Dilation determines the Riesz-potential target exponent

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].