Young's inequality for convolutions Let's assume that the convolution $f * g$ is continuous with $\lim_{|x| \to \infty}(f*g)(x) = 0$ and that $f, g \in L^2$. Then the following inequality holds
$$ \| f * g \|_{\infty} \leq \| f \|_2 \|g\|_2.$$
Can someone give me a hint in order to prove this? The left side can be written as
$$ \| f * g \|_{\infty} = \max_{x \in \mathbb R} \left| \int_{-\infty}^{\infty} f(x-u)g(u) \operatorname{d} u\right|$$
But know I'm stuck.. I think the Cauchy Schwarz inequality could be used here because what's inside the absolute value looks similar to the inner product.
 A: Let $1 \leqslant p \leqslant +\infty$. For $x\in \mathbb{R}$ [it works with the same argument in $L^p(\mathbb{R}^n)$ for any $n > 0$], let $\tau_x \colon L^p(\mathbb{R}) \to L^p(\mathbb{R})$ be given by
$$\tau_x(f) \colon u \mapsto f(u-x).$$
By the translation-invariance of the Lebesgue measure, $\tau_x$ is an isometry of $L^p(\mathbb{R})$ for every $x$.
Further, let $\rho \colon L^p(\mathbb{R}) \to L^p(\mathbb{R})$ be the reflection, $\rho(f) \colon u \mapsto f(-u)$. Then $\rho$ is also an isometry of $L^p(\mathbb{R})$ since the Lebesgue measure is also invariant under reflection.
Thus for any $f,g\in L^2(\mathbb{R})$ we can write
$$(f\ast g)(x) = \int_\mathbb{R} f(x-u)g(u)\,du = \int_\mathbb{R} \bigl(\rho(\tau_x(f))\bigr)(u)g(u)\,du.$$
For real-valued functions, we then see
$$(f\ast g)(x) = \langle \rho(\tau_x(f)), g\rangle_{L^2},\tag{1}$$
and for complex-valued functions, we need to conjugate one of the arguments of the inner product, and then note that complex conjugation is also an isometry of $L^p(\mathbb{R}^n)$ for all $1\leqslant p \leqslant +\infty$.
Now we can apply the Cauchy-Schwarz inequality to the right hand side of $(1)$ to obtain
$$\lvert (f\ast g)(x)\rvert \leqslant \lVert \rho(\tau_x(f))\rVert_2\cdot \lVert g\rVert_2\tag{2}.$$
But since $\rho$ and $\tau_x$ are isometries, the right hand side of $(2)$ is independent of $x$, and we have
$$\lvert (f\ast g)(x)\rvert \leqslant \lVert f\rVert_2\cdot\lVert g\rVert_2.\tag{3}$$
By $(3)$, the convolution $f\ast g$ is uniformly bounded, and taking the supremum over all $x\in\mathbb{R}$ yields
$$\lVert f\ast g\rVert_\infty = \sup_{x\in\mathbb{R}} \lvert (f\ast g)(x)\rvert \leqslant \lVert f\rVert_2\cdot\lVert g\rVert_2.\tag{4}$$
Since $f\ast g$ is continuous and $(f\ast g)(x) \to 0$ for $\lvert x\rvert \to \infty$ - note that these properties need not be assumed, they can be proved under the assumptions - the supremum in $(4)$ is actually attained, and can be replaced by $\max$ if one so wishes.

Basically the same argument, using the more general Hölder inequality instead of the Cauchy-Schwarz inequality shows that for $1\leqslant p,q\leqslant +\infty$ with $\frac{1}{p} + \frac{1}{q} = 1$ we have
$$\lVert f\ast g\rVert_\infty \leqslant \lVert f\rVert_p\cdot \lVert g\rVert_q$$
for all $f\in L^p(\mathbb{R})$ and $g\in L^q(\mathbb{R})$.
For $1\leqslant p < +\infty$, the continuous functions with compact support are dense in $L^p(\mathbb{R})$, and from that follows that for any fixed $f\in L^p(\mathbb{R})$ the function $x \mapsto \tau_x(f)$ is continuous, which shows that $f\ast g$ is continuous - if $p = \infty$, then $q = 1 < \infty$, and we note that we have
$$(f\ast g)(x) = \int_\mathbb{R} f(u)g(x-u)\,du,$$
so the continuity of $x\mapsto \tau_x(g)$ yields the continuity of $f\ast g$.
For $1 < p,q < +\infty$, we have
$$\lim_{\lvert x\rvert\to\infty} (f\ast g)(x) = 0,$$
as one can again see by using the denseness of $C_c(\mathbb{R})$ in $L^p(\mathbb{R})$ and $L^q(\mathbb{R})$ then, but that need not hold in the $\{ p,q\} = \{1,\infty\}$ case.
