I have an operator $ Ax(t) = \int_{0}^{t^2} x(s)ds $ in $ C [0;1] $. I need to find spectrum of this operator.
$ A $ is a compact operator, so the spectrum consists of 0 and eigenvalues. As I understand there is no eigenvalues, so spectrum = 0. But I don't know how to prove that there is no eigenvalues in spectrum.
Would be very glad for any help!
On
Idea for a solution:
Assume $\lambda$ is a non-zero eigenvalue of $f$. So:
$$Af(t)=\int_{0}^{t^2}f(s)ds=\lambda f(t)$$
First, note by taking $t=0$ that you must have $f(0)=0$ (unless $\lambda=0$, which is against the assumption). Now, since $f$ is continuous, the LHS is differentiable (and thus so is the RHS), and by differentiating we get:
$$\lambda f'(t)=2tf(t^2)$$
Taking $t=0$ we get: $$\lambda f'(0)=0$$ If $\lambda\neq0$ then $f'(0)=0$.
We can differentiate the equation above again (try to see why, by a similar arguement to the one above) and get: $$\lambda f''(t)=2f(t^2)+4t^2 f'(t^2)$$ Taking $t=0$ we get: $$\lambda f''(0)=2f(0)=0\Rightarrow f''(0)=0$$
Continuing like so (noticing that we can always differentiate), we see that $f^{(n)}(0)=0$ for all $n$. I'm not sure why this means that $f=0$, since you don't know that $f$ is analytic. But maybe this would help you in showing that $f=0$, which will show that there are no non-zero eigenvalues. On the other hand, I could be mistaken and there can still be a smooth but non-analytic eigenfunction.
Hints: Suppose $\int_0^{t^{2}} x(s)ds=\lambda x(t), \lambda \neq 0$. Without loss of generality assume $|x(t)| \leq 1$ for all $t$. Let $u_1=1$ and $u_{n+1}=2(u_n+1)$. Let $N_1=1$ and $N_{n+1} =u_{n+1} N_n$. Verify, by induction that $|x(t)| \leq \frac 1 {|\lambda|^{n}} \frac {t^{u_n}} {N_n}$. Observe that $N_n \geq {N!}$ and $u_n \geq n$. Letting $n \to \infty$ conclude that $x(t)=0$ for all $t$.