Finding the spectrum of a compact operator

Question

Task

Given $$T : L_2[−1, 1] \to L_2[−1, 1]$$ is an operator such that:

$$Tf(x)=\int_{-1}^{1} \max(0,x+y)f(y)dy$$

Find the spectrum of $T$.

The idea that $T$ is compact is given.

Tried

From the idea that T is compact $\implies$ $0 \in \sigma(T)$

We have to solve the equation: $$Tf(x)=\lambda f(x)$$

We can first make it look better:

$$Tf(x)=\int_{-1}^{1} \max(0,x+y)f(y)dy=x \int_{-1}^{1}f(y)dy+y \int_{-1}^{1}f(y)dy=\lambda f(x)$$

Lets differentiate two times:

$$\int_{-x}^1 f(y)dy=\lambda f'(x)$$ $$f(-x)=\lambda f''(x)$$

Are these correct. Maybe I have to find a connection between these equations? Am I missing something?

Let $x=0$, then: $$f(1)-f(0)=\lambda f'(0)$$

Let $x=1$, then: $$\int_{-1}^{1} f(y) dy =f(1) - f(-1) = \lambda f'(1)$$

Maybe if I find that the function f is odd/even, I could use it's properties? But can't seem to proove it

Lets substitute: $$\mu = \sqrt{\frac{1}{|\lambda|}}$$

Let $\lambda>0$, then $$f''(x)=-\mu f(-x)$$ It's given to us that: $$f(x)=c_1 \cos(\mu x) + c_2 \frac{e^{\mu x}-e^{-\mu x}}{2}$$

From here we can differentiate: $$f'(x)=-c_1 \sin(\mu x) + c_2 \frac{\mu e^{\mu x}+ \mu e^{-\mu x}}{2}$$

And from here maybe we have to find $c_1$ and $c_2$ but no working ideas were found, tried: $$f(1)-f(-1)=\lambda f'(1)$$ $$f(1)-f(0)=\lambda f'(0)$$

I stuck from here on.

Question

Are my calculations correct? What is the spectrum, how can I calculate it ?

Can you help me?

Answer 1

Note that $$Tf(x)=\int_{-1}^1\max\{0,x+y\}\ f(y)\ dy=\int_{-x}^1 (x+y)\ f(y)\ dy.$$ Therefore, $Tf$ has a weak derivative $(Tf)'$ given by $$(Tf)'(x)=\int_{-x}^1f(y)\ dy+xf(-x)-xf(-x)=\int_{-x}^1f(y)\ dy.\tag{1}$$ It follows then that $(Tf)'$ has a weak derivative $(Tf)''$ given by $$(Tf)''(x)=f(-x).\tag{2}$$

Suppose that $\lambda \in \sigma(T)$. Therefore if $f$ is an eigenfunction of $T$ wrt eigenvalue $\lambda$, then $$\lambda f(x)=Tf(x).\tag{3}$$ If $\lambda=0$, then $Tf=0$ so that $(Tf)'=0$, and $(Tf)''=0$. This means $f(-x)=0$ for almost every $x$. Hence, $f=0$ almost everywhere, making $\lambda=0$ not an eigenvalue of $T$.

If $\lambda \ne 0$, then by $(3)$, $f$ is weakly differentiable. If $f'$ is a weak derivative of $f$, then $$\lambda f'(x)=(Tf)'(x).\tag{4}$$ That is, $f'$ is also weakly differentiable. If $f''$ is a weak derivative of $f'$, then $$\lambda f''(x)=(Tf)''(x)=f(-x).$$ This shows that $f$ has all weak derivatives, and $$\lambda^2 f''''(x)=\lambda f''(-x)=f(x).$$ Hence, if $\frac1\lambda=\omega^2$, then $$f(x)=C_1\cosh(\omega x)+S_1\sinh(\omega x)+C_2\cos(\omega x)+S_2\sin(\omega x)$$ for almost every $x$.

