Commuting Operator If we consider the operator $T(f) = \int_0^x f(t)dt$ on the Banach space $C[0,1]$, is it possible to classify all linear operators $A$ such that $AT=TA$? Intutively, I feel like $AT=TA$, then it must be the case that either $A=T$ or $A$ is given by covolution.  Does anyone have any ideas if my hunch might be true? 
 A: Let $u_n$ be the function $u_n(x) = x^n$ (with $u_0(x) = 1$).  Then $T(u_n) = u_{n+1}/(n+1)$, and $T^{n}(u_0) = u_n/n!$.
If $A$ is any bounded linear operator commuting with $T$, we must have 
$A(u_n) = n! A T^n u_0 = n! T^n A u_0$.  Since polynomials are dense in $C[0,1]$, $A$ is uniquely determined by its value on $u_0$.  In particular, if $A u_0$ is a polynomial $\sum_{i=0}^d c_k u_k = \sum_{i=0}^d c_k k! T^{k} u_0$, then we must have $A = \sum_{i=0}^d c_k k! T^k$.
EDIT:  It can be shown by induction that $$T^n(f)(x) = \int_0^x \frac{(x-t)^{n-1}}{(n-1)!} f(t)\; dt$$
Thus if $A u_0$ is a polynomial $p(x)$, 
$$A f(x) = p(0) f(x) + \int_0^x p'(x-t) f(t)\; dt $$
On the other hand, for any constant $c$ and continuous signed Borel measure $\mu$ on $[0,1]$, you can consider the operator
$$A f(x) = c f(x) + \int_{[0,x]} f(x-t)\; d\mu(t)$$
which will commute with $T$.  We have $A u_0(x) = c + \mu([0,x])$.  Now I suspect that for any bounded linear operator $A$ commuting with $T$,  $v = A u_0$ must have bounded variation, which implies $A$ is of that form.
