Function $f$ receives a maximum in $[0,\infty)$

Question

Let $f$ be a continuous function in $[0,\infty)$ such that $\lim\limits_{x \to \infty} f(x)=0$.

We assume that there is a $c \geq 0$ such that $f(c)>0$.

Prove or disprove Does $f$ has a global maximum in $[0,\infty)$.

Usually with these questions I use the limit definition to find an interval to use the Weierstrass theorem. However here I find it harder and couldn't figure this out. I do think it is a true statement.

Thanks!

Answer 1

Let $\varepsilon=\frac{1}{2}f(c)>0$. Since $\lim_{x\to\infty}f(x)=0$, there exists some $x_0>c$ such that if $x>x_0$ then $|f(x)|<\varepsilon$.

Now because $f$ is continuous it has a maximum on $[0,x_0]$, call it $M$. Then $M\geq f(c)$ by the choice of $x_0$, and since $f(x)<\frac{1}{2}f(c)$ for all $x>x_0$ it follows that $M$ is a global maximum for $f$ on $[0,\infty)$.

Answer 2

It does. To see this, there must be a whole number $n$ such that for $x > n$ it follows that $f(x) < \dfrac{f(c)}{2}.$ On the compact interval $[0, n],$ $f$ attains a maximum $f(x_0) \geq f(c).$ QED

Answer 3

If the limit is zero then there's some interval $(N,\infty)$ on which the function is strictly less than one in absolute value, by definition.

Now think about what you can say about the function's behavior on $[0,N]$.