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.


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.

  • $\begingroup$ 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? $\endgroup$ – Rakka Feb 13 '13 at 5:06
  • $\begingroup$ 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, $f(x_0)>0$. $\endgroup$ – Clayton Feb 13 '13 at 5:07

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