Suppose $f:[0,1]\to\Bbb R$ is a continuous function and $\int_0^xf=\int_x^1f$ for all $x\in[0,1]$. Show that $f(x)=0$ for every $x\in[0,1]$.

Question

Suppose $f : [0, 1] \to \mathbb{R}$ is a continuous function with the property that $$\int_{0}^{x}f = \int_{x}^{1}f $$ for all $x \in [0, 1]$. Show that $f(x) = 0$ for every $x \in [0, 1]$.

I am new to such undergrad analysis, can someone please provide any hint to approach the problem, as it seems very intuitive.

Answer 1

Take the limit as $x \to 1$: $$ \int_{0}^{1}f = \lim_{x \to 1}\bigg(\int_{0}^{x}f + \int_{x}^{1}f \bigg) $$ By continuity of $f$ we can take the limit, and the second integral equal to $0$, and so is the first, by the equality, which is valid $\forall x \in [0,1]$: $$ \int_{0}^{1}f=\int_{1}^{1} f \equiv 0 $$ EDIT: Therefore, $F(x) = \int_{0}^{x}f \equiv 0 \ \forall \ x$. Using the definition of antiderivative, $$ f(x) = \lim_{h \to 0}\frac{F(x+h)-F(x)}{h} = \frac{0-0}{h} = 0 $$

Answer 2

The derivative of the left side is $f(x)$ and that of the right side is $-f(x)$. Hence $f(x)=-f(x)$ and $f(x)=0$.