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Suppose $f:E \rightarrow \mathbb{R}$ is continuous at $p$. Prove that if $f(p) > 0$, then there is $\delta>0$ s.t. $f(x) \geq f(p)/2$ $\forall x$ in $E$ s.t. $|x-p| \leq \delta$.

I couldn't find other questions like it and I can't quite figure it out. I recognize that I am given that $f$ is continuous and so that allows me to play with the $\delta$ and $\epsilon$ to try and make the desired result pop up but i'm not quite sure how to. Any help would be appreciated.

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2 Answers 2

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Take $\epsilon=f(p)/2$. By the definition of continuity at point $p$ there exists $\delta>0$ so that for all $x\in E $ with $|x-p|<\delta$ we have $|f(x)-f(p)|<\epsilon=f(p)/2$.

In particular $f(x)-f(p)>-f(p)/2\implies f(x)>f(p)/2$

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  • $\begingroup$ Okay, that made it look simple. Playing with the absolute values gives the desired result. What would I need to do to show that it is $f(x) \geq f(p)/2$? Is there a difference? $\endgroup$
    – badeconomist
    Apr 6, 2016 at 2:53
  • $\begingroup$ What I did shows that $f(x)\geq f(p)/2$. Notice that $f(x)>f(p)/2$ implies $f(x)\geq f(p)/2$. $\endgroup$
    – Asinomás
    Apr 6, 2016 at 2:58
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By definition, if $f$ is continuous at $p$, then $\forall \varepsilon>0, \exists \delta>0$ such that

$$ f(p)-\varepsilon<f(x)<f(x)+\varepsilon $$

$\forall x\in (p-\delta,p+\delta)$. Can you come up with a suitable $\varepsilon$ that helps you prove the statement?

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