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.
to show $f(t)=g(t)$ for some $t\in [0,1]$
2 Answers
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!
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0onish pandu name change korechhis? :-o – 2014-12-11
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