$f: [0,1] \to \mathbb R$ is continuous and $\int_0^x f(t) dt = \int_x^1 f(t)dt$ for all $x \in [0,1]$, then $f(x) = 0$ for all $x \in [0,1]$.

Question

A function $f: [0,1] \to \mathbb R$ is continuous on $[0,1]$ and $\int_0^x f(t) dt = \int_x^1 f(t)dt$ for all $x \in [0,1]$, then $f(x) = 0$ for all $x \in [0,1]$.

My Try: Let us assume that $f(x) \neq 0$ at one point say $\alpha$.

Without loss of generality we take $f(\alpha) > 0$.

Let us take an $x$ such that $0< \alpha < x$.

Then using the fact that $\int_0^x f(t) dt = \int_x^1 f(t)dt$ for all $x \in [0,1]$ and that $f(x) = 0$ at all points other than $\alpha$ we have a contradiction. I am stucked here please help!

Answer 1

Let $g(x) = \int_0^x f(t)\, dt - \int_x^1 f(t)\, dt$, we have $g(x) = 0$, for all $x \in [0,1]$ by assumption, hence $g'=0$. On the other hand, by the fundamental theorem of calculus (note that $f$ is continuous, $$ g'(x) = f(x) + f(x) = 2f(x) $$ So $f(x) = 0$ for all $x$.

Answer 2

Why don't you just derive the given equality? You will get $f(x) = -f(x)$, hence $f=0$.