{\mid }f(y)\mathrm{\mid }$ and $(x,y)\mapsto (\mathrm{Re}f{)}^{\pm }(y)$, $(\mathrm{Im}f{)}^{\pm }(y)$ product measurable. Products of measurable finite-valued functions are measurable by [F9], so the five functions $(x,y)\mapsto K_{\alpha }(x-y)\mathrm{\mid }f(y)\mathrm{\mid }$ and $K_{\alpha }(x-y)(\mathrm{Re}f{)}^{\pm }(y)$, $K_{\alpha }(x-y)(\mathrm{Im}f{)}^{\pm }(y)$ are product measurable and take values in $[0,\mathrm{\infty })$.

The maximal function is finite almost everywhere. The function $\mathrm{\mid }f\mathrm{\mid }$ is real, measurable and lies in $L^p$, so the maximal bound of [F5] applied to $\mathrm{\mid }f\mathrm{\mid }$ gives $\mathrm{\parallel }Mf{\mathrm{\parallel }}_p=\mathrm{\parallel }M(\mathrm{\mid }f\mathrm{\mid }){\mathrm{\parallel }}_p\le C_{n,p}\mathrm{\parallel }\mathrm{\mid }f\mathrm{\mid }{\mathrm{\parallel }}_p=C_{n,p}\mathrm{\parallel }f{\mathrm{\parallel }}_p<\mathrm{\infty }$. The power $(Mf{)}^p$ is nonnegative and measurable by the threshold criterion [F7]: for $a<0$ its superlevel set is all of ${\mathbb{R}}^n$, and for $a\ge 0$ it equals $\{Mf>a^{1\mathrm{/}p}\}$. Applying [F6] to $(Mf{)}^p$ gives $(Mf{)}^p<\mathrm{\infty }$ almost everywhere, that is, $Mf<\mathrm{\infty }$ almost everywhere.

Tonelli and measurability of the integral. By [F8] the measure space $({\mathbb{R}}^n,\mathcal{L}({\mathbb{R}}^n),{\lambda }_n)$ is sigma-finite, so Tonelli applies to each product-measurable function of step 1.2: the functions $G(x):=\int K_{\alpha }(x-y)\mathrm{\mid }f(y)\mathrm{\mid }dy,G_1(x):=\int K_{\alpha }(x-y)(\mathrm{Re}f{)}^{+}(y)dy,\dots ,G_4(x):=\int K_{\alpha }(x-y)(\mathrm{Im}f{)}^{-}(y)dy$ are measurable $[0,\mathrm{\infty }]$-valued functions of $x$. The set $E:=\{G<\mathrm{\infty }\}={\bigcup }_{m\ge 1}\{G\le m\}$ is measurable, and $0\le G_j\le G$ on all of ${\mathbb{R}}^n$ for $j=1,\dots ,4$; hence on $E$ the four numbers $G_j(x)$ are finite and $h(x):=(G_1(x)-G_2(x))+i(G_3(x)-G_4(x))$ is a well-defined complex number. Since each $G_j$ is measurable and $E$ is measurable, the product ${\tilde{I}}_{\alpha }f:=h\cdot 1_E$ (with value $0$ off $E$) is a measurable complex-valued function; and at every $x\in E$ it equals $\int K_{\alpha }(x-y)f(y)dy$ by componentwise integration [F2] and the definition [F1], because $K_{\alpha }(x-\cdot )\mathrm{\mid }f\mathrm{\mid }$ has finite integral there.

