finding a function satisfiying a certain equation I want to know that if there exists any function $f\in L^2[0,1]$ satisfying
$$f(x) = \int^x_0 f(y)\,dy$$
I do not know the value of $f(0)$ and if $f$ is differentiable.
 A: Let be $f \in L^2([0, 1])$ such that $\forall x \in [0, 1], f(x) = \int_0^x f(t) \textrm{d}t$.
The right-hand is continuous, so the left-hand must be so.
Now, the right-hand becomes continuously differentiable, so the left-hand must be so too.
Let be $x \in [0, 1]$.
Thus, by the fundamental theorem of calculus:
$f(x) = \int_0^x f'(t) \textrm{d}t + f(0)$.
So: $\int_0^x f(t) \textrm{d}t = \int_0^x f'(t) \textrm{d}t + f(0)$.
By differentiating, we get: $f(x) = f'(x)$.
So, necessarily: $f(x) = C\exp(x)$.
But, by plugging in the initial equation: $C\exp(x) = C\exp(x) - C$.
So: $C = 0$.
So: $f(x) = 0$.
As this is true for all $x \in [0, 1]$.
We must conclude: $f = 0$.
Remark : 
If we denote:
$\begin{array}{lcl}
T_f : & [0, 1] & \to \mathbb{R} \\
& x & \mapsto \int_0^x f(t) \textrm{d}t
\end{array}$
$\begin{array}{lcl}
T : & L^2([0, 1]) & \to L^2([0, 1]) \\
& f & \mapsto T_f
\end{array}$
Your equation becomes $f = T(f)$, and it asks whether or not $1$ is in the spectrum of the linear operator $T$ known as the Volterra operator.
Also known as having no eigenvalues at all: $T(f) = \lambda f$ has no solution other than the null function for all $\lambda \in \mathbb{R}$.
