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Say $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$, what's the most concise way you know of to show that it's bounded?

I was thinking let $A=\{u : f(x) \text{ is bounded on }x<u\}$ Is there a way to show that $\sup(A)=b$?

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I think you should use a different variable in your set $A$ other than $a$, since that's already the left endpoint of the given interval. – Quinn Culver Jun 3 '11 at 14:10
Possible duplicate:… -- see my answer… – lhf Jun 3 '11 at 14:27
Assume $f$ is not bounded then take the sequence $x_k$ with $f(x_k) \rightarrow \infty$. Certainly $x_k$ has a convergent subsequence as $[a,b]$ is compact but $f(x_{k_n})$ doesn't converge against $f(x^*)$. Done – Listing Jun 3 '11 at 14:30
up vote 1 down vote accepted

Yes, this is a good idea. Since $A \subseteq [a,b]$, it has a supremem, say $c = \sup A$. If $c<b$, then since $f$ is continuous at $c$, there is $\delta>0$ such that $|f(x)-f(c)|< \frac{1}{2}$ whenever $x \in (c, c+\delta)$. Then $|f(x)| < |f(c)| + \frac{1}{2}$ for every whenever $x \in (c, c+\delta)$. But that means $f$ is bounded on $[a, c+ \delta]$, contrary to $c = \sup A$.

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Aha, yes, this makes far more sense now :) – Freeman Jun 3 '11 at 14:32

Here is a "from scratch" attempt. Suppose $f$ were to be unbounded on $[a,b]$ yet continuous there. Then $f$ is unbounded on one of $[a,(a + b)/2]$ or $[(a + b)/2, b]$. Denote this interval by $I_1$. Keep subdividing in this fashion to obtain a sequence of intervals $I_n$ so that $I_{n+1}\subseteq I_n$ for all $n$ and so that the length of $I_n$ is $$(b-a)/2^n.$$

Now write $I_n = [a_n, b_n]$ for each $n$. The sequence $a_n$ is increasing and bounded by $b_1$ so it converges to a limit $l$. Since the lengths of the $I_n$ converge to zero, we have $b_n\rightarrow l$. By continuity, $f(l) = \lim f(a_n) = \lim f(b_n).$ By continuity, we can choose $\delta > 0$ so that $f(x) < f(l) + 1$ for $l - \delta < x < l + \delta$.

Pick $n$ so that $I_n \subseteq (l - \delta, l + \delta)$. The function $f$ must be bounded on $I_n$, a contradiction of our construction.

The arabesque executed here has a feel very similar to that of the proof of Heine-Borel theorem.

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Here's the most concise proof I know: The continuous image of a compact set is itself compact. Now, $[a,b]$ is compact so $f([a,b])$ is compact. By the Heine-Borel theorem, an interval in $\mathbb{R}$ is compact if and only if it is closed and bounded. Therefore, $f([a,b])$ is bounded.

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Well, here the difficulty is transferred to proving the Heine-Borel theorem, which is really nontrivial. – Mark Jun 3 '11 at 15:48
@Mark Agreed, but the OP did request "concise"... – ItsNotObvious Jun 3 '11 at 16:43

The distance (function) of points from the set $\{0\}$ is uniformly continuous, so it is bounded on any compact set, so the compact set can be enclosed in a ball with a radius of that distance. Consider the compact set $f([0,1])$.

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