# Inequality using continuity of the function

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!

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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 $$\epsilon=\frac{f(a)}{2}.$$