Almost-everywhere absolute convergence and the pointwise bound. By [F3] every representative of $f$ is locally integrable, so $Mf$ is defined everywhere; by the Hedberg inequality [F4], at every $x$ with $Mf(x)<\mathrm{\infty }$ the defining integral converges absolutely and $\mid I_{\alpha }f(x)\mid \le C_{n,\alpha ,p}^H(Mf(x){)}^{1-\theta }\mathrm{\parallel }f{\mathrm{\parallel }}_p^{\theta }$, where $C_{n,\alpha ,p}^H$ denotes the constant of [F4], renamed to avoid a clash with the constant claimed in the Statement. Let $S:=\{Mf<\mathrm{\infty }\}$; step 1.3 makes $S$ conull, and $S\subseteq E$ by the definition of $E$ in step 2.1. Hence ${\tilde{I}}_{\alpha }f=h\cdot 1_E$ agrees with the pointwise potential $I_{\alpha }f$ of [F1] at every point of $S$ and differs from it only on the null set ${\mathbb{R}}^n\setminus S$; in particular the defining integral converges absolutely almost everywhere, and ${\tilde{I}}_{\alpha }f$ is a measurable representative of the almost-everywhere defined integral. At every point of $S$ the displayed inequality is exactly the Hedberg bound. At a point with $Mf(x)=+\mathrm{\infty }$: if $\mathrm{\parallel }f{\mathrm{\parallel }}_p>0$ then the right-hand side is $+\mathrm{\infty }$, because $1-\theta \in (0,1)$ and $\theta \in (0,1)$ are applied to $+\mathrm{\infty }$ and to the strictly positive number $\mathrm{\parallel }f{\mathrm{\parallel }}_p$, so the inequality holds trivially against the finite value $\mathrm{\mid }{\tilde{I}}_{\alpha }f(x)\mathrm{\mid }$; and if $\mathrm{\parallel }f{\mathrm{\parallel }}_p=0$ then $f$ is the zero class with $Mf=0$ everywhere by [F4], so the case $Mf(x)=+\mathrm{\infty }$ cannot occur. In every case the pointwise bound holds at every $x\in {\mathbb{R}}^n$ in the form $\mathrm{\mid }{\tilde{I}}_{\alpha }f(x)\mathrm{\mid }\le C_{n,\alpha ,p}^H(Mf(x){)}^{1-\theta }\mathrm{\parallel }f{\mathrm{\parallel }}_p^{\theta }\mathrm{.}$

The $L^q$ estimate. Raising the bound of step 3.1 to the $q$-th power and using $(1-\theta )q=p$ from step 1.1 gives, at every $x$, $\mathrm{\mid }{\tilde{I}}_{\alpha }f(x){\mathrm{\mid }}^q\le (C_{n,\alpha ,p}^H{)}^q(Mf(x){)}^p\mathrm{\parallel }f{\mathrm{\parallel }}_p^{\theta q},$ an inequality between nonnegative measurable functions. Monotonicity of the integral [F10], the identity $\mathrm{\parallel }Mf{\mathrm{\parallel }}_p^p=\int (Mf{)}^p$, and the maximal bound of step 1.3 give $\mathrm{\parallel }{\tilde{I}}_{\alpha }f{\mathrm{\parallel }}_q^q=\int \mathrm{\mid }{\tilde{I}}_{\alpha }f{\mathrm{\mid }}^q\le (C_{n,\alpha ,p}^H{)}^q\mathrm{\parallel }f{\mathrm{\parallel }}_p^{\theta q}\mathrm{\parallel }Mf{\mathrm{\parallel }}_p^p\le (C_{n,\alpha ,p}^H{)}^q{C}_{n,p}^p\mathrm{\parallel }f{\mathrm{\parallel }}_p^{\theta q+p}=(C_{n,\alpha ,p}^H{)}^q{C}_{n,p}^p\mathrm{\parallel }f{\mathrm{\parallel }}_p^q,$ where $\theta q+p=q$ by step 1.1. Hence $\mathrm{\parallel }{\tilde{I}}_{\alpha }f{\mathrm{\parallel }}_q\le C_{n,\alpha ,p}\mathrm{\parallel }f{\mathrm{\parallel }}_p$ with $C_{n,\alpha ,p}:=C_{n,\alpha ,p}^H{C}_{n,p}^{p\mathrm{/}q}=C_{n,\alpha ,p}^H{C}_{n,p}^{1-\theta }<\mathrm{\infty }$, a constant depending only on $n,\alpha ,p$.

