Talk:Young's convolution inequality

Should we consider replacing $\mathbb {R} ^{d}$ with $G$ for any locally compact abelian group $G$? Given the fact that Young's inequality is used not only for $\mathbb {R} ^{d}$ but also $\mathbb {Z} ^{d}$ and sometimes even $\mathbb {T} ^{d}$, I think the theorem would be better formulated with a higher level of generality than it currently is. — Preceding unsigned comment added by 79.68.238.151 (talk) 12:22, 19 August 2017 (UTC)

$C_{0}(\mathbb {R} ^{d})$ and $L^{\infty }(\mathbb {R} ^{d})$

I think it should be mentioned in the article that, let $L_{p}(\mathbb {R} ^{d}):={\begin{cases}L^{p}(\mathbb {R} ^{d}),&1\leq p<\infty \\C_{0}(\mathbb {R} ^{d}),&p=\infty \end{cases}}$ where $C_{0}(\mathbb {R} ^{d})$ is the space of continue functions that vanish at infinity (note that the dual space of $L_{p}(\mathbb {R} ^{d})$ is $L^{p'}(\mathbb {R} ^{d})$, where $p'$ is the dual index of $p$; I believe that $(C_{0}(X))^{*}=L^{1}(X)$ for any $X$ whose measure and topology cooperate well), then we actually have $1\leq p,q,r\leq \infty$, ${\dfrac {1}{p}}+{\dfrac {1}{q}}=1+{\dfrac {1}{r}}$ implying $L_{p}(\mathbb {R} ^{d})*L_{q}(\mathbb {R} ^{d})\subset L_{r}(\mathbb {R} ^{d})$ (see here). For $L^{1}(\mathbb {R} ^{d})$ and $L^{\infty }(\mathbb {R} ^{d})$ we only know that the convolution is continuous and bounded (take $g$ to be constant, for example).

This link may be relevant, but I'm not sure. 129.104.241.214 (talk) 22:11, 15 February 2024 (UTC)

r and its conjugate exponent

It's confusing that the symbol r is used twice, both for an exponent and for its conjugate exponent. In the "duality formulation" (the second in the article), the r that appears there is the conjugate exponent of the r in the first inequality. I propose changing the second r to s. Ulrigo (talk) 14:36, 20 March 2024 (UTC)