Representation of a closed ideal in $L^1(T)$.

Question

Let $T$ be the unit circle. Given that $I$ is a closed ideal in $L^1(T)$. I conjecture that there exists a set $S \subseteq \mathbb{Z}$ such that \begin{equation} I=\{f \in L^1(T) : \widehat{f}(n)=0, n \in S\}. \end{equation} What I have shown so far is that if $I$ has this form, then it is indeed a closed ideal in $L^1(T)$. What should I consider next?

Answer 1

TRUE.

Assume that for every $k\in\mathbb Z$, there exists an $f_k\in I$, with $\hat f(k)\ne 0$. Then $$ \mathrm{e}^{ikx}=\left(\frac{1}{\hat f(k)}\mathrm{e}^{ikx}\right)*f(x)\in I, $$ and hence all trigonometric polynomials belongs to $I$, and as $I$ is closed, then $I=L^1(T)$.

Note. I have assume the the multiplication is the ring $L^1(T)$ is the standard convolution.