a seemingly simple functional equation Is it possible to find a nonzero function $f(x)$ such that $$(1-a)\int_0^axf(x)dx+a\int_a^1(1-x)f(x)dx=0$$
for $a\in[0,1]$ such that $f(x)$ is independent of $a$?
My own answer is that it is not possible, since if we take two derivatives from the original equation, we end up with $f(x)=0$. Is this correct? 
Many thanks for your hints.

When you say that $f$ has to be independent of $a$, I interpret it by considering your equation as a family of equations for $f$ holding for any $a$, i.e.
$$
\Phi(a):=(1-a)\int_0^axf(x)dx+a\int_a^1(1-x)f(x)dx=0\qquad \forall a\in[0,1]
$$
I rewrite $\Phi$ as
$$
\Phi(a)=\int_0^axf(x)dx-a\int_0^1 xf(x)dx+a\int_a^1f(x)dx
$$
Then (under hypothesis of derivability, to be verified a posteriori) I derive this equation obtaining
$$
\Phi'(a):=af(a)-\int_0^1 xf(x)dx+\int_a^1f(x)dx-af(a)=-\int_0^1 xf(x)dx+\int_a^1f(x)dx=0.
$$
By deriving again $ \Phi''(a)=-f(a)=0$. So $f$ has to be the null function.
