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Hello folks,

Suppose we're working in ${\mathbb T}^q$, the $q$-times Cartesian product of $[-\pi,\pi]$ with the endpoints identified. So, $ L^p := L^p({\mathbb T}^q) $.

The Young inequality may be written in the following form: For $p \in [1,\infty]$, $f \in L^p$, $g \in L^1$, we have $f * g \in L^p$ and

$\|f*g\|_p \le \|f\|_p \|g\|_1$.

Could we extend this to an arbitrary measure $\mu$ on ${\mathbb T}^q$ by:

Let $p \in [1,\infty]$, $f \in L^p$, $g \in L^1(\mu)$. Then $ f*_\mu g := \int_{{\mathbb T}^q} f(\circ-t)g(t)d\mu(t) \in L^p$ and $\|f*_\mu g\|_p \le \|f\|_p \|g\|_{\mu;1}$ ?

I'm working on a related problem in approximation theory with a barely-Rudin level understanding of measure theory, so I'm not at the level to prove it myself or find a counterexample, but the estimate (or one like it) is necessary to connect some dots in my head. Any inference as to why it should or shouldn't be true?

Sincere thanks for any help.

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It's not true in general. One of the ways that Young's inequality is motivated (and essentially proved) is by using a scaling argument, which isn't possible for arbitrary measures. – Christopher A. Wong Mar 9 '12 at 16:45

1 Answer 1

The most general version of Young's convolution inequality that I know of is as follows: Let $(X,\mu)$ be a $\sigma$-finite measure space. For $1 \leq p, r \leq \infty$,

$\left\|f \ast g\right\|_{L^{s}} \leq \left\|f\right\|_{L^{r}}\left\|g\right\|_{L^{p}}$

where $\frac{1}{p} + \frac{1}{r} = 1 + \frac{1}{s}$.

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