Let $f:\mathbb R\to\mathbb R$ be a continuous and not identically zero function such that $$\lim_{x\to\pm\infty} f(x)=0.$$
Under which other extra assumptions I can say that $f$ has an absolute maximum which is strictly positive? I mean that there exists $\bar{x}\in\mathbb R$ such that $f(\bar{x})\ge f(x)$ for all $x\in\mathbb R$ and $f(\bar{x})>0$.
I am pretty convinced that assuming $f$ nonnegative, I can get the result. Anyone could please tell me if the existence of a strictly positive absolute maximum can be deduced assuming any other (possibly "lighter") assumption?
You could say:
This is both necessary and sufficient. the necessity is obvious so let's prove the sufficiency: To prove it you could take S to be the set of all such $f(x)$ and then prove $sup(S) \in S$, which is the same thing as saying $f$ has a positive maximum:
If $\sup(S)$ is in S then we are done. If not, $\sup(S)$ is a limit point of $S$, so you can construct a sequence in $S$ that converges to $sup(S)$:
$\lim_{i \to \infty}f(x_i) = \sup(S)$
Now take the sequence of their inputs:
$x_1, x_2 ,\dots$
If it converges to $\alpha$ then $sup(S) = f(\alpha)$ (because $f$ is continuous) and we are done.
If it diverges you have 2 cases:
$1.$It diverges to infinity, which means $sup(S) = 0$ (because of the assumption: $f(x)$ goes to zero as $x$ goes to infinity) which is not possible, because all member of S are greater than zero.
$2.$it diverges but is bounded, so you have a subsequence that converges(Bolzano-Weierstrass Theorem):
$\lim_{i \to \infty}u_i = U$
this would mean(Because $f$ is continuous):
$\lim_{i \to \infty}f(u_i) = f(U)$
So since the sequence:
$f(x_1),f(x_2),...$
Is convergent to $f(S)$ and since any subsequence must converge to the same limit, we have $sup(S) = f(U) \in S$ ($sup(S)>0$)