# Supremum norm for convolution in sequence spaces

Question:

Suppose that $$1 \leq p \leq \infty$$, and the convolution $$x \ast y$$ exists. For sequences $$x \in \ell^p(\mathbb{Z})$$ and $$y \in \ell^q(\mathbb{Z})$$, we have

$$||x \ast y ||_{\infty} \leq ||x||_p||y||_q,$$

and $$x \ast y \in \ell^{\infty}$$.

We know $$p=1$$ and $$p=\infty$$ follow by Young's Inequality ($$||x \ast y||_p \leq ||x||_p||y||_1$$).

For $$1 < p < \infty$$, I'm trying take the supremum over what Hölder's inequality gives. We can define the $$n$$-th "convolution" as

$$(x \ast y)_n = \sum_{i=-\infty}^{\infty} x_iy_{n-i},$$

and so Hölder's gives

$$||(x \ast y)_n||_1 \leq ||x||_p\left(\sum_{i=-\infty}^{\infty} |y_{n-i}|^q\right)^{\frac{1}{q}}.$$

Taking the sup over $$n$$,

$$||(x \ast y)_n||_{\infty} \leq ||x||_p \sup_{n \in \mathbb{Z}} \left(\sum_{i = -\infty}^{\infty} |y_{n-i}|^q\right)^{\frac{1}{q}} \stackrel{?}{=} ||x||_p||y||_q.$$

Can I just reindex this in some manner, $$i \mapsto n - i$$? Is this equivalent to the translation invariant that one would do for the convolution in function spaces?

This follows immediately from Hölder's inequality, just note that if $$\mu$$ is the counting measure in $$\mathbb{Z}$$ then
\begin{align*} \|f*g\|_\infty &=\sup_y\left|\int f(x)g(y-x)\mathop{}\!d \mu(x)\right|\\ &\leqslant \sup_y\int|f(x)g(y-x)|\mathop{}\!d \mu(x)\\ &\leqslant\sup_y\|f\|_p\|g(y-\cdot )\|_q\\&=\|f\|_p\|g\|_q \end{align*}