Function as a convolution product of other two

Question

I need help with this:

I have to prove that a function $f\in L_{2}(T)$ can be expressed as $f=g*h$ (convolution product) for some functions $g,h\in L_{2}(T)$ if and only if $(\hat{f}(n))_{n}\in l_{1}(Z)$

Thanks for any help.

Answer 1

Let $f \in L^2(T)$, where $T$ is the 1-torus, aka unit circle.

Assume that $f = g * h$ for $g,h \in L^2(T)$. Notice that $\hat{f} = \hat{g} \cdot \hat{h}$ elementwise on $\mathbb{Z}$ by the convolution theorem in this setting. Note $g,h \in \ell^2(\mathbb{Z})$. By the Holder Inequality, $$ \|\hat{f}\|_{\ell^1}= \|\hat{g} \hat{h}\|_{\ell^1} \leq \|\hat{g}\|_{\ell^2} \|\hat{h}\|_{\ell^2} < \infty $$

So much for the "easy" part. For the other direction, we basically want to define $\hat{g}$ and $\hat{h}$ by "hand". One such construction is as follows: \begin{align*} \hat{g}(n) &= \sqrt{|\hat{f}(n)|} \\ \hat{h}(n) &= \begin{cases} \frac{\hat{f}(n)}{|\hat{f}(n)|} \sqrt{|\hat{f}(n)|} & \hat{f}(n) \neq 0 \\ 0 & \text{else}\end{cases} \end{align*}

We now have $\hat{f} = \hat{g} \cdot \hat{h}$ elementwise on $\mathbb{Z}$, and both $\hat{g}, \hat{h} \in \ell^2(\mathbb{Z})$ whenever $\hat{f} \in \ell^1$ (just look at how the norms are defined). Now recall that the fourier transform on $T$ is an isometry $L^2(T) \mapsto \ell^2(T)$, and so the inverse transforms of $\hat{g},\hat{h}$ are in $L^2(T)$.