Hölder inequality with Lebesgue measure

Question

Let $(\Omega, \mathcal A, \mu)$ be a measure space.

Let $p, q, r \in [1, \infty]$ with $1/p + 1/q + 1/r = 2$. Let $f \in L^p(\mathbb R^n), g \in L^q(\mathbb R^n), h \in L^r(\mathbb R^n)$.

Why is then: $$\int_{\mathbb R^n \times \mathbb R^n} | f(x) g(y-x) h(y) | d \lambda_{2n} (x, y) \leq \|f\|_p \|g\|_q \|h\|_r$$

Answer 1

I give a sketch of proof for the case $1/p + 1/q + 1/r = 1$

Without loss of generality you can assume $\|f\|_p=1$,$\|g\|_q=1$ and $\|h\|_r=1$. Now you can use the fact that for non negative $a$, $b$ and $c$, you have $abc \leq \frac{a^p}{p}+\frac{b^q}{q}+\frac{c^r}{r}$ to prove it.

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for non negative $a$, $b$ and $c$ and $1/p + 1/q + 1/r = 1$, we have that, by convexity of log, \begin{align*} \log\left( \frac{a^p}{p}+\frac{b^q}{q}+\frac{c^r}{r} \right)&\geq \frac{1}{p} \log(a^p)+\frac{1}{q}\log(b^q)+\frac{1}{r}\log(c^r)\\ &= \log(a)+\log(b)+\log(c)\\ &= \log(abc) \end{align*} which proves that $abc\leq \frac{a^p}{p}+\frac{b^q}{q}+\frac{c^r}{r}$