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$ Problem $

Let $ f $ be a real uniform continuous function on the bounded set $E$ in $\mathbb{R}$ . Prove that $f$ is bounded on $E$.

$ Proof $

Let $f$ be a real uniformly continuous function on the bounded set $E$ in $\mathbb{R}$. We want to show that $f$ is bounded in $E$ ,i.e. $f(E)$ is bounded .

Since $E$ is bounded , we see that $\overline E$ is bounded and closed in $\mathbb{R}$ , which implies $\overline E$ is compact .

Therefore $f(\overline E)$ is compact ,which implies $f(\overline E)$ is bounded , which implies $f(E)$ is bounded because $f(E)\subset f(\overline E)$.

$ Doubt$

What is the use of uniform continuity ? In my proof i have used the results for continuity only.

Useful results suggested by Henno Brandsma--

Continuous extension theorem

Suppose f is uniformly continuous on a dense subset $B$ of $A$. Then there is a unique function $F$ continuous on $A$ such that $F(b) = f(b)$ for every $b\subset B$

In special case of $\mathbb{R}$ ,this theorem can be written as --

$f$ is uniformly continuous on $(a,b)$ if and only if it can be extended at the endpoints $a$ and $b$ such that the extended function $\overline f$ is continuous on $[a,b]$.

Jack J.
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Bluey
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1 Answers1

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The proof is correct provided you know the following fact:

If $f: E \to \mathbb{R}$ is uniformly continuous, then it can be continuously extended to $\overline{f}: \overline{E} \to \mathbb{R}$.

This is a non-trivial fact you should prove first or have a citation for from the text you're using.

There also is a direct proof from $E$ being totally bounded: Let $\varepsilon = 1$ and find $\delta >0$ such that

$$|x-y| < \delta \implies |f(x) - f(y)| < 1$$

by uniform continuity.

Then there are finitely many $x_1,\ldots x_N \in E$ such that

$$\forall x \in E: \exists 1 \le i \le N: |x_i - x| < \delta$$

by $E$ being totally bounded. Then note that $f[E]$ is bounded by $-1 + \min_{1 \le i \le N} f(x_i)$ from below and $1+\max_{1 \le i \le N} f(x_i)$ from above, say.

Henno Brandsma
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  • Thanks for answering. I didn't know about the continuous extension. I will try to prove it by myself. I think there is a question on it in Rudin's exercise. – Bluey Jun 30 '18 at 08:19
  • @blueboy You need the continuous extension in your idea as otherwise $f[\overline{E}]$ makes no sense. The direct proof is easier, and just uses the definition. – Henno Brandsma Jun 30 '18 at 08:21
  • Ok . Thanks.... – Bluey Jun 30 '18 at 08:22