Show the existence of a global maximum of a continuous function with unbounded domain I am given a function $f(t) \in \mathbb{R}$ which is continuous; bounded above by $M$ and below by $0$. $f$ is differentiable everywhere except at $f=0$. Also, $\lim_{t \to \infty} f  = 0$ and $t \in [0, \infty)$.
How do I show that a global maximum/supremum exists and where it exists?
I thought of dividing this into 3 cases: when function is non-decreasing, non-increasing and when there exists a local extremum. 
$(i) $ $f$ can not be non-decreasing since it converges to $0$, except when $f(0) = 0$ which will be a trivial case. 
$(ii)$ If $f$ is non-increasing then, $f(0)$ is the global maxima as $f(0) \ge f(t) \forall \ t$. 
$(iii)$ That leaves us with the case when there exists at least $1$ local extremum. Now, if the domain were bounded, I could have used Extreme value theorem and say that global maximum exists either at $t=0$ or at a local maximum, but the domain is unbounded in my case, so that can't be used. However I do have information about the limit, which could be helpful.
Intuitively, I still think the result will be same as that of Extreme value theorem, but I want to prove it using proper results. How should I modify my process?
 A: If $f:[0,\infty)\to[0,\infty)$ is continuous and $\lim_{t\to\infty}f(t)=0$, then you can prove the existence of a global maximum. 
If $f\equiv 0$, the statement is trivial. Otherwise, there exists a point $t_0\in\mathbb R$ such that $f(t_0)>0$. Further, from the limit you can conclude the existence of $R>0$ such that $f(t)<f(t_0)$ for all $t>R$. Since $I:=[0,R]$ is compact and $f$ continuous there exists a global maximum of $f\mid_I$ which is a global maximum of $f$.
To find the extreme point, you have to do it in two steps.
1) Consider the extreme points of $f\mid_{(0,\infty)}$ using $f'$.
2) Compare the maxima on $(0,\infty)$ with the value at $0$.
You can have 3 cases:
a) If there are no maxima on $(0,\infty)$, then the global maximum has to be at $0$.
b) If there exist maxima on $(0,\infty)$ but their heights are below $f(0)$, then the global maximum is at $0$.
c) If there exist maxima on $(0,\infty)$ and the highest value is above $f(0)$, then the maxima with the highest value are the global maxima.
A: If $f$ is the null function, then it attains its maximum everywhere. Otherwise, tak $t_0>$ such that $f(t_0)>0$. Then there is a $R>0$ such that $t>R\implies f(t)<f(t_0)$. By the extreme value theorem, the restriction of $f$ to $[0,R]$ has a maximum in some $t_1$ in that interval and since outside the interva $f$ always takes values smaller that $f(t_0)$, $f$ attains its global maximum at $t_1$. Therefore, either $f'(t_1)=0$ or the maximum is attained at $0$.
