Spectrum of operator on $L^2([0,1])$

Question

I need to calculate the spectrum of the operator $T$ for $f \in L^2([0,1])$ defined by:

\begin{equation} (Tf)(x) = \int_0^1 (x+y)f(y)dy. \end{equation}

I know that $T$ is compact and self-adjoint so the residual spectrum is empty and the eigenvalues are real and a closed subset of $[-||T||, ||T||]$.

So I let $\lambda$ be an eigenvalue so I know that $\lambda f(x) = (Tf)(x)$. By differentiating twice, I found that $\lambda f''(x) = 0$ but I don't really know how I can continue.

Answer 1

Notice that the range of $T$ is the span of $\{1,x\}$, that is is the space of linear functions on $[0,1]$. So if $f(y)=ay+b$ is an eigenvector of $T$ corresponding to $\lambda\in\mathbb{C}\setminus\{0\}$, $$ (\frac12 a+ b)x + (\frac13 a + \frac12b) = \lambda a x + \lambda b $$ Thus $$ \begin{pmatrix} \frac12-\lambda & 1\\ \frac13 & \frac12-\lambda \end{pmatrix}\begin{pmatrix} a\\ b \end{pmatrix}=\begin{pmatrix} 0\\0\end{pmatrix} $$

Answer 2

To continue on your path, note that once you have justified that $f$ must in fact be smooth to be an eigenvector for nonzero eigenvalue $\lambda$, you can differentiate twice as you did and note that $$ \lambda f''(x)=0 $$ In general, you are a long way from being allowed to differentiate an arbitrary element of $L^2$, so beware in the future.

So that, for $\lambda\ne 0$, by solving the ode $f(x)=ax+b$. Now proceed as in the other (hint, when does that system have a nontrivial solution?).

The point $\lambda=0$ is also in your spectrum however (this is of course obvious when noting that we took an infinite dimensional space to a finite dimensional one). For an explicit example, note that $Tf(x)=0$ where $f(x)=6x^2-6x+1$. This answer was arrived at by looking for a quadratic polynomial orthogonal to both $1$ and $y$.