So, am I stupid or isn't this kind of trivial?
I'm having this problem:
Let $T$ be the Volterra operator on $L^2([0,1])$ defined by \begin{equation} Tf(s) = \int^s_0 \, f(t) \, \text{d}t. \end{equation} For $0 \leq \alpha \leq 1$ set \begin{equation} M_{\alpha} = \{f \in L^2([0,1]) : f(t) = 0, 0 \leq t \leq \alpha\}. \end{equation}
Prove that each $M_{\alpha}$ is an invariant subspace for $T$.
My solution is this:
$M_{\alpha}$ is an invariant subspace for $T$ if $f(t) \in M_{\alpha} \Rightarrow Tf(s) \in M_{\alpha}$. So if $f(t) \in M_{\alpha}$, then
\begin{equation*} \begin{aligned} Tf(s) &= \int^s_0 \, f(t) \, \text{d}t\\ &= \begin{cases}\displaystyle\int^s_0 \, 0 \, \text{d}t &\quad \text{if } 0 \leq t \leq \alpha\\\displaystyle\int^s_0 \, f(t) \, \text{d}t &\quad \text{if } \alpha < t \leq 1\\\end{cases}\\ &= \begin{cases}0 &\quad \text{if } 0 \leq t \leq \alpha\\\displaystyle\int^s_0 \, f(t) \text{d}t &\quad \text{if } \alpha < t \leq 1\\\end{cases}\\ &\in M_{\alpha} \end{aligned} \end{equation*}
Am I wrong or is it this simple?
Your proof is right if you indeed wishes to prove $M_\alpha$ is invariant subspace. However, it will be nontrivial if you want to prove the converse, namely, every invariant subspace of $T$ is of this form.
References:
D. Sarason, A remark on the Volterra operator
W. Donoghue, The Lattice of invariant subspaces of a completment continuous quasinilpotent transformation