L2 norm inequality respect to convolution on $S^1$

Question

I am trying to show if $f,g\in C^0(S^1)$, then $$||f\ast g||^2_{L^2(S^1)}\leq ||f\ast f||_{L^2(S^1)}||g\ast g||_{L^2(S^1)}.$$ But nothing comes out in mind, any ideas?

Answer 1

By Young's convolution inequality we know that if $f\in L^2(S^1)$ and $g\in L^2(S^1)$ then $$ ||f*g||_{\infty}\leq||f||_2||g||_2 $$

and $$ ||f*g||_{2}\leq||f*g||_{\infty}\ $$ since $m(S^1)=1$. This tells us that the function $f*g\in L^2$ and we can use the Parseval's identity which gives us

$$ ||f*g||^2_{2}=\displaystyle\sum_{n=-\infty}^{\infty}|\widehat{f*g}(n)|^2. $$

Next we use the fact that if $f,g\in L^1$ then $\widehat{f*g}(n)=\hat{f}(n)\hat{g}(n)$ to get $$ \displaystyle\sum_{n=-\infty}^{\infty}|\widehat{f*g}(n)|^2=\displaystyle\sum_{n=-\infty}^{\infty}|\hat{f}(n)\hat{g}(n)|^2=\displaystyle\sum_{n=-\infty}^{\infty}|\hat{f}(n)|^2|\hat{g}(n)|^2. $$ Now we use the Cauchy-Schwartz inequallity

$$ \displaystyle\sum_{n=-\infty}^{\infty}|\hat{f}(n)|^2|\hat{g}(n)|^2\leq\sqrt{\displaystyle\sum_{n=-\infty}^{\infty}|\hat{f}(n)|^4}\sqrt{\displaystyle\sum_{n=-\infty}^{\infty}|\hat{g}(n)|^4}=\sqrt{\displaystyle\sum_{n=-\infty}^{\infty}|\widehat{f*f}(n)|^2}\sqrt{\displaystyle\sum_{n=-\infty}^{\infty}|\widehat{g*g}(n)|^2}. $$

And lastly we use the Parseval's identity again to get.$$ \sqrt{\displaystyle\sum_{n=-\infty}^{\infty}|\widehat{f*f}(n)|^2}\sqrt{\displaystyle\sum_{n=-\infty}^{\infty}|\widehat{g*g}(n)|^2}=||f*f||_2||g*g||_2. $$