If $f:[0,1] \to \mathbb{R}$ is continuous and $\int^{x}_{0} f = \int^{1}_xf,$ then $f(x) = 0, \forall x\in [0,1].$

Question

If $f:[0,1] \to \mathbb{R}$ is continuous and $\int^{x}_{0} f = \int^{1}_xf,$ then $f(x) = 0, \forall x\in [0,1].$

May I verify if my proof is valid? Thank you:)

Proof: $\int^{c}_{0} f = \int^{1}_cf \implies\int^{1}_{0}f=2\int^{c}_{0}f=2\int^{d}_{0}f \implies \int^{d}_{c}f=0, \forall c,d \in [0,1]. $

Suppose $\exists c \in (0,1)$ such that $f(c)>0.$ Since $f$ is continuous at $c,$ given $\epsilon = \frac{f(c)}{2},$ $\exists \delta >0$ such that $f(x)>\frac{f(c)}{2}, \forall x\in (c-\delta,c+\delta)\subseteq [0,1].$

It follows that $\int^{c+\frac{\delta}{2}}_{c}f \geq \frac{f(c)}{4}\delta>0$ (Contradiction).

If $f(0)>0, $ given $\epsilon = \frac{f(0)}{2},$ $\exists \delta >0$ such that $f(x)>\frac{f(c)}{2}, \forall x\in [0,0+\delta)\subseteq [0,1].$

It follows that $\int^{0+\frac{\delta}{2}}_{0}f \geq \frac{f(0)}{4}\delta>0$ (Contradiction). Similary, $f(1)=0$ results in contradiction.

Answer 1

I think you can be satisfied with the conclusion that $\int_\alpha^\beta f(x)\, dx=0$ for all $\alpha$ and $\beta$. As you noticed, it is standard to deduce that $f=0$ everywhere.

Another proof goes as follows. Let $F$ be a primitive of $f$ (which exists since $f$ is continuous). Then $F(x)-F(0)=\int_0^x f(t)\, dt$ and $F(1)-F(x)=\int_x^1 f(t)\, dt$. Hence $F'(x)=-F'(x)$ for every $x$, and therefore $2f(x)=0$ for every $x$.