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

$$\mathbf {The \ Problem \ is}:$$ 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 .

• You may consider $f(x)=e^{2\pi ix}$ and $g(x)=1$. In general, choose $(a_n)_{n\in\mathbb{Z}}$ such that $\sum_n |a_n|<\infty$ and then set $$f(x)=\sum_{n\text{ odd}}a_n e^{2\pi i n x}\qquad\text{and}\qquad g(x)=\sum_{n\text{ even}}a_n e^{2\pi i n x}.$$ Jan 11 at 11:57
• Easy example (special case of the above) - if we set $\mathbb T=[0,1]$, $f(x)=\sin(2\pi x)$, $h(x)=\sin (4\pi x)$, then $\int f(s)h(s-x)\ dx=0$. quick verification on Desmos: desmos.com/calculator/rqmudg7azm Jan 11 at 12:00
• The similar result that I know, due to Titchmarsh: $C([0,\infty))$ under convolution is a division algebra. Jan 11 at 12:08