Show that a function with a limit either has maximum or supremum equals limit

Question

Not sure exactly why this statement from a (real analysis) textbook is true:

Let $f:[0,+\infty)\to\mathbb{R}$ be continuous and let $\lim_{x\to+\infty}f(x)=L\in\mathbb{R}$. Then $f$ either attains maximum* or $\sup_{x\in[0;+\infty)}f(x)=L$.

*: Equivalent to saying there exists $y\in[0,+\infty)$ such that $f(y) = \sup_{x\in[0;+\infty)} f(x) = \max_{x\in[0,+\infty)}f(x)$.

What would the proof for this statement look like? The extreme value theorem doesn't apply since the interval isn't bounded. It's not entirely clear to me why it is true for examples like f(x) = 1/x. Any insight is appreciated!

Answer 1

Hint :

1. Prove that $f$ is bounded on $[0,+\infty)$. Deduce that $f$ has a finite supremum over $[0, +\infty)$.

2. If $f$ does not attains its supremum, show that the supremum has to be the limit $L$.

Answer 2

For the supremum to be defined, $f$ needs to be bounded above, so we need to establish this first. Your original thought to use the Extreme Value Theorem was on the right track; we can partition the domain of $f$ into a closed interval and open interval, use the Extreme Value Theorem to show $f$ is bounded above on the closed interval, and use the fact that $\lim_{x\to\infty}f(x)=L\in \mathbb{R}$ to show $f$ is bounded above on the open one. Specifically, by the epsilon-delta criterion of a limit, for say $\epsilon=1$ we know there exists some $x_\epsilon\in[0,\infty)$ such that for all $x>x_\epsilon$, we have $|f(x)-L|<\epsilon = 1$, hence the restriction of $f$ to $(x_\epsilon,\infty)$ is bounded above by $|L|+1$. Apply the Extreme Value Theorem to show the restriction of $f$ to $[0,x_\epsilon]$ is bounded above by some $M\in \mathbb{R}$. Thus $f$ is bounded above by $\max (|L|+1,M)$.

Since the image of $f$ is bounded above, its supremum exists, and the image of $f$ either contains its supremum or does not contain its supremum.

• If $\sup f([0,\infty))\in f([0,\infty))$, then $f$ has a global maximum by definition.
• If $\sup f([0,\infty))\notin f([0,\infty))$, then $f$ does not have a global maximum, so we need to show that $\sup f([0,\infty))=L$. It suffices to prove the existence of some sequence $\{x_i\}\subset [0,\infty)$ such that $x_i\to \infty$ and $f(x_i)\to \sup f([0,\infty))$, because then the equality $L=\sup f([0,\infty))$ follows immediately from the uniqueness of limits. We now argue such a sequence exists. Since the supremum is the $\textit{least}$ upper bound, there exists a sequence $y_i\in f([0,\infty))$ such that $y_i\to\sup f([0,\infty))$. By construction, each $y_i=f(x_i)$ for some $x_i\in[0,\infty)$. All that's left to do is show $x_i\to \infty$. Suppose for the sake of contradiction that $x_i\not\to \infty$, then this implies $\{x_i\}$ has a bounded subsequence, (hint: any of its subsequences will automatically be bounded below by $0$, use the definition of $x_i\not\to \infty$ to obtain one that's bounded above), and so $\{x_i\}$ also has a convergent subsequence by the Bolzano-Weierstrass Theorem. Call this convergent subsequence $\{x_{i_k}\}$ and call its limit $a_0$. Since $[0,\infty)$ is closed and each $x_{i_k}\in [0,\infty)$, we have $a_0\in[0,\infty)$. Observe by the continuity of $f$ that we have $f(x_{i_k})\to f(a_0)$. Since every subsequence of a convergent sequence converges to the same limit as the parent sequence, we also have $f(x_{i_k})\to \sup f([0,\infty))$. By the uniqueness of limits, $\sup f([0,\infty))=f(a_0)\in f([0,\infty))$. But this is a contradiction to the fact that $\sup f([0,\infty))\notin f([0,\infty))$, hence our supposition that $x_i\not\to \infty$ was incorrect.

Also, if you are interested in examples, $f(x)=-\frac{x}{x^2+1}$ illustrates that it's possible for both $f$ to have a global maximum and for $\sup f([0,\infty))=L$, and $f(x)=\frac{1}{x-1}$ illustrates why we would require $f$ to be continuous in our hypothesis.

Answer 3

First, as pointed by TheSilverDoe, prove that $f$ is bounded on $[0,+\infty)$. Since this is easy, we omit the details. So f has a finite supremum over $[0,+\infty)$. Now write $\sup f([0,+\infty))=A$.

If $A<L$, then by $f(x)\le A$, we get $L=\lim_{x\to+\infty}f(x)\le A<L$. A contradiction.

If $A=L$, then the conclusion is true.

If $A>L$, then there exists a point $x_0$ such that $f(x_0)>L$. Since $L=\lim_{x\to+\infty}f(x)$, there exists a sufficiently large $X>x_0$ such that for $x>X$, $f(x)<L+\frac{f(x_0)-L}{2}<f(x_0)$. Now $f$ is continuous on $[0,X]$, so it has the maximum $M$ on $[0,X]$. Since for $x>X$, $f(x)<f(x_0)\le M$, the value $M$ is the whole maximum of $f$ on $[0,+\infty)$.