Show $\sup_{x\in[0,1]} | x -f(x)|\geq \frac{1}{2}$ for any $f$ with $f(0)=0$ and $\int^1_0 f(t)dt=0$.

Question

Let $f:[0,1]\to \mathbb R$ be a continuous function with $f(0)=0$ and $\int^1_0 f(t)dt=0$.
I need to show that $$\sup_{x\in[0,1]} | x -f(x)|\geq \frac{1}{2}$$
I'm not sure how to approach this problem. Any tips are greatly appreciated!

Answer 1

Hint: $\displaystyle\int_0^1x-f(x)\,\mathrm dx=\frac12$

Answer 2

Suppose instead that $\lvert x - f(x) \rvert < 1/2$ for all $x \in [0,1]$. Then we see that $$\frac 1 2 = \left \lvert \int^1_0 (x-f(x)) dx\right \rvert \le \int^1_0 \lvert x - f(x)\rvert dx < \int^1_0 \frac 1 2 dx = \frac 1 2,$$ a contradiction. Thus we must have $\lvert x - f(x) \rvert \ge 1/2$ for some $x \in [0,1]$ and the conclusion follows.

Unless I've made a mistake, it seems the assumption that $f(0) = 0$ is unnecessary.