a seemingly simple functional equation

Is it possible to find a nonzero function f(x)f(x) such that (1−a)∫a0xf(x)dx+a∫1a(1−x)f(x)dx=0(1-a)\int_0^axf(x)dx+a\int_a^1(1-x)f(x)dx=0
for a∈[0,1]a\in[0,1] such that f(x)f(x) is independent of aa?

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)=0f(x)=0. Is this correct?

Many thanks for your hints.



1 Answer


When you say that ff has to be independent of aa, I interpret it by considering your equation as a family of equations for ff holding for any aa, 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 Φ\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 Φ″(a)=−f(a)=0 \Phi”(a)=-f(a)=0. So ff has to be the null function.



Many thanks for hints on the question (+1), I was a bit sloppy in posing the question. Your answer is pretty much what I did as a proof that f(x)=0f(x)=0.
– Math-fun
2 days ago



@Math-fun. I read the re-editing of the question just after I wrote the solution; so I am confident that this is the right answer, a part from the initial derivability hypothesis. So we can actually say that there are no continuous non-vanishing functions f(x)f(x).
– guestDiego
2 days ago