Is Banach algebra $L^1(\mathbb T)$ under convolution a division algebra?

Question

$\mathbf {The \ Problem \ is}:$ If $f$ & $h$ are $L^1(\mathbb T)$ functions with $f\star h=0$ identically on $\mathbb T.$ Then can we say either $f\equiv 0$ or $h\equiv 0 ?$

$\mathbf {My \ approach}:$ By convolution theorem, $\widehat f(n)$.$\widehat h(n)=0$ for all $n\in \mathbb N.$

But, can we find $f,h \in L^1(\mathbb T)$ with $\widehat f$ is supported on odd integers & $\widehat h$ on even integers ?

I couldn't find anything . Thanks in adv. for a hint .

Answer 1

Note that a complex division Banach algebra is isomorphic to the complex numbers. Even if you consider the real scalars, the complexification gives the answer immediately.