# If f is a uniformly continuous function on a bounded set S, then f is a bounded on S

• Mar 17th 2010, 05:32 PM
Pinkk
If f is a uniformly continuous function on a bounded set S, then f is a bounded on S
Here's my attempt:

Assume to the contrary that $\displaystyle f$ is a uniformly continuous function on a bounded set $\displaystyle S$ but $\displaystyle f$ is unbounded on $\displaystyle S$. Then there exists a sequence $\displaystyle (s_{n})\in S$ such that $\displaystyle |f(s_{n})|\ge n$ for all $\displaystyle n\in \mathbb{N}$. Since $\displaystyle (s_{n})$ is a bounded sequence, there exists a subsequence $\displaystyle (s_{n_{k}})$ that is a Cauchy sequence. Since $\displaystyle f$ is uniformly continuous and $\displaystyle (s_{n_{k}})$ is a Cauchy sequence in $\displaystyle S$, then $\displaystyle (f(s_{n_{k}}))$ is a Cauchy sequence. So $\displaystyle (f(s_{n_{k}}))$ converges to some real number. But $\displaystyle |f(s_{n_{k}})|\ge n_{k}$ for all $\displaystyle n_{k}\in \mathbb{N}$ as well, which is a contradiction. Q.E.D.

Is this correct? Thanks.
• Mar 17th 2010, 06:02 PM
Drexel28
Quote:

Originally Posted by Pinkk
Here's my attempt:

Assume to the contrary that $\displaystyle f$ is a uniformly continuous function on a bounded set $\displaystyle S$ but $\displaystyle f$ is unbounded on $\displaystyle S$. Then there exists a sequence $\displaystyle (s_{n})\in S$ such that $\displaystyle |f(s_{n})|\ge n$ for all $\displaystyle n\in \mathbb{N}$. Since $\displaystyle (s_{n})$ is a bounded sequence, there exists a subsequence $\displaystyle (s_{n_{k}})$ that is a Cauchy sequence. Since $\displaystyle f$ is uniformly continuous and $\displaystyle (s_{n_{k}})$ is a Cauchy sequence in $\displaystyle S$, then $\displaystyle (f(s_{n_{k}}))$ is a Cauchy sequence. So $\displaystyle (f(s_{n_{k}}))$ converges to some real number. But $\displaystyle |f(s_{n_{k}})|\ge n_{k}$ for all $\displaystyle n_{k}\in \mathbb{N}$ as well, which is a contradiction. Q.E.D.

Is this correct? Thanks.

Looks good to me.

Another way to look at it might be as follows:

Since $\displaystyle f$ is unif. cont. we may choose a $\displaystyle \delta$ such that $\displaystyle |x-y|<\delta\implies |f(x)-f(y)|<1$.

Now notice that since $\displaystyle S$ is bounded we have that $\displaystyle S\subseteq [-M,M]$ for some $\displaystyle M>0$. But, this latter set is compact and thus we may cover it with finitely many intervals of the form $\displaystyle I_1,\cdots,I_n$ such that $\displaystyle \text{diam }I_k<\frac{\delta}{2}$. Let $\displaystyle I_{\alpha_1},\cdots,I_{\alpha_m}$ be those that intersect $\displaystyle S$. Choose $\displaystyle x_k\in I_{\alpha_k}\cap S$ then $\displaystyle I_{\alpha_k}\subseteq B_{\frac{\delta}{2}}(x_k)$. It follows that $\displaystyle B_{\frac{\delta}{2}}(x_1),\cdots,B_{\frac{\delta}{ 2}}(x_m)$ cover $\displaystyle S$.

So, let $\displaystyle x\in S$. Then $\displaystyle x\in B_{\frac{\delta}{2}}(x_k)$ for some $\displaystyle k$ and so $\displaystyle |f(x)|=|f(x)-f(x_k)+f(x_k)|\leqslant |f(x)-f(x_k)|+f(x_k)\leqslant 1+\max_{1\leqslant j\leqslant m}f(x_j)$
• Mar 17th 2010, 06:08 PM
Pinkk
Hmm, thanks. The teacher's assistant for my real analysis class talks about using compactness in proofs for uniform continuity but our professor and curriculum doesn't cover compactness (at least we haven't touched on it at all). Thanks for the feedback! (Hi)
• Mar 17th 2010, 06:12 PM
Drexel28
Quote:

Originally Posted by Pinkk
Hmm, thanks. The teacher's assistant for my real analysis class talks about using compactness in proofs for uniform continuity but our professor and curriculum doesn't cover compactness (at least we haven't touched on it at all).

That's a shame. Compactness is one of the most important parts of analysis. A lot of important theorems are just statements about compactness and continuous maps (e.g. the extreme value theorem)

Quote:

Thanks for the feedback! (Hi)
Any time :)