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I have this question: If $f:[0,\infty) \to \mathbb R$ is continuous and $$\lim _{x\to \infty }\:f\left(x\right)\:=\:L$$ then $f$ has a maximum or a minimum on the interval.

I've already proved that this function (on the same terms and limit) is bounded on the interval. I understand why this is true - because the limit can be below the function and then it won't have minimum or above the function and then it wont have maximum. I don't know how to write it formally.

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    Maybe by contradiction? Assume it doesn't have (at least) a global minimum or maximum, which would mean it diverges on $x \rightarrow \infty$. – 2017-01-14
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    @OpenBall Doesn't that happen because the interval is open on $0$? OP has a closed interval. Sorry, I'm not that well-acquainted with formality. – 2017-01-14
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    @O.VonSeckendorff I removed my example because it's technically incorrect, but you get the idea. Of course, OP's interval is closed, but I was objecting to your suggestion that not having global extrema would mean that the function diverges as $x\to \infty$. – 2017-01-14
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    OP, is $\mathbb{R}$ the image? – 2017-01-14
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    $[0, \infty)$ is a closed interval, not half-closed. – 2017-01-14

2 Answers 2

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If $f$ is not a constant function there exists $a \geqslant 0$ such that $f(a) \neq L$. Let $M=f(a)$ and let us suppose that $M>L$ (an analoguous reasoning would apply if $M0$.

Since $\lim_{x\to \infty } f(x)=L$ you have the existence of $A > 0$ such that for all $x \geqslant A$, $f(x) \in [L-\epsilon, L+\epsilon]$. Then for all $x \geqslant A$, $f(x) \leqslant L+\frac{M-L}{2} < L + M -L = M$.

On $[0, A]$, $f$ is continuous then has a maximum $B \geqslant M$ (because $a \in [0, A]$). Consequently $f$ has a maximum on $[0, \infty)$.

If $f$ is constant, it should be equal to $L$ and $f$ has a maximum and a minimum.

If $M

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Take $\phi\colon[0,1) \to [0,+\infty)$, $\phi(x)=1/(1-x)-1$ (any continuous increasing unbounded function would work) and define the function $g\colon [0,1]\to \mathbb R$ by $$ g(t) = \begin{cases} f(\phi(t)) & \text{if $t\in[0,1)$}\\ L & \text{if $t=1$} \end{cases}. $$ Clearly $g$ is continuous on $[0,1]$ hence by Weierstrass theorem has minimum and maximum. Either the minimum or the maximum is not achieved in $t=1$. Being $\phi$ increasing the corresponding point $\phi(t)$ is a minimum or maximum for $g$.