D bounded => f is bounded on D

**dannyboycurtis** · November 1st 2009, 08:42 PM

I need some help with a proof
Assume $f<img src=$ \to \mathbb{R}" alt="f

\to \mathbb{R}" /> is uniformly continous on D.
Show that D being bounded implies f is bounded on D.

All I can think of is to assume f is unbounded and show that it leads to a contradiction, but not sure how to proceed with the proof.
any suggestions? tHanks

**Enrique2** · November 2nd 2009, 02:09 AM

Precompactness is the key. Let $f$ be uniformly continuous on $D\subseteq \mathbb{R}^n$ . You get d>0 such that the oscilation of $f$ in each ball of radius d is smaller than 1. Every bounded subset in $\mathbb{R}^n$ is precompact, hence you can find $x_1,\ldots, x_n\in D$ such that $D\subset\cup_{i=!}^{n}B(x_i,d)$ . Now you can easily get a bound for $|f|$ .

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