Spectrum of $Tu=\int^1_0 (x+y)u(y)dy$

Question

Given the operator

$$Tu(x)=\int^1_0 (x+y)u(y)dy$$

on $L^2(0,1)$, find the spectrum of $T$. For all eigenvalues, find their multiplicities and the eigenfunctions.

The kernel is Hilbert Schmidt and symmetric, so we know $T$ is compact and self adjoint.

The self adjointness tells us that $\sigma_r(T)=\emptyset$.

I have found so far that for $\lambda_\pm=1/2\pm 1/\sqrt 3$ we have eigenfunctions $u_\pm(x)=c(x\pm 1/\sqrt 3)$ respectively- both with multiplicity one. (Do you agree?)

$\lambda=0$ is giving me some trouble. By compactness, it is either in the point spectrum or continuous spectrum, but which? I answered this question with what I think so far. Please comment.

Answer 1

Consider $\lambda=0$.

$$\int^1_0 (x+y)u(y)dy = 0 $$

or

$$x\int^1_0 u(y)dy+\int^1_0 yu(y)dy = 0 $$

So $u(x)$ is an eigenfunction if both $<y,u>=0$ and $<1,u>=0$ (by linear independence). It seems me that this implies $0$ is an eigenvalue of infinite multiplicity because any polynomial of degree greater than 2 can be made orthogonal to both $1$ and $y$.

Answer 2

The subspace $M$ spanned by $\{ 1, x\}$ is invariant under this operator. And, if $u \perp M$, then $Tu=0$. So this completely reduces to a $2\times 2$ matrix problem. You could use an orthonormal basis for $M$, but there's no real need that I can tell. Using the ordered basis $b=\{ 1,x\}$, $$ Tu = x(u,1)+1(u,x),\;\;\; (f,g) = \int_{0}^{1}f(t)g(t)\,dt.\\ T1 = x(1,1)+1(1,x) = \frac{1}{2}1+x,\\ Tx = x(x,1)+1(x,x) = \frac{1}{3}1+\frac{1}{2}x. $$ So $T$ has the following matrix representation with respect to the basis $\{ 1,x\}$: $$ [T]_{b,b} = \left[\begin{array}{cc}\frac{1}{2} & \frac{1}{3} \\ 1 & \frac{1}{2}\end{array}\right] $$ That definitely has characeristic polynomial $(\lambda-1/2)^{2}-1/3$ with roots as you stated. So I agree that those are eigenvalues. The remaining eigenvalue is $0$ because $T=0$ on $M^{\perp}$.

To find the eigenvectors of $T$ in $M$, first consider $$ [T]-\left(1/2-\frac{1}{\sqrt{3}}\right)I= \left[\begin{array}{cc} \frac{1}{\sqrt{3}} & \frac{1}{3} \\ 1 & \frac{1}{\sqrt{3}} \end{array}\right], $$ whose null space is spanned by $$ \left[\begin{array}{c} \frac{1}{\sqrt{3}} \\ -1\end{array}\right] $$ In terms of polynomials, $\frac{1}{\sqrt{3}}-x$ is an eigenvector of $T$ with eigenvalue $1/2-1/\sqrt{3}$. Similarly $\frac{1}{\sqrt{3}}+x$ is an eigenvector of $T$ with eigenvvalue $1/2+1/\sqrt{3}$. It can be checked that these eigenvectors are orthogonal because $$ (1/\sqrt{3}-x,1/\sqrt{3}+x) = \int_{0}^{1}\left(\frac{1}{3}-x^{2}\right)\,dx=0. $$