It can be shown that if $V(f) = \int_0^x f(t)\,dt$ denotes the Volterra operator on $C([0,1])$, then the closed subspace $A_b = \{f : f|_{[0,b]} =0\}$ satisfies $ V(A_b) \subset A_b$, that is, $A_b$ is invariant under the Volterra operator. Is it true that $A(A_b) \subset A_b$ whenever $A$ is a bounded operator satisfying $AV=VA$?
Edit: Here's a potential solution, however, I'm not sure how to bring it full circle. Any comments would be helpful! : )
Inductively we can show the following three properties: $$ (i) \quad V x^m = \frac{x^{m+1}}{m+1} $$ $$ (ii) \quad V^m e = \frac{x^m}{m!} $$ $$ (iii) \quad V^m f(x) = (x^{m-1} \star f)(x), $$ where $e \equiv 1$ on $[0,1]$. Applying these observations to $A$, we see that $Ax^m = (x^m \star A e)'$, the derivative of the convolution. Given $f \in C[0,1]$ and $\epsilon>0$, we can find a polynomial $p(x) = \sum_{k=0}^ma_k x^k$ such that $||f-p|| < \epsilon/ \lambda$, where $\lambda > 0$ is a constant to be determined later. Then, by the triangle inequality, we must have $$ ||Vf-Af|| \leq ||Vf-Vp|| + ||Af-Ap|| + ||Vp-Ap|| \leq (M_1+M_2) \epsilon + ||Vp-Ap||. $$ The using linearity and a rule for differentiating a convolution, we get $$ ||Vp-Ap|| \leq m \max |a_k| ||Vx^k - Ax^k|| \leq M_3 ||V x^k - (kx^{k-1} \star Ae)||. $$
My fight is with getting this very last estimate arbitrarily small.