The question :

Let $f:[1,\infty)\to \mathbb{R}$ be a continuous function such that $\underset{x\rightarrow\infty}{\lim}f(x)=L$

Prove that the function is bounded.

My try :

By definition a continuous function $f:[1,\infty)\rightarrow\mathbb{R}$ 1. $f$ is continuous in $(1,\infty)$ i.e $\forall x_{0}>1\underset{x\rightarrow x_{0}}{\lim}f(x)=f(x_{0})$

  1. $f$ continuous at $1^{+}$ i.e $\underset{x\rightarrow1^{+}}{\lim}f(x)=f(1)$

as stated in the question's contitions :$ \underset{x\rightarrow\infty}{\lim}f(x)=L\iff\forall\varepsilon>0\,\exists M\in\mathbb{R}: x>M\rightarrow\left|f(x)-L\right|<\varepsilon$

Set $\varepsilon=\left|f(1)-L\right|$

Then exists $M$ such that $x>M\rightarrow\left|f(x)-L\right|\leq\left|f(1)-L\right|$ including when $x=1$

Set $\left|f(1)-L\right|=K$

This also implies that$ M\geq1$

Thus we get $x\geq1\rightarrow\left|f(x)-L\right|\leq K \iff x\geq1\rightarrow L-K\leq f(x)\leq L+K$

Thus $f$ is bounded.

However I feel that this proof doesnt work and I do not fully understand what I have done here actually (just tried to replicate the lecture notes of my professor)

I want to understand this question and the correct way to answer it. Assistance will be greatly appreciated.

  • 1
    $\begingroup$ You may already know that a continuous functoin on a compact (closed and bounded) interval would be bounded? $\endgroup$ – Hagen von Eitzen Apr 8 '16 at 19:28
  • $\begingroup$ OK, funny but 100% true story... I had EXACTLY this problem on a term paper (trimester exam) in eleventh grade in high school, somewhere in Eastern Europe where I grew up. Several years later I had the exact same problem on a PhD qualifying exam at one of the top universities in the U.S. ("qualifying exams" are given at the end of PhD courses, right before you are allowed to start working on a dissertation). The kicker? My masters' degree in math from my old country was NOT RECOGNIZED in the U.S.! Go figure. $\endgroup$ – mathguy Apr 8 '16 at 19:40
  • $\begingroup$ That is very interesting what you tell, which country if I could please ask ? also as an experienced person with this question could you please give me a hint on how to solve it ? $\endgroup$ – Pavel Penshin Apr 8 '16 at 19:48
  • $\begingroup$ @mathguy: That's an astonishingly gentle question for an analysis qualifying exam. $\endgroup$ – Bungo Apr 8 '16 at 19:49
  • $\begingroup$ Yep. I actually knew who wrote the problems for the exam, so I asked. The professor wanted to identify the candidates who didn't belong in the PhD program (which he did). Still seemed mighty odd to me - there ought to be other ways to achieve the same purpose. $\endgroup$ – mathguy Apr 8 '16 at 19:51

Since the comments seem to indicate that the OP is not familiar with the theorem that a continuous function $f$ is bounded on a compact set, here is a simple proof for the special case of a closed bounded interval $[a,b] \subset \mathbb R$.

Suppose for a contradiction that $f$ is not bounded on $[a,b]$. Therefore it must be either unbounded above or unbounded below. Without loss of generality, assume $f$ is unbounded above. (Otherwise replace $f$ with $-f$.)

Now divide the interval into two subintervals, $[a + (a+b)/2]$ and $[(a+b)/2, b]$. Now $f$ must be unbounded on one of these subintervals. Repeating this procedure, we identify a sequence of closed bounded intervals $[a,b] = I_0 \supset I_1 \supset I_2 \supset \cdots$ such that the length of $I_n$ is $(b-a)/2^n$, and $f$ is unbounded on each of these intervals.

This means that we can choose points $x_0 \in I_0$, $x_1 \in I_1$, $x_2 \in I_2$, etc. such that $f(x_n) > n$ for every $n$.

Now each $I_n$ contains every $x_k$ for $k \geq n$, and from this we can easily conclude that $x_k$ converges to some limit $x$. Since each $x_n$ is in $[a,b]$ and $[a,b]$ is closed, it contains all of its limit points, hence $x\in [a,b]$.

By continuity of $f$, we must have $$\lim_{n \to \infty}f(x_n) = f(x)$$ but this is impossible since $f(x_n) > n$ for every $n$.

Our assumption that $f$ is unbounded on $[a,b]$ is untenable, so $f$ must be bounded after all.


Take $\epsilon=1$ then there is $N$ so that if $x>N$ we have $|f(x)-L|<1$.

So the function is bounded on $[N,\infty)$.

On the other hand, every continuous function defined on a closed bounded interval is bounded. So $f$ is bounded on $[1,N]$.

Since $f$ is bounded on $[1,N]$ and $[N,\infty)$ we have $f$ is bounded on $[1,\infty)$ as desired.

  • $\begingroup$ so the first thing that the function is bounded on the interval $[N,\infty)$ is understood because of the limit definition. The second part of $f$ being bounded on a closed interval $[1,N]$ is it from the definition of a continuous function ? and the last part is just a union $[1,N]\cup[N,\infty)$ Please let me know if I got it right. $\endgroup$ – Pavel Penshin Apr 8 '16 at 19:44
  • $\begingroup$ @user313448: The fact that a continuous function $f$ is bounded on $[1,N]$ is not a consequence of the definition of continuity. It is a consequence of two important theorems: a closed, bounded set (such as $[1,N]$) is compact, and a continuous function is bounded on any compact set. $\endgroup$ – Bungo Apr 8 '16 at 19:55
  • 1
    $\begingroup$ If a function is bounded on $A$ and $B$ then it is also bounded on $A \cup B$ - just use the higher of the two bounds as the bound over the whole union. Then: closed bounded intervals are compact, the image of a compact set under a continuous function is compact, and compact sets in $\mathbb R$ are bounded. These statements must all be covered in your calculus course, or else the problem is unfair. (Perhaps all they show in class/in the book is that the image of a closed and bounded interval is closed and bounded...) $\endgroup$ – mathguy Apr 8 '16 at 19:57
  • $\begingroup$ Sorry, I was away, thanks to the other guys for covering all that other stuff so well. $\endgroup$ – Jorge Fernández Hidalgo Apr 8 '16 at 20:02

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