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?


1 Answer 1


This follows immediately from Hölder's inequality, just note that if $\mu$ is the counting measure in $\mathbb{Z}$ then

$$ \begin{align*} \|f*g\|_\infty &=\sup_y\left|\int f(x)g(y-x)\mathop{}\!d \mu(x)\right|\\ &\leqslant \sup_y\int|f(x)g(y-x)|\mathop{}\!d \mu(x)\\ &\leqslant\sup_y\|f\|_p\|g(y-\cdot )\|_q\\&=\|f\|_p\|g\|_q \end{align*} $$

where in the second inequality we used Hölder's inequality.


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