Spectrum of integral operator

599 Views Asked by At

Given $g\in C^1([0,1]\times[0,1])$, consider the operator $$Tu(x) = \int_0^1 g(x,t) u(t) dt$$ defined on $u\in C([0,1])$. Discuss the spectrum of T.

My attempt:

First I can show that $T$ is a compact operator. Given a sequence $u_n$ bounded in $C([0,1])$$\|u_n \|_\infty \leq M$, and let $\|g\|_\infty = N$, we have $$|(Tu_n)(x)| \leq M\cdot N$$ And given $x_0$ we have $$|(Tu_n)(x_0) - (Tu_n)(x)| \leq \int_0^1 |g(x_0,t) - g(x,t)| M dt$$ by continuity of $g$, we see that $(Tu_n)(x)$ is equicontinuous.

Now we know that $0$ is in the spectrum of $T$. To look for other eigenvalues, let $\lambda \neq 0$ and $f\neq 0$. $$\lambda f(x) = \int_0^1 g(x,t)f(t) dt$$ using the Lebesgue differential theorem, I get that $$\lambda f'(x) = \frac{d}{dx} \int_0^1 g(x,t)f(t) dt = \int_0^1 \frac{\partial}{\partial x} g(x,t)f(t) dt.$$

How would I continue from here? Thank you very much!

1

There are 1 best solutions below

0
On

In order to show that your operator is compact, show that it maps a bounded sequence $\{ f_{n} \}_{n=1}^{\infty}\subset C[0,1]$ to an equicontinuous sequence of functions. So, let $\{ f_{n} \}_{n=1}^{\infty}$ satisfy $\|f_{n}\|_{C[0,1]}\le M$ for all $n$ and some fixed $M$; then, for every $\epsilon > 0$, show that there is a $\delta > 0$ such that $$ |Tf_{n}(x)-Tf_{n}(y)| < \epsilon $$ holds for all $n$ whenever $|x-y| < \delta$. Differentiability is a stronger condition than what you need, but it makes the proof a little easier: $$ \begin{align} |Tf_{n}(x)-Tf_{n}(y)| & \le \int_{0}^{1}|g(x,t)-g(y,t)||f_{n}(t)|\,dt \\ & \le M \int_{0}^{1}|g(x,t)-g(y,t)|\,dt \\ & \le M \int_{0}^{1}\left|\int_{x}^{y}\frac{\partial g}{\partial u}g(u,t)\,du\right|\,dt. \end{align} $$ Let $L$ be a bound for $\frac{\partial g}{\partial u}$ on $[0,1]\times[0,1]$, and you get $$ |Tf_{n}(x)-Tf_{n}(y)| \le LM|x-y|. $$ For the $\epsilon > 0$ which was given, choose $\delta = \frac{\epsilon}{2LM}$ and you get the desired equicontinuity of $\{ f_{n} \}_{n=1}^{\infty}$, i.e., that $|Tf_{n}(x)-Tf_{n}(y)| < \epsilon$ for all $n$ whenever $|x-y| < \delta$.

As you suggest, once you know $T$ is compact, you should be able to say more about the spectrum.