Suppose $X_1, X_2$ are independent random variables, with the same support $[0,1]$, on the same probability space with densities $f_1,f_2$ respectively.
By support I mean, $f_i$'s are $0$ outside $[0,1]$.
We have that $$\int_{x} f_1(x)f_2(z-x) \,dx = \int_{x} g_1(x)g_2(z-x) \,dx$$ for all $z \in [0,2]$ and $g_i(y) = f_i(1-y)$ for both $i=\{1,2\}$.
We also have that $$ f_i(x) \leq f_i(1-x), \forall x \in [0.5,1]$$ for both $i=\{1,2\}$.
I want to conclude that $$ f_1(x)f_2(z-x) = f_1(1-x)f_2(1-(z-x)) $$ for some $z$.
My try:
Rewrite (1) as $$\int_{x} (f_1(1-x)f_2(1-(z-x)) - f_1(x)f_2(z-x)) \,dx = 0. \quad (*)$$ For $z=1.5$, since we can restrict our interest to $0\leq z-x \leq 1$, we'll have both $x$ and $(z-x)$ exceeding $0.5$. Now from (2) we'll have $$ f_1(x) \leq f_1(1-x) ~\text{and} \\ f_2(z-x) \leq f_2(1-(z-x)).$$ So $$f_1(1-x)f_2(1-(z-x)) - f_1(x)f_2(z-x) \geq 0.$$ With $(*)$ we can in fact conclude $$ f_1(x)f_2(z-x) = f_1(1-x)f_2(1-(z-x)).$$
Now if the above conclusion holds can we say more? That is, from (2) I want to further conclude that $$ f_1(x) = f_1(1-x) $$ and $$f_2(z-x) =f_2(1-(z-x))$$ for some $z$.
Please comment on both the above conclusions I made. Thanks in advance for any help! Please feel free to make any further conclusions from these facts too, it'd be interesting to know them.
In (1) both sides are indeed $0$. the set $\{(x_1,x_2):x_1+x_2=z\}$ is a Nullset in $\mathbb{R}^2$ (w.r.t. Lebeguesmeasure), since it is the graph of the continuous function $x_1 \to z-x_1$. Therefore, (1) is meaningless and the only assumption left is $f_i(x) \le f_i(1-x)$ for $x \ge 0.5$. Now you can find counterexamples for both conclusions (f.e. $f_1(x)=f_2(x)=0.5x$ is a counterexample for conclusion 2).