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?

  • 2
    $\begingroup$ When you write $\lim_{t\to\infty}$, do you mean $\lim_{t\to\pm\infty}$? $\endgroup$ Jul 6, 2018 at 9:54
  • $\begingroup$ I would claim that the supremum exists (be careful, but it works since you are bounded from above and you converge to $0$). Then take a sequence converging to this supremum and by continuity this makes sure that the supremum and the maximum coincide $\endgroup$ Jul 6, 2018 at 9:57
  • $\begingroup$ @JoséCarlosSantos No just positive one. I think I should mention it, $t \in [0, \infty)$ $\endgroup$
    – Manish
    Jul 6, 2018 at 9:57
  • $\begingroup$ @StanTendijck Can you please elaborate a little, especially the argument regarding supremum and maximum coinciding? Also, it should be same as local maximum, right? $\endgroup$
    – Manish
    Jul 6, 2018 at 9:59
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    $\begingroup$ You can attain the supremum as a limit of a sequence, i.e., there exist $a_n$ such that $\lim f(a_n) = S$ (with $S$ the supremum). You can argue by convergence to $0$ that the supremum is 'attained' at a bounded number, i.e. $a_n$ does not converge to $\infty$. Hence, there exist an $a$ which occurs at the limit of $a_n$. Moreover, by continuity you can interchange the limit and $f$ such that $\lim f(a_n) = f(a)$. Is this better? $\endgroup$ Jul 6, 2018 at 10:11

2 Answers 2


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.


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 $0$.

  • $\begingroup$ I can understand the logic properly now. I think both the answers convey the same explanation. It's sad that both can't be tagged accepted :( $\endgroup$
    – Manish
    Jul 6, 2018 at 10:17
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    $\begingroup$ I accept the one which has more appeal to you. It's as simple as that. $\endgroup$ Jul 6, 2018 at 10:21

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