Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a result (or source) that says if I have a function $f$ defined on some set $S$ and that $f(y)>0~\forall y\in S$, then there exists a constant $K>0$ such that $f(y)\geq K~\forall y\in S$?

Intuitively I would take $\min_{y\in S} f(y)>0$ and take $\epsilon$ away, but I'm not too sure.

share|cite|improve this question

No: consider the function $f(x)=e^{x}$ and $S=\Bbb R$. Then $e^{x}$ does not have a minimum. Note that if you know $S$ is compact and $f$ is continuous, then you can say there is a minimum and it has to be greater than zero.

share|cite|improve this answer
Thank you @Clayton. What/where is the result that if $S$ is compact and $f$ continuous then $f$ is bounded below by some non-negative constant? – Rakka Feb 13 '13 at 5:06
Ah, that statement is true given your assumptions, in particular, $f(y)>0$ for all $y\in S$. Then we know it exists since $f$ takes a minimum, say $f(x_0)$, and by hypothesis, $f(x_0)>0$. – Clayton Feb 13 '13 at 5:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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