I have a function $f(x)$ positive for $x>0$ and $f(0)=0$. I also have functions $g,h,k$ and a constant $a$ s.t. for all $t$:
$$\int_0^t f(t-x) g(x)dx=- k(t)\int_0^t f(t-x) h(x)dx$$
So I propose that, if $k(t)\neq 0$, this is equal to
$$\int_0^t f(t-x) \left(\dfrac{g(x)}{k(t)}+h(x) \right) dx=0$$
Once $f(t-x)\geq 0$, so I must have $\dfrac{g(x)}{k(t)}+h(x)\equiv 0$ for all $x\leq t$. In particular, $g(t)=-h(t)k(t)$ for all $t$.
I did not see error on this calculation. However, it seems a little naive, so I 'd like to understand if this calculaton is correct. Thank you!
Just consider the integral $\int_{-1}^1 x^2 \cdot x \,dx$. You have that $x^2 \ge 0$, the integral is zero, but $g(x)= x$ is not identically zero.
In this case, maybe you can get something out of differentiating both sides on the relation...
$$ \frac{d}{dt} \int_0^t f(t-x) g(x) dx = f(t-t)g(t) + \int_0^t f'(t-x)g(x) dx= \int_0^t f'(t-x)g(x) dx $$
$$ \frac{d}{dt} \left(-k(t) \int_0^t f(t-x) h(x) dx\right) = -k'(t) \int_0^t f(t-x) h(x) dx -k(t) \int_0^t f'(t-x) h(x) dx $$