by Random-generator
Last Updated October 09, 2019 21:20 PM

Suppose $f\in C[0,1]$ and $f(a)>0$ for an $a\in [0,1]$. To show that there is a closed interval $[c,d]\subseteq [0,1]$(which contains $a$) such that $f(x)\geq f(a)/2$ for all $x\in [c,d].$

My try:

From continuity of $f$ at $a$, for all $\epsilon>0$, there exists a $\delta>0$ such that $$ |x-a|<\epsilon \implies |f(x)-f(a)|<\delta. $$ Can we say that there is some $\epsilon>0$ for which $0<\delta< f(a)/2$, and hence $f(x)\geq f(a)/2$ the statement is true?

Thanks in advance for any help!

First of all, it should be $$|x-a|<\delta\implies |f(x)-f(a)|<\epsilon$$ (your $\epsilon$ and $\delta$ were switched). Now, see what happens if you take $\epsilon=\frac{f(a)}{2}.$

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