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?

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.