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Thread: D bounded => f is bounded on D

  1. #1
    Sep 2009

    D bounded => f is bounded on D

    I need some help with a proof
    Assume $\displaystyle f\to \mathbb{R}$ 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
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  2. #2
    Jun 2009
    Precompactness is the key. Let $\displaystyle f$ be uniformly continuous on $\displaystyle D\subseteq \mathbb{R}^n$. You get d>0 such that the oscilation of $\displaystyle f$in each ball of radius d is smaller than 1. Every bounded subset in $\displaystyle \mathbb{R}^n$ is precompact, hence you can find $\displaystyle x_1,\ldots, x_n\in D$ such that $\displaystyle D\subset\cup_{i=!}^{n}B(x_i,d)$. Now you can easily get a bound for $\displaystyle |f|$.
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