intotoanalysis

**sisa** · December 9th 2008, 07:12 PM

Let K be a bounded non-empty subset of R.Let f:K->R be a uniformly continous function.Show that f is bounded.

**Mathstud28** · December 9th 2008, 09:57 PM

Originally Posted by sisa

Let K be a bounded non-empty subset of R.Let f:K->R be a uniformly continous function.Show that f is bounded.

Ok, so lets just establish a few things, the metric $d(p,p')$ in $\mathbb{R}^n$ is $|p-p'|$ . Since assuming that $K\subset\mathbb{R}$ we have that $\phi:K\to\mathbb{R}$ is a real function. So for $\phi$ to be uniformly continous we must have that for every $\varepsilon>0$ there exists a $\delta>0$ such that $|p-p'|<\delta\implies|\phi(p)-\phi(p')|<\varepsilon$ .

1. Now assume the converse and state that $\phi(p)$ is unbounded. By the fact that we are mapping to the reals let us fix $p'$ and let $\phi(p')$ be finite. Now since $\phi(p)$ is unbounded in $K$ for every $M\in\mathbb{R}$ we may find a $p\in K$ such that $M<|\phi(p)|$ , so for every $M_0\in\mathbb{R}$ we can find a $p\in K$ such that $M_0<|\phi(p)-\phi(p')|$ which violates $|\phi(p)-\phi(p')|<\varepsilon$

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