Representative independence. Let $f$ and $g$ be measurable representatives of the same class, so that $f=g$ almost everywhere; then $\mathrm{\mid }f\mathrm{\mid }=\mathrm{\mid }g\mathrm{\mid }$ almost everywhere, and both $\mathrm{\mid }f\mathrm{\mid }{1}_B$ and $\mathrm{\mid }g\mathrm{\mid }{1}_B$ are integrable for every ball $B$ because the representatives are locally integrable by [F3] and balls have finite measure, so the almost-everywhere-equality clause of [F10] gives ${\int }_B\mathrm{\mid }f\mathrm{\mid }={\int }_B\mathrm{\mid }g\mathrm{\mid }$ for every ball $B$. Hence $Mf=Mg$ as extended-real functions and the two conull sets coincide: writing $S:=\{Mf<\mathrm{\infty }\}=\{Mg<\mathrm{\infty }\}$, step 1.3 makes $S$ conull. At every $x\in S$ and every $R>0$ the splitting lemma [F3] gives that $N_R$ and $F_R$ are finite for each of the two representatives, and its representative-independence clause gives that the total potential is defined at $x$ and takes the same value for $f$ and for $g$; thus $I_{\alpha }f(x)=I_{\alpha }g(x)$ for every $x\in S$. By step 3.1 both ${\tilde{I}}_{\alpha }f$ and ${\tilde{I}}_{\alpha }g$ agree with these potentials at every point of $S$; since $S$ is conull, the two measurable functions of step 2.1 define the same class in $L^q$. Therefore the class $[{\tilde{I}}_{\alpha }f]\in L^q$ depends only on the class $[f]\in L^p$, and step 4.1 gives the bound $\mathrm{\parallel }{\tilde{I}}_{\alpha }[f]{\mathrm{\parallel }}_q\le C_{n,\alpha ,p}\mathrm{\parallel }[f]{\mathrm{\parallel }}_p$ for this well-defined assignment.

Linearity. Let $f,g$ be measurable representatives of classes in $L^p$ and let $c\in \mathbb{C}$; then $f+g$ and $cf$ are measurable representatives of the corresponding classes. At every point $x$ of $S_f\cap S_g\cap S_{f+g}\cap S_{cf}$, all of the complex functions $K_{\alpha }(x-\cdot )f$, $K_{\alpha }(x-\cdot )g$, $K_{\alpha }(x-\cdot )(f+g)$ and $K_{\alpha }(x-\cdot )(cf)$ are integrable, and the linearity of the Lebesgue integral on $L^1$ [F10] gives $\int K_{\alpha }(x-y)(f+g)(y)dy=\int K_{\alpha }(x-y)f(y)dy+\int K_{\alpha }(x-y)g(y)dy$ and $\int K_{\alpha }(x-y)(cf)(y)dy=c\int K_{\alpha }(x-y)f(y)dy$. The four sets $S_f,S_g,S_{f+g},S_{cf}$ are conull by step 1.3 and their intersection is conull by the null-union clause of [F10]; on that intersection, where each ${\tilde{I}}_{\alpha }$ is the corresponding integral by step 3.1, the a.e.-equal functions ${\tilde{I}}_{\alpha }(f+g)$ and ${\tilde{I}}_{\alpha }f+{\tilde{I}}_{\alpha }g$ define the same class in $L^q$, and likewise ${\tilde{I}}_{\alpha }(cf)$ and $c{\tilde{I}}_{\alpha }f$. So the assignment of step 5.1 is complex-linear.

