I am trying to solve the following problem:
Let $H=L^2[0,1]$, $T=C_f$ (convolution) where $f \in C[0,1]$ (continuous function).
Define $M_g(h) = gh$.
Prove that if $TM_{e^{2\pi ix}}=M_{e^{2\pi ix}}T$ then $f=0$.
I feel close but I'm unable to complete the proof.
What I've done is I transitioned to the matrix representation of both so that the hypothesis becomes: $[T][M_{e^{2\pi ix}}]=[M_{e^{2\pi ix}}][T]$. Observing the a convolution's matrix representation is the fourier coefficients of the function on the diagonal, and $0$ otherwise, we obtain:
$$ \big([T][M_{e^{2\pi ix}}]\big)_{kj}=\langle e^{2\pi ix}e_j,e_k\rangle\hat{f}(k)$$
$$ \big([M_{e^{2\pi ix}}][T]\big)_{kj}=\langle e^{2\pi ix}e_j,e_k\rangle\hat{f}(j)$$
So, given the equality, I concluded that for $k\neq j$ we must have:
$$ \langle e^{2\pi ix}e_j,e_k\rangle\hat{f}(k)=\langle e^{2\pi ix}e_j,e_k\rangle\hat{f}(j)$$
which yields: $$\hat{f}(j)=\hat{f}(k)$$
but I don't see why this necessarily means that $f=0$. So either I'm wrong altogether, or I'm missing something.
Any help would be much appreciated.