TITLE: a problem in functional analysis or operator theory
QUESTION [6 upvotes]: I need a small hint for solving this problem please. 
Problem:
Consider the operator
$ T: C([ 0,1 ])\to C([ 0,1 ]) $
which is defined as follows 
$$ T(u)(x)=\int_{0}^{x^{2}}u(\sqrt s)ds. $$
Prove that for each
$\lambda\in \mathbb{R},$
there does not exist 
$ u\in C([ 0,1 ]) $
different from null function, such that $ Tu=\lambda u. $
What I have done:
I tried to test the proposition with some examples. For instance, I put 
$ u(x)=x $ 
or 
$ u(x)=2x^{2}. $
And I found it without having any problems which contradicts with the above. Please let me know if there is something which is helpful.

REPLY [2 votes]: I would like to suggest a solution that is more functional-analysis oriented, which is based on analysis of the spectrum of Volterra operators. This can be finished in a sentence:
It turns out $T$ is Volterra operator from $C([0,1])$ to itself, so the spectrum is $\{0\}$, but $0$ is not an eigenvalue.
Ofcourse, this is not a hint, nor it is easy, but I decided it is reasonable to show it, since you already got many good hints and answers. I will outline most of the proof, a major part (Neumann series THM) is omitted, but it is not difficult if you are familiar with some operator theory over Banach spaces.
Proof: Note $Tf(x) = \int_0 ^{x^2} f(\sqrt{y}) \mathrm{d}y = \int_0 ^x 2 f(\tau)\tau \mathrm{d}\tau$ which is to say that $T$ is a Volterra operator (and we can say alot about these). Recall that Neumann THM about operator series: Suppose $T\colon X \rightarrow X$ is an operator defined on a Banach space $X$, then if  $\sum_{n=0}^\infty\| T^n \| < \infty$ we have $I-T$ is invertible, and its inverse is $\sum_{n=0}^\infty T ^n$.
Let us evaluate $\|T^n\|$, (recall $x\in(0,1)$):
$$|Tf(x)|=\left| \int_0 ^x 2\tau f(\tau) \mathrm{d}\tau  \right| \leq  \int_0 ^x |2 f(\tau)| \mathrm{d}\tau \leq \int_0^x2\|f\|_\infty \mathrm{d}\tau=2\|f\|_\infty x $$
which gives a point-wise bound for $Tf(x)$ and also shows $\|T\| \leq 2$. Let us evaluate $T^2$ in a similar manner:
$$|T^2f(x)|= \left| \int_0^x 2\tau \cdot Tf(\tau) \mathrm{d}\tau \right| \leq \int_0^x |2\tau \cdot Tf(\tau) |\mathrm{d}\tau\leq 2^2\|f\|_\infty \int_0^x\tau \mathrm{d}\tau = 2^2 \|f\|_\infty \frac{x^2}{2}$$
which again gives a point-wise bound $|T^2f(x)| \leq 2\|f\|_\infty x^2$ and $\|T^2\| \leq 2\|f\|_\infty$. Inductive reasoning shows
$$|T^nf(x)| \leq 2^n \|f\|_\infty \cdot \frac{x^{n}}{n!}$$
that gives $\|T^n\| \leq 2^n/(n-1)!$, so the series converge and $I-T$ is invertible, and $T-I$ is invertible.
Take $\lambda \in \mathbb{C} \setminus{0}$, and look at $T-\lambda I$. Is it invertible? Yes. $T-\lambda I = \lambda ( \lambda^{-1} T -I)$, so it is enough to show $ \lambda^{-1} T -I$ is invertible, but it is (since $\lambda^{-1}T$ is Volterra operator).
Since the spectrum $\sigma(T)$ is not empty - it must contain the point $\{0\}$, and we are left to determine if $0$ is an eigenvalue.
Although not immediate, showing $0$ is not an eigenvalue is easy, and I take the privilige to stop here.