So I solved the integral equation $\int_{-\infty}^\infty f(t)f(x-t)\,dt = f(x)$ using the Fourier transform and convolution.
With $F$ as $f$'s Fourier transform, I found that $F²=F,$ then by using $F$'s continuity and integrability I concluded that $F$ was the null function on $\Bbb R$ and so was $f$. Now I've got a very similar equation to solve :
$$\int_{0}^x f(t)f(x-t)\,dt = f(x).$$
Now it still looks like a convolution product but from Laplace's transform point of view, which use is strongly recommended in order to solve this. I fail to see why the conclusion would be any different, though; can someone push me in the right direction? Thanks a lot!
You have \begin{align*} \mathcal{L}\bigg[\int_{0}^x f(t)f(x-t)\,dt &= f(x)\bigg]\\ F(s)\,F(s)&=F(s)\\ F(s)&=0\;\text{or}\;1\\ f(t)&=0\;\text{or}\;\delta(t). \end{align*} Here I'm using the Laplace Transform to go from the $x$ domain to the $s$ domain. You would need additional information to pick one of these answers.