# 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.

• Note that $f\in L^1[0,1]$, so the function $f$ must be absolutely continuous. It has a derivative almost everywhere – ncmathsadist Apr 5 '18 at 21:59
• What about $f(x)=e^x$? – Zachary Apr 5 '18 at 22:05
• Well, the null function works, but I guess you want an non-trivial solution. I think, if you try to consider a certain sequence of functions $f_n \in L^2([0, 1])$ which converges at least pointwise to a certain function $f$ over $[0, 1]$ and try to build it so that $f_{n + 1}(x) = \int_0^x f_n(y) \textrm{d}y$, then you can see if whether you can show an uniform convergence over $[0, 1]$, eventually, taking limits would give you $f(x) = \int_0^x f(y) \textrm{d}y$. Unsure which $f_n$ you could consider. – Raito Apr 5 '18 at 22:05
• @Messney $\int_0^x e^y \textrm{d}y = e^x - 1$ as far as I know. – Raito Apr 5 '18 at 22:06
• @Raito I also thought of $f=0$. I also want to know if it is the only solution to the problem or not, to be exact. – user1292919 Apr 6 '18 at 0:40

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 $\lambda \in \mathbb{R}$.