Can the Volterra operator have nonzero eigenvalues?

[Chris L T521]

Here's this week's problem.

-----

Problem: Let $V:L^2([0,1])\rightarrow L^2([0,1])$ be the operator defined by $Vf(x) = \displaystyle\int_0^x f(t)\,dt$ for all $f\in L^2([0,1])$. This is known as the Volterra operator.

(a) Show that $V$ has no nonzero eigenvalues.
(b) Compute $V^{\ast}$, the adjoint of $V$.

-----

 

[Chris L T521]

This week's question was correctly answered by girdav. You can find his solution below.

Spoiler

a) If $\lambda$ was an eigenvalue, and $f$ an eigenvector for $\lambda$, then for each $x\in \Bbb R$, we would have $$\int_0^xf(t)dt=\lambda f(x).$$
If $\lambda$ was equal to $0$, we would have $f\equiv 0$, which is not allowed (indeed, we would have $\int_I f(t)dt=0$ for all interval $I\subset [0,1]$, and the same would be true for the finite disjoint unions of intervals; then we extend by dominated convergence to Borel sets to get that $f\equiv 0$) . So $\lambda\neq 0$, and since $f$ is continuous (use Cauchy-Schwarz inequality to see it's actually $1/2$-Hölderian), $f$ is $C^1$, as a primitive of a continuous function. So we have $f(x)=\lambda f'(x)$ for all $x$ and $f(0)=0$. This gives $$\left(\frac 1{\lambda}f(x)-f'(x)\right)e^{-\frac x{\lambda}}=0,$$
hence $f(x)=Ce^{\frac x{\lambda}}$. This gives that $f(0)=C=0$, hence $f\equiv 0$. b) Let $f,g\in L^2[0,1]$. We have
$$\langle f,Tg\rangle=\int_{[0,1]}\int_{[0,x]}f(x)g(t)dtdx=\int_{0\leq t\leq x\leq 1}f(x)g(t)=\int_{[0,1]}\int_{[t,1]}f(x)g(t)dxdt,$$
which gives $T^*f(t)=\int_{[t,1]}f(x)dx$ (reversing integration order is allowed as the integrand is (absolutely) integrable).