Convolution inequality $||f*g||_{L^1}\le||f||_{L^2}||g||_{L^2} $ in $L^2$

Question

I am having a hard time proving that if f,g$\in{L^2}$, then the convolution g*f(x)= $\int_{y\in \mathbb R}$g(y)f(x+y)dy is continuous.

I am stuck trying to show that $||f*g||_{L^1}$$\le||f||_{L^2}||g||_{L^2} $. I am able to show it for $L^1$ all around but not for $L^2$.

Answer 1

This follows from the Cauchy-Schwarz inequality for functions and the fact that the $L_2$ norm is invariant under translations.

Answer 2

This is a special case for Young's inequality.

Theorem (Young's) Suppose $1 \leq p, q, r \leq \infty$, and $\frac{1}{r}+1 = \frac{1}{p} + \frac{1}{q}$. If $f \in L^p$ and $g \in L^q$, then $f * g \in L^r$ and $\|f * g\|_r \leq \|f\|_p \|g\|_q$.

So your result would be taking $r = 1$, $p = q = 2$. You could go for a direct proof by splitting the exponents, but I think a faster proof would be using Riesz-Thorin interpolation.

The case where $r = \infty$ follows immediately from Holder's inequality. When $r = q$ and $p = 1$, by Minkowski's inequality for integrals,

$$ \|f * g\|_q = \left[ \int \big| \int f(y)g(x-y)dy \big|^q dx \right]^{1/q} \leq \int \left[\int |f(y)g(x-y)|^q dx \right]^{1/q} dy = \|f\|_1 \|g\|_q.$$

The remaining is to apply Riesz-Throin. Let $g \in L^q$ be fixed and define $Tf := f * g$. Then $Tf$ is strong type $(p,1)$ and $( \frac{q}{q-1},\infty)$. Now let $q = 2$ and interpolating yields the result.