If $f(x)$ is uniformly continuous at $(0,1)$ then is it bounded at $(0,1)$?

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Uniform continuity

I have a question which is:

If ${f(x)}$ is uniformly continuous at ${(0,1)}$ then is it bounded at ${(0,1)}$?

This sound like it's correct to me but I can't see why exactly (Or maybe it's wrong :P).

Could someone help me figure out the truth here? :) Thanks!

Hint: fix $\epsilon>0$ and make $n\delta$ steps from $\frac12$ until you reach any bound of $(0,1)$. — SBF, Dec 02 '11 at 16:25

score 8 · Accepted Answer · answered Dec 02 '11 at 17:00

8

By uniform continuity there is an $n\in{\mathbb N}_{\geq 1}$ such that $0<x\leq y<x+{1\over n}<1$ implies $|f(y)-f(x)|<1$. Put $x_k:={k\over n+1}$ $\ (1\leq k\leq n)$. Then any $y \in \ ]0,1[\ $ is at distance ${}<{1\over n}$ from an $x_k$. Therefore we have

$$|f(y)| < \max_{1\leq k\leq n} |f(x_k)| + 1\ =:\ M$$

for all $y \in \ ]0,1[\ $.

answered Dec 02 '11 at 17:00

Christian Blatter

226,825

why we can be sure that max |f(xk)| exists? – user10024395 Apr 24 '14 at 07:58
@user136266: There are only $n$ points $x_k$ involved, where $n$ has been chosen in the first sentence of my answer. – Christian Blatter Apr 25 '14 at 19:13
Why is it $y\in [0,1]$ ? or why does uniform continuouity implies $x,y\in [0,1]$ ? – GinKin Jun 17 '14 at 18:55
@GinKin: All points $x$, $y$ where $f$ is evaluated lie in the open interval $>]0,1[>$. Uniform continuity implies something about the values of $f$, not about $x$ and $y$. – Christian Blatter Jun 18 '14 at 07:34
Right, it was in the question... – GinKin Jun 18 '14 at 09:01

David Mitra · Answer 2 · 2011-12-02T16:48:37.330

7

$f$ would indeed be bounded.

$f$ is continuous on $(0,1)$; thus, if it were not bounded on $(0,1)$, there would exist $x_n\nearrow 1$ or $x_n\searrow 0$ with $f(x_n)\rightarrow\infty$ or $f(x_n)\rightarrow-\infty$. But, none of these options can happen due to the following facts: 1) uniformly continuous functions map Cauchy sequences to Cauchy sequences. 2) Cauchy sequences are bounded.

Fact 1) is easily proven: Let $\{x_n\}$ be Cauchy and $\epsilon>0$. Let $\delta$ be such that $|f(x)-f(y)|\lt\epsilon$ whenever $|x-y|<\delta$. Now choose $N$ so that $m,n> N$ implies $|x_n-x_m|<\delta$. We then have for any $n, m>N$ that $|f(x_n)-f(x_m)|<\epsilon$. Thus, $\{f(x_n)\}$ is Cauchy.

Fact 2) is easy to prove also, but I'll leave this to the interested reader.

edited Dec 02 '11 at 16:48

answered Dec 02 '11 at 16:42

David Mitra

74,748

I don't really understand why this proof works. My concern is what happens if u plug in a number that is not part of the cauchy sequence but within the interval (0,1), how can u be sure that it is continuous at that point? – user10024395 Apr 24 '14 at 08:20
I read about 4-5 proofs with $\epsilon=1$, $N\delta$, etc, I didn't get them. But your first fact is what made it finally click. Thanks. – GinKin Jun 17 '14 at 19:05

If $f(x)$ is uniformly continuous at $(0,1)$ then is it bounded at $(0,1)$?

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