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I have to prove or find an counterexample of the statement: if $f$ is uniformly continuous on $D$ then $f$ is bounded on $D$.

I think this statement is not true since if $f(x)=x$ is uniformly continuous on $D=\Bbb R$ but not bounded on it. Am I right to assume that $D$ can be the whole $\Bbb R$? Are there any other classical examples?

Thank you, Klara

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Well, does the original question put any restrictions on D? – Mike B Oct 28 '12 at 2:12
@ Mike No it does not – Klara Oct 28 '12 at 2:13
@Klara: What kind of counterexample are you looking for? As it stands, you've already answered the given question. – wj32 Oct 28 '12 at 3:57
@ WJ32 I was looking for other counterexamples and If it legit to have D=R, I think that D=R is the only scenario when uniform continuity is not bounded.Do you agree – Klara Oct 28 '12 at 16:11
up vote 0 down vote accepted

For $f(x)=x$ and $D=(-\infty,a) \cup (b,+\infty), \ a<b,$ $f$ is uniformly continuous on $D$ and not bounded on it.

Maybe you are looking for a bounded set $D$ s.t. $f$ is uniformly continuous and unbounded on $D$. You cannot find such $f$.

Or (equivalently) if you are asking:

What are the conditions on $D$ such that if $f$ is uniformly continuous on $D$ then $f$ is bounded?

then the answer is $$\text{$D$ must be bounded.}$$

For the proof we will use the following

If f is uniformly continuous on $D$ then f is Cauchy-continuous on D.

Or, if you want, if $f$ is uniformly continuous on $D$ then sends Cauchy sequences on $D$ to Cauchy sequences.

Now suppose that $D$ is bounded and $f$ is uniformly continuous and unbounded on $D$. Then $\forall n \in \mathbb{N}$ we can find an $x_n \in D$ such that $|f(x_n)|>n$. (Obviously) $f(x_n)$ does not have a Cauchy subsequence. Since $D$ is bounded and $(x_n)_{n \in \mathbb{N}} \subseteq D \Rightarrow (x_n)_{n \in \mathbb{N}}$ is bounded so from Bolzano–Weierstrass theorem it has a Cauchy subsequence, lets say $(x_{k_n})_{n \in \mathbb{N}} $. So we found a Cauchy sequence $(x_{k_n})_{n \in \mathbb{N}} $ such that $f(x_{k_n})_{n \in \mathbb{N}} $ is not Cauchy. Contradiction.

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