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

Results 1 to 2 of 2

- Dec 9th 2008, 07:12 PM #1

- Joined
- Dec 2008
- Posts
- 13

- Dec 9th 2008, 09:57 PM #2
Ok, so lets just establish a few things, the metric $\displaystyle d(p,p')$ in $\displaystyle \mathbb{R}^n$ is $\displaystyle |p-p'|$. Since assuming that $\displaystyle K\subset\mathbb{R}$ we have that $\displaystyle \phi:K\to\mathbb{R}$ is a real function. So for $\displaystyle \phi$ to be uniformly continous we must have that for every $\displaystyle \varepsilon>0$ there exists a $\displaystyle \delta>0$ such that $\displaystyle |p-p'|<\delta\implies|\phi(p)-\phi(p')|<\varepsilon$.

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