Young's convolution inequality under Countable Choice

Under the given Countable Choice premise, choose finite-valued Borel representatives $\tilde{f},\tilde{g}$ using [L1], applying it to real and imaginary parts. Alter values on their Borel null exceptional sets to zero. By [L2], $H(x,y)=\tilde{f}(x-y)\tilde{g}(y)$ and $\mathrm{\mid }H\mathrm{\mid }$ are product-measurable. Translation and reflection preserve the $L^p$ norms of $\tilde{f}(x-\cdot )$ for every $x$, including essential suprema. If either norm is zero, the corresponding representative vanishes almost everywhere; for every $x$, $H(x,\cdot )=0$ almost everywhere by [L5]. Its integral and the claimed inequality are then zero. Hence assume both norms are positive.

Suppose $r<\mathrm{\infty }$. The exponent identity implies $p,q<\mathrm{\infty }$ and $r\ge p,q$. Tonelli and translation invariance give ${\int }_x{\int }_y\mathrm{\mid }\tilde{f}(x-y){\mathrm{\mid }}^p\mathrm{\mid }\tilde{g}(y){\mathrm{\mid }}^{q}dydx=\mathrm{\parallel }f{\mathrm{\parallel }}_p^p\mathrm{\parallel }g{\mathrm{\parallel }}_q^q<\mathrm{\infty }\mathrm{.}$ Therefore $A_x:={\int }_y\mathrm{\mid }\tilde{f}(x-y){\mathrm{\mid }}^p\mathrm{\mid }\tilde{g}(y){\mathrm{\mid }}^{q}dy$ is finite for almost every $x$.

If $r=\mathrm{\infty }$, then $1\mathrm{/}p+1\mathrm{/}q=1$. For each $x$, [L4] applied to $\tilde{f}(x-\cdot )$ and $\tilde{g}$ gives $\int \mathrm{\mid }H(x,y)\mathrm{\mid }dy\le \mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{\parallel }g{\mathrm{\parallel }}_q<\mathrm{\infty }\mathrm{.}$ Thus convolution exists for every $x$ for these representatives, is measurable as an integral of product-measurable sections, and its essential supremum obeys the bound.

Fix such an $x$. Factor $\mathrm{\mid }H\mathrm{\mid }$ as $A(y)B(y)C(y)$ with $A=(\mathrm{\mid }\tilde{f}(x-y){\mathrm{\mid }}^p\mathrm{\mid }\tilde{g}(y){\mathrm{\mid }}^q{)}^{1\mathrm{/}r},B=\mathrm{\mid }\tilde{f}(x-y){\mathrm{\mid }}^{1-p\mathrm{/}r},C=\mathrm{\mid }\tilde{g}(y){\mathrm{\mid }}^{1-q\mathrm{/}r}\mathrm{.}$ If $r=p$ or $r=q$, the corresponding zero-power factor means the constant $1$. The three Holder exponents are $r$, $pr\mathrm{/}(r-p)$ and $qr\mathrm{/}(r-q)$, with $\mathrm{\infty }$ at a zero denominator; their reciprocal sum is $1$. Applying generalized Holder to $BC$ and then Holder to $A(BC)$ (including endpoint forms) yields $\int \mathrm{\mid }H(x,y)\mathrm{\mid }dy\le A_x^{1\mathrm{/}r}\mathrm{\parallel }f{\mathrm{\parallel }}_p^{1-p\mathrm{/}r}\mathrm{\parallel }g{\mathrm{\parallel }}_q^{1-q\mathrm{/}r}\mathrm{.}$ This proves absolute convergence for almost every $x$. Tonelli applied to the positive and negative parts of the real and imaginary components of $H$ makes $x\mapsto \int H(x,y)dy$ measurable on this full-measure domain; define it as zero elsewhere. Raise the inequality to the $r$th power, integrate in $x$, and use step 1.2 to obtain $\mathrm{\parallel }f\ast g{\mathrm{\parallel }}_r^r\le \mathrm{\parallel }f{\mathrm{\parallel }}_p^{r-p}\mathrm{\parallel }g{\mathrm{\parallel }}_q^{r-q}\mathrm{\parallel }f{\mathrm{\parallel }}_p^p\mathrm{\parallel }g{\mathrm{\parallel }}_q^q=\mathrm{\parallel }f{\mathrm{\parallel }}_p^r\mathrm{\parallel }g{\mathrm{\parallel }}_q^r,$ interpreting a zero exponent as a factor $1$.

If another pair of Borel representatives is chosen, each differs from the first on a Borel null set. For every fixed $x$, the first-factor difference is confined to a reflected translate of that null set and the second-factor difference to its original null set. By [L5] the two section products agree almost everywhere in $y$. Both integrals exist on a common full-measure set by the preceding bounds, and [L5] makes them equal there. Thus $f\ast g$ is a well-defined $L^r$ class, and steps 2.1 and 2.2 prove the asserted bound.