spectrum/compactness of operator

Question

Question: Consider $T: L^2[0,1] \to L^2[0,1] : f \mapsto f(x) + \displaystyle\int_0^x \sin(t^2) f(t) \; dt$.

(a) Is $T$ compact?

(b) Classify the spectrum of $T$ (i.e. $\sigma_p(T), \sigma_r(T), \sigma_c(T)$)

Solution: (a) Write $T = I + T_1$. Note that $\displaystyle \int_0^x \sin(t^2) f(t) \; dt = \displaystyle \int_0^1 k(t,x) f(t) \; dt$, where $k(t,x) = \chi_{[0,x]} \sin(t^2)$. Since $k \in L^2([0,1]^2)$, we see that $T_1$ is Hilbert - Schmidt and hence compact.

Therefore $T$ cannot be compact, less the identity $(I+T_1) - T_1=I$ is compact, which is impossible in infinite dimensions.

Alternatively, I could note that for $\|f\|_{L^2}\le 1$, by Cauchy Schwarz: $$ \ |T_1(f(s)) - T_1(f(t))| \le \displaystyle\int_s^t |f(u)| \; du \le \sqrt{t-s}\|f\|_{L^2} \le \sqrt{t-s}.$$ so by Arzela-Ascoli, $T_1(B(0;1))$ is an equicontinuous and bounded family of function, meaning if $(f_n) \subset B(0;1)$, there exists a uniformly convergent subsequence of $T_1(f_n)$ which is also an $L^2$ convergent subsequence.

(b) We have $(\lambda I - T) = (\lambda I - (I+T_1) = ((\lambda -1) I - T_1)$, so $\sigma(T) = \{1\}+ \sigma(T_1)$.

Then, since $-1 < \sin(t^2) < 1$ on $[0,1]$, I can say $\displaystyle\int_0^x -f(t) \; dt < \displaystyle\int_0^x \sin(t^2) f(t) \; dt < \displaystyle\int_0^x f(t) \; dt$.

Thus $\|T_1\|_{op} \le \|V\|_{op} =0$, where $V:L^2[0,1] \to L^2[0,1]: f \mapsto \displaystyle\int_0^x f(t) \; dt$ is the Volterra operator. (See Spectral radius of the Volterra operator)

So $0 \in \sigma_p(T_1)$ or $0 \in \sigma_c(T_1)$ by spectral theory, as $T_1$ is compact.

It is clear that $0 \notin \sigma_p(T_1) \implies 0 \in \sigma_c(T_1) \implies 1 \in \sigma_c(T) = \sigma(T)$.

Remarks: (First off, are my calculations correct?) Is there any way to do this without looking at the Volterra operator? This is from an old test . . . I knew $T_1$ looked familiar, so I stumbled upon the Volterra operator, but I don't think I could have shown that $\|V\|_{op}=0$ without looking here (or in a reasonable amount of time on my own)

If you have any comments/better ways to do things let me know. I am trying to get better at classifying compactness and spectra. Anything helps!

Answer 1

You can construct the resolvent. For $\lambda \ne 1$, the resolvent equation is $(T-\lambda I)f=g$, which is a first order differentiable equation that can solved by using an integrating factor. The function $s(t)=\sin(t^2)$ can be rather general: $$ \int_{0}^{x}s(t)f(t)dt+(1-\lambda) f(x)=g(x) \\ s(x)\int_{0}^{x}s(t)f(t)dt+(1-\lambda)s(x)f(x)=s(x)g(x) \\ s(x)\int_{0}^{x}s(t)f(t)dt+(1-\lambda)\frac{d}{dx}\int_{0}^{x}s(t)f(t)dt=s(x)g(x) \\ \frac{d}{dx}\int_{0}^{x}s(t)f(t)dt +\frac{s(x)}{1-\lambda}\int_{0}^{x}s(t)f(t)dt=\frac{s(x)}{1-\lambda}g(x) $$

Then, if you solve for $\int_{0}^{x}s(t)f(t)dt$, and plug this back into the first equation, you obtain an expression for $f=(T-\lambda I)^{-1}g$, which is the resolvent: $$ (T-\lambda I)^{-1}g = f = \frac{g}{1-\lambda}-\frac{1}{(1-\lambda)^2}e^{-\int_{0}^{x}\frac{s(u)}{1-\lambda}du}\int_{0}^{x}e^{\int_{0}^{v}\frac{s(u)}{1-\lambda}du}s(v)g(v)dv $$ Then you see that $\sigma(T)={1}$ because the resolvent is defined and bounded for all $\lambda\ne 1$.