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I'm trying to show that if we have a continuous function $f:[a,b]\to\mathbb{R}$ with $f(x)>0$ on it's entire domain, then there exists an $\epsilon>0$ such that $f(x)\geq\epsilon$ for all $x\in[a,b]$.

I've figured that f being continuous on a closed interval means it's uniformly continuous, and this must have something to do with it, because I can think of functions such as $e^x$ which if defined from $\mathbb{R}\to\mathbb{R}$ do not satisfy this condition (since it's not uniformly continuous).

I'm struggling with the rigour and how to write this up in particular - any ideas please?

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    $\begingroup$ Extreme Value Theorem. $\endgroup$ – Hans Lundmark Oct 3 '17 at 13:46
  • $\begingroup$ Can it be done without using the Extreme Value Theorem? We haven't covered this yet so am not sure about it $\endgroup$ – Analysis is fun Oct 3 '17 at 13:53
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    $\begingroup$ You could do it from scratch, but that would most likely mean reinventing a proof of the EVT, more or less. $\endgroup$ – Hans Lundmark Oct 3 '17 at 13:56
  • $\begingroup$ As an aside, I was wondering if continuity is necessary: so as an example I constructed $f: [0, 1] \to \mathbb R$ defined by $f(x) = x $ for $x \ne 0$; $f(0) = 1$. $\endgroup$ – Tom Collinge Oct 3 '17 at 14:02
  • $\begingroup$ but $\inf (f)=0$ $\endgroup$ – Stu Oct 3 '17 at 14:05
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Suppose there is no such $\epsilon$. Then for every $n \geq 1$ we can find $x_n \in [a,b]$ such that $f(x_n) < 1/n$. The sequence $x_n$ is bounded and hence by Bolzano-Weierstrass has a convergent subsequence $y_n$, whose limit $y$ lies in $[a,b]$ since $[a,b]$ is closed. Then by continuity of $f$:

$$ f(y) = f(\lim _{n\to \infty}y_n) = \lim_{n\to \infty} f(y_n) = 0 \,,$$ which is a contradiction.

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  • $\begingroup$ Thanks! Since you used the fact that the limit of the subsequence lies in [a,b], this wouldn't work if the domain was the reals right? $\endgroup$ – Analysis is fun Oct 3 '17 at 14:02
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    $\begingroup$ Both $\mathbb{R}$ and $[a,b]$ are closed, so that's not the important distinguishing feature. What's relevant is that $[a,b]$ is bounded, which allows us to conclude that there is a convergent subsequence in the first place. We couldn't do this if the domain was $\mathbb{R}$. Exercise: what goes wrong in the proof if the domain is $(a,b)$? $\endgroup$ – gj255 Oct 3 '17 at 14:04
  • $\begingroup$ If the domain is (a,b) we can't conclude that the limit of the convergent subsequence lies in (a,b) right? cause it could be a or b themselves? $\endgroup$ – Analysis is fun Oct 3 '17 at 14:10
  • $\begingroup$ Yes, that's right. $\endgroup$ – gj255 Oct 3 '17 at 14:11

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