The condition $\lambda f''(x)=f(-x)$ implies that $$f(x)=C_1\cosh(\omega x)-S_1\sinh(\omega x)-C_2\cos(\omega x)+S_2\sin(\omega x).$$ Therefore, $S_1=0$ and $C_2=0$. Now $f(-1)=0$ implies $$C_1\cosh\omega-S_2\sin\omega=0.$$ Hence, we may assume that $S_2=\cosh \omega$, so $C_1=\sin\omega$. That is $$f(x)=\sin\omega \cosh(\omega x)+\cosh\omega\sin(\omega x).$$ From $(2)$, $f'(-1)=0$, we get $$0=- \sin\omega \sinh\omega+\cosh\omega\cos\omega.$$ Therefore, $$\tan\omega \tanh\omega =1.\tag{5}$$ This seems to be the only requirement. There are infinitely many real roots to $(5)$. The smallest positive real root of $(5)$ is $\omega\approx 0.93755$. Therefore, the largest positive real eigenvalue $\lambda$ is $\lambda\approx 1.13765$. There are also infinitely many negative real eigenvalues $\lambda$ (corresponding to purely imaginary roots of $(5)$), the largest (in magnitude) of which is $\lambda\approx-0.181534$ (arising from $\omega\approx 2.34705i$).

Finally, let $$\Lambda=\left\{\frac{1}{\omega^2}\ \Big|\ \omega\in\Bbb C\wedge \tan\omega\tanh\omega=1\right\}.$$ Since $T$ is self-adjoint, $\Lambda\subseteq \Bbb R$, so $\Lambda=\Lambda_+\cup\Lambda_-$, where $$\Lambda_+=\left\{\lambda>0\ \Big|\ \tan\frac{1}{\sqrt{\lambda}}\tanh\frac{1}{\sqrt{\lambda}}=1\right\}$$ and $$\Lambda_-= \left\{\lambda<0\ \Big|\ \tan\frac{1}{\sqrt{|\lambda|}}\tanh\frac{1}{\sqrt{|\lambda|}}=-1\right\}.$$ Finally, we have $$\sigma(T)= \Lambda=\Lambda_-\cup\Lambda_+.$$ The spectral radius of $T$ is approximately $1.13765$.

Note that $T:L_2[-1,1]\to \operatorname{im}T$ has an unbounded inverse $T^{-1}:\operatorname{im} T\to L_2[-1,1]$. The inverse of $T$ is given by $$T^{-1}f(x)=f''(-x).$$ To see this, we let $Sf(x)=f''(-x)$ for $f\in \operatorname{im}T$. Then by $(2)$, $$STf(x)=(Tf)''(-x)=f(x)$$ for all $f\in L_2[-1,1]$.
On the other hand, for $f\in \operatorname{im}T$, $$TSf(x)=\int_{-x}^1(x+y)f''(-y)\ dy=\int_{-1}^x(x-y)f''(y)\ dy.$$ Recall that $f(-1)=0$ and $f'(-1)=0$ for all $f\in\operatorname{im}T$. Hence, integration by parts yields $$TSf(x)=\big((x-y)f'(y)\big)\Big|^{y=x}_{y=-1}+\int_{-1}^x f'(y) dy=f(x)-f(-1)=f(x)$$ for almost every $x$. This means $\operatorname{im}T$ consists of continuously differentiable functions $f\in L_2[-1,1]$ which has an essentially bounded weak second derivative $f''\in L_2[-1,1]$ with $f(-1)=f'(-1)=0$.

Answer 2

Though it is given that $T$ is compact, it doesn't hurt to prove it. The kernel of $T$ is $K(x,y) = \max(x+y,0)$, and since $[-1,1]^2$ is compact and $\sup_{x,y}|K(x,y)|=2<\infty$, by Fubini's theorem we have $$ \iint_{[-1,1]^2} |K(x,y)|^2\ \mathsf dm(x,y) < \infty, $$ and hence $$ \iint_{[-1,1]^2]} K(x,y)^2\ \mathsf dm(x,y) = \int_{-1}^1\int_{-1}^1\max(x+y,0)\ \mathsf dx\ \mathsf dy = \frac43<\infty, $$ so that $K\in L^2([-1,1]^2)$. This implies that $T$ is compact.

I wasn't able to find the eigenvalues, but I did find a bound on the norm. Let $f_n(x) = \sqrt n\mathsf 1_{(1-1/n,1)}$. Then $$\|f\|_2 = \int_{1-1/n}^1 n\ \mathsf dx = 1, $$ and $$ \|Tf_n\|_2 = \left(\int_{-1}^1\int_{1-1/n}^1 \max(x+y,0)n\ \mathsf dx\ \mathsf dy\right)^{\frac12} = \left(\frac{32 n^3-24 n^2+8 n-1}{12 n^3}\right)^{\frac12} \stackrel{n\to\infty}\longrightarrow \frac43, $$ so $\|T\|\geqslant\frac43$ (and hence the spectral radius is at least $\frac43$).