# Positive continuous functions on closed intervals always greater than epsilon?

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?

• Extreme Value Theorem. – Hans Lundmark Oct 3 '17 at 13:46
• Can it be done without using the Extreme Value Theorem? We haven't covered this yet so am not sure about it – Analysis is fun Oct 3 '17 at 13:53
• You could do it from scratch, but that would most likely mean reinventing a proof of the EVT, more or less. – Hans Lundmark Oct 3 '17 at 13:56
• 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 $x \ne 0$; $f(0) = 1$. – Tom Collinge Oct 3 '17 at 14:02
• but $\inf (f)=0$ – Stu Oct 3 '17 at 14:05

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.
• 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 $(a,b)$? – gj255 Oct 3 '17 at 14:04