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Find the spectrum of operator $(Af)(s) = \int _0 ^1 K(s,t)f(t)dt $ on $L_2([0,1])$, where $K(s,t)=\max\{s,t\}$ and prove it is adjoint.

I did manage to show it is self-adjoint for $\overline{K(s, t)}=K(t, s)$. I saw a solution claiming because K is compact, the spectrum is made of $\{0\}$ and an at most countable set $A$ of eigenvalues, and that when $A$ is infinite, $0$ is the only limit point of $A$. In this regard, I manage to calculate it and prove that for any $\lambda \neq 0$ there exsits $f$ s.t $Af=\lambda f$. But that does not align with the claim for any $\epsilon$ there is a finite number of eigenvalues s.t $|\lambda |>\epsilon$

My calculation for the spectrum involves deriving $Af=\lambda f$ twice and solving.

Mittens
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Its me
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1 Answers1

3

Here is more or less what you one can do (I leave the details to the OP):

Presumably the space of definition is $L_2([0,1])$.

  • As for the self-adjointness of $A$: $$\langle Af,g\rangle =\int^1_0\Big(\int^1_0 (t\vee s)\,f(t)\,dt\Big)\overline{g(s)}\,ds\int^1_0 f(t)\overline{\Big(\int^1_0 (t\vee s)g(s)\,ds\Big)}\,dt=\langle f, Ag\rangle$$

  • Observe that $A$ is in fact a compact operator (it is a Hilbert-Schmidt operator, see also here).

  • As $L_2([0,1])\subset L_1([0,1])$, it follows from $$Af(s)=\int^s_0+\int^1_s (t\vee s) f(t)\,dt= s\int^s_0 f(t)\,dt+\int^1_st\,f(t)\,dt $$ that $Af$ is a.s. differentiable on $[0,1]$. If $\lambda\in\sigma(A)\setminus\{0\}$, then $\lambda\in\mathbb{R}\setminus\{0\}$ \begin{align} \lambda f(s)=s\int^s_0f(t)\,dt+\int^1_stf(t)\,dt\tag{1}\label{one} \end{align} It follows that $f$ is twice differentiable and $$\lambda f''(s)=f(s)$$ The general solution to this differential equation is $$ f_\lambda(s)=\left\{\begin{array}{lcr} c_1 e^{\tfrac{s}{\sqrt{\lambda}}}+c_2e^{-\tfrac{s}{\sqrt{\lambda}}} &\text{if} & \lambda>0\\ c_1 \cos\big(\tfrac{s}{\sqrt{|\lambda|}}\big)+c_2\sin\big(\tfrac{s}{\sqrt{|\lambda|}}\big) &\text{if} & \lambda<0 \end{array} \right. $$

  • Plug this into \eqref{one} and see what possibles solutions for $\lambda$ (if any) yield a solution $f_\lambda$ to the equation $(A-\lambda)f_\lambda=0$.

  • $\lambda<0$. Since $$s \int^s_0 \cos(\tfrac{t}{\sqrt{|\lambda|}}) \,dt + \int^1_s t \cos(\tfrac{t}{\sqrt{|\lambda|}})\, dt - \lambda \cos(\tfrac{s}{\sqrt{|\lambda|}}) = \sqrt{|\lambda|}\sin(\tfrac{1}{\sqrt{|\lambda|}}) -\lambda\cos(\tfrac{1}{\sqrt{|\lambda|}})$$ and $$ s \int^s_0 \sin(\tfrac{t}{\sqrt{|\lambda|}}) dt + \int^1_s t \sin(\tfrac{t}{\sqrt{|\lambda|}}) dt - \lambda \sin(\tfrac{s}{\sqrt{|\lambda|}}) = \sqrt{|\lambda|} (\sqrt{|\lambda|} \sin(\tfrac{1}{\sqrt{|\lambda|}}) - \cos(\tfrac{1}{\sqrt{|\lambda|}})) + \sqrt{|\lambda|}s $$ we see that in this case, solutions to \eqref{one} are of the form $f_\lambda(s)= a\cos(\tfrac{s}{\sqrt{|\lambda|}})$ with $$ \sqrt{|\lambda|}\sin(\tfrac{1}{\sqrt{|\lambda|}}) -\lambda\cos(\tfrac{1}{\sqrt{|\lambda|}})=0$$ that is, all $\lambda<0$ such that $$\cot\big(\frac{1}{\sqrt{|\lambda|}}\big)=-\frac{1}{\sqrt{|\lambda|}}$$ which yields a sequence $\lambda_n<0$ with $\lambda_n\nearrow0$,

  • Similarly, for $\lambda>0$, it can be seen that all solutions to \eqref{one} are of the form $$f_\lambda(s)=a\cosh\big(\frac{s}{\sqrt{\lambda}}\big)$$ where $$ \sinh\big(\tfrac{1}{\sqrt{\lambda}}\big)-\sqrt{\lambda}\cosh\big(\tfrac{1}{\sqrt{\lambda}}\big)=0$$ This has a unique solution $a\approx 0.694817\ldots$.

  • I leave it to the OP to check that $\lambda=0$ is not an eigenvalue.

Mittens
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