to show $f(t)=g(t)$ for some $t\in [0,1]$

Question

Let $f,g:[0,1] \rightarrow \mathbb{R}$ be non-negative, continuous functions so that $$\sup_{x \in [0,1]} f(x)= \sup_{x \in [0,1]} g(x).$$ We need To show $f(t)=g(t)$ for some $t\in [0,1].$ Thank you for help.

Answer 1

If $f$ is nowhere equal to $g$, then by continuity $f-g$ has to be of uniform sign. Without Loss of Generality say $f-g>0$

Hence $\displaystyle\frac1{f-g}$ is continuous. Since $[0,1]$ is compact, $\displaystyle\frac1{f-g}\le M$ that is $\displaystyle f-g\ge\frac1M$ uniformly on $[0,1]$

This shows that$$\sup f(x)\ge f(x)\ge g(x)+\frac1M\quad\forall x\in[0,1]$$so that $\displaystyle\sup f(x)\ge\sup g(x)+\frac1M$ , a contradiction to data!

Answer 2

Let $$ f(x_1)=\max_{0 \leq x \leq 1} f(x), \quad g(x_2) = \max_{0 \leq x \leq 1} g(x). $$ If $x_1=x_2$, we are done. Otherwise, suppose that $x_1<x_2$. Look at $f(x_2)$: clearly $f(x_2) \leq g(x_2)$. If $f(x_2)=g(x_2)$, we are done. Otherwise $f(x_2)<g(x_2)$. Swapping $f$ and $g$, we can conclude that $f(x_1)>g(x_1)$. By the intermediate value theorem, there must exist a point $\bar{x} \in (x_1,x_2)$ such that $f(\bar{x})=g(\bar{x})$.