The smooth core. Let $g\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n;\mathbb{C})$. Then $\mathrm{\mid }g\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$ everywhere, so every ball average of $\mathrm{\mid }g\mathrm{\mid }$ is at most $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $Mg(x)\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $x$; also $g\in L^p$ because it is bounded and supported in a bounded measurable set of finite measure by [F8]. By the near/far bounds [F3] the near and far integrals of $g$ are finite at every $x$, so the defining absolute integral is finite everywhere and [F1] defines $I_{\alpha }g$ everywhere as the pointwise integral $x\mapsto \int K_{\alpha }(x-y)g(y)dy$. Hence $S={\mathbb{R}}^n$ and $E={\mathbb{R}}^n$, and ${\tilde{I}}_{\alpha }g$ equals this pointwise integral at every point. Let $D\subseteq L^p({\mathbb{R}}^n;\mathbb{C})$ be the image of $C_c^{\mathrm{\infty }}({\mathbb{R}}^n;\mathbb{C})$; the assignment of steps 4.1, 5.1 and 6.1 restricts to a bounded linear map $T:D\to L^q$ which is exactly the pointwise-integral map, with $\mathrm{\parallel }T\mathrm{\parallel }$ at most the constant of step 4.1.

Density and the abstract extension. By [F11] the subspace $D$ is dense in the normed space $L^p$ and $L^q$ is complete, hence a Banach space; the map $T$ of step 7.1 is bounded and linear. The extension theorem [F12] therefore produces a unique bounded linear map $\tilde{T}:L^p({\mathbb{R}}^n;\mathbb{C})\to L^q({\mathbb{R}}^n;\mathbb{C})$ with $\tilde{T}{\mathrm{\mid }}_D=T$ and $\mathrm{\parallel }\tilde{T}\mathrm{\parallel }=\mathrm{\parallel }T\mathrm{\parallel }$.

Identification of the extension with the integral operator. Let $\mathcal{I}:L^p\to L^q$ denote the bounded linear almost-everywhere integral map of steps 4.1, 5.1 and 6.1. Let $x\in L^p$. By density [F11] there are $d_k\in D$ with $\mathrm{\parallel }{d}_k-x{\mathrm{\parallel }}_p\to 0$. Both $\mathcal{I}$ and $\tilde{T}$ are bounded linear, hence continuous on the normed space $L^p$ by [F12], and they agree on $D$ because $\mathcal{I}{\mathrm{\mid }}_D=T=\tilde{T}{\mathrm{\mid }}_D$ by steps 7.1 and 8.1. Therefore $\mathcal{I}(x)={\lim }_k\mathcal{I}(d_k)={\lim }_{k}T(d_k)={\lim }_k\tilde{T}(d_k)=\tilde{T}(x),$ the two outer equalities by continuity and the middle one because $d_k\in D$; limits in the normed space $L^q$ are unique by [F12]. Hence the unique bounded dense-core extension of the pointwise-integral map on $C_c^{\mathrm{\infty }}$ is precisely the almost-everywhere integral operator $\mathcal{I}$, and it satisfies the bound $\mathrm{\parallel }\mathcal{I}(f){\mathrm{\parallel }}_q\le C_{n,\alpha ,p}\mathrm{\parallel }f{\mathrm{\parallel }}_p$ of step 4.1.

Conclusion. For $1<p<n\mathrm{/}\alpha$ and the exponent $q$ of the statement, steps 1.1, 4.1, 5.1 and 6.1 prove that the defining integral of $I_{\alpha }f$ converges absolutely almost everywhere for every $f\in L^p$ and that its class obeys the $L^q$ bound with a constant depending only on $n,\alpha ,p$, and that this gives a well-defined bounded linear map on the quotient classes; and steps 7.1, 8.1 and 9.1 prove that on the dense smooth core the map is the pointwise integral and that its unique bounded dense-core extension is this same almost-everywhere integral operator. Countable Choice is spent exactly through the maximal-function, Tonelli and sigma-finiteness, density, completeness and extension interfaces [F5], [F8], [F11], [F12]; no full Axiom of Choice is invoked.