Find all continuous (and not necessarily differentiable) funtions satisfying $[f(x)]^2=\int_0^t f(s)ds,\quad t\ge0$

Question

Can someone please tell me if my approach is right or has errors?

First, we need to use Gronwall - Bellman's inequality to show that all non-negative, continuous functions $f(t)$ that satisfy $$f(t)\le\int_0^t f(s)ds,\quad t\ge0$$ are equal to 0. Since we want to show that $[f(t)]^2=\int_0^t f(s)ds$, we know that $\int_0^t f(s)ds\ge 0$, while $f$ is continuous. Hence, the functions that satisfy the equality above are all $f(t)=0,\quad t\ge0\quad$(since we don't know if the functions are differentiable or not).

Answer 1

There is no relation between your inequality and your original equality.

If $f $ is continuous and satisfies the equality, then $f^2$ is differentiable. If $f (x_0)>0$ for some $x_0 $, then $f (x)=[f (x)^2]^{1/2} $ is differentiable on that interval. Similarly where $f $ is negative.

So, on any interval where $f $ is nonzero we can differentiate to get $$2f (x)f'(x)=f (x). $$ Thus on such interval $f'(x)=1/2$. This gives $f (x)=x/2+c $. Plugging this into the equation gives $$cx+c^2=cx, $$ so $c=0$.

Then a continuous $f $ that satisfies the equation is $f (x)=x/2$ for every $x $ such that $f (x)\ne0$. By continuity, $f (x)=x/2$. This is the only solution other than the zero function.

As mentioned by A.G. below, if the domain is the whole real line, then f is zero for negative x.

Answer 2

There is a typo in the integral: it should be $x$ instead of $t$.

It is clear that $f(0)=0$. Proving with a polynomial you have $$\left(\sum a_nx^n\right)^2=\sum \frac{a_nx^{n+1}}{n+1}\Rightarrow f(x)=\frac x2$$ Taking now $f(x)=\dfrac x2+g(x)$ again $g(0)=0$ and $$\int_0^x\left(\frac s2+g(s)\right)ds=\int_0^x\frac s2ds+\int_0^x g(s)ds=\frac{x^2}{4}+xg(x)+(g(x))^2$$ hence$$\int_0^x g(s)ds=xg(x)+(g(x))^2$$ Since $xg(x)$ is the area of the rectangle of sides $x$ and $g(x)$ it is easy to show that $g(x)=0$ because if not then there is a point where $\int_0^x g(s)ds\lt xg(x)$.

Thus $f(x)=\dfrac x2$ is the only continuous solution.