Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi.

I have come across a proof which I understand almost completely, except for one part:

THEOREM: If $f$ is uniformly continuous on a bounded interval $I, [a,b]$ then $f$ is also bounded on $I$.

PROOF: Fix an $\epsilon > 0$, for instance $\epsilon = 1$. Since $f$ is uniformly continuous, there is a $\delta > 0$ such that:

$|f(x_1) - f(x_2)| < \epsilon = 1$ when $x_1, x_2 \in I$ and $|x_1 - x_2| < \delta$

Divide $I$ into $N$ intervals, $I_1, . . ., I_N$, where $N$ is chosen so that $\frac{b-a}{N} < \delta$.

Let $z_i$ be the center point of $I_i$. For each $i$ and $x \in I_i$, $|x - z_i| < \delta$, and then we have:

$|f(x)| = |f(x) - f(z_i) + f(z_i)| \leq |f(x) - f(z_i)| + |f(z_i)| \leq 1 + |f(z_i)|$. Then for $x \in I_i$,

$|f(x)| \leq 1 + max_{1 \leq i \leq N}\{|f(z_i)|\}$.

Let $M = max_{1 \leq i \leq N}\{|f(z_i)|\}$. Then $|f(x)| \leq 1 + M$

QED

OK, so the one thing I am a bit unsure of here, is when we write:

Let $M = max_{1 \leq i \leq N}\{|f(z_i)|\}$.

How is it that we know for sure that each $|f(z_i)|$ is also bounded? I see how the presence of a maximum value completes the proof, but why is it not possible that we have an $|f(z_i)|$ which is unbounded?

If anyone could explain this to me I would greatly appreciate it!

Also, for what it's worth, I tried to solve this my own way, but I am not sure if the proof is rigorous enough (it's much simpler!). It goes as follows:

PROOF BY CONTRADICTION

Suppose $f$ is not bounded on $I$. Then, for each $M > 0$, we have $|f(x)| > M$ for some $x \in I$. However, since $f$ is uniformly continuous, for every $\epsilon > 0$ there exists a $\delta > 0$ such that

$|f(x) - f(y)| < \epsilon$ when $x, y \in I$ and $|x - y| < \delta$

And it follows from this that:

$|f(x)| < \epsilon + f(y)$

Which is a contradiction if $|f(x)|$ is greater than any $M > 0$.

QED

If anyone also can let me know if my proof is OK, I would also be very grateful!

share|improve this question

closed as off topic by GH from MO, Yemon Choi, Gerald Edgar, fedja, S. Carnahan Dec 4 '11 at 15:10

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

2  
This site is for research level questions, please read the FAQ. –  GH from MO Dec 4 '11 at 9:31
1  
GH - you're right. Thanks. I didn't read the FAQ - guess I should have. I was recommended this site by a friend, and he didn't inform me that this site was primarily for research level questions. I will keep that in mind from now on. –  krje1980 Dec 4 '11 at 11:09
2  
krje1980: I think such a question would be a very good fit for math.stackexchange.com. I'm sure that you'd get a much warmer welcome and much more extensive answers there. –  Theo Buehler Dec 4 '11 at 12:08
    
Thank you very much for the tip, Theo. I will check that out. –  krje1980 Dec 4 '11 at 13:27
    
A version of this question is now also on math.SE: math.stackexchange.com/questions/88257 –  Theo Buehler Dec 4 '11 at 13:38
add comment

2 Answers 2

The theorem you mention is kind of strange. You don't need to assume uniform continuity, it is enough to suppose that your function $f$ is continuous: every continuous function on a compact subset of $\mathbb R$ is automatically uniformly continuous. Then, what you are trying to prove is that continuity on a compact $\Rightarrow$ boundedness (so called, extreme value theorem, see http://en.wikipedia.org/wiki/Extreme_value_theorem where a the standard proof is outlined).

share|improve this answer
1  
Thanks. But in this case, we only know the set is bounded, not compact. –  krje1980 Dec 4 '11 at 9:54
    
[a,b] is compact, end of story. –  Antoine Levitt Dec 4 '11 at 10:51
    
You wrote $[a,b]$, this is closed and bounded $\Rightarrow$ compact. –  Andrei MF Dec 4 '11 at 10:52
    
Oh, yeah. That's right! Ha ha. Thanks a lot :) –  krje1980 Dec 4 '11 at 11:07
    
I assumed that "$I,[a,b]$" should be read as "$I \subseteq [a,b]$", rather than as "$I = [a,b]$". –  Toby Bartels Dec 24 '11 at 6:40
add comment

I'm afraid that I don't like your proposed proof. You derive a bound on $f(x)$, namely $\epsilon + f(y)$, but this is not fixed. Although you may choose any positive $\epsilon$ you wish (which then gives you $\delta$), $y$ must be chosen to be within $\delta$ of $x$. So as you vary $M$, you vary $x$ (to keep $f(x) > M$), but then (to keep it close enough to $x$) you vary $y$, and it seems possible that $f(y)$ would grow fast enough that $\epsilon + f(y) > M$ would be maintained.

share|improve this answer
    
Thanks. Appreciate the input! –  krje1980 Dec 4 '11 at 11:08
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.