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 $$L_p(\mathbb{R}^d):=\begin{cases}L^p(\mathbb{R}^d),&1\le p<\infty\\ C_0(\mathbb{R}^d), &p=\infty\end{cases}$$ 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 $$1\le p,q,r\le\infty$$, $$\dfrac{1}{p}+\dfrac{1}{q}=1+\dfrac{1}{r}$$ 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)