showing that two functions are the same

Question

suppose I have two arbitrary probability density functions $f(\varepsilon)$ and $h(\varepsilon)$.

If $\int_{\mathbb{R}}\int^{z+\varepsilon_1}_{-\infty}f(\varepsilon_1)f(\varepsilon_2)~d\varepsilon_2d\varepsilon_1 = \int_{\mathbb{R}}\int^{z+\varepsilon_1}_{-\infty}h(\varepsilon_1)h(\varepsilon_2)~d\varepsilon_2d\varepsilon_1$

for all $z\in \mathbb{R}$, then $f(\varepsilon)=h(\varepsilon)$ for all $\varepsilon$.

I feel like this is true, but I am not sure how I am supposed to approach...

Answer 1

If $h(x)=f(x-1)$ for all $x$ then the condition holds. [If $\{X,Y\}$ is i.i.d with density $f$ and if $\{U,V\}$ is i.i.d with density $h$ then the condition is simply that $X-Y$ has same distribution as $U-V$. This does not imply that $X$ and $U$ have the same distribution: we may have $U=X+1$ for example].

Answer 2

This is not true: consider $f$ being the indicator function of $[0,1]$ and $h$ being the indicator of $[1,2]$.