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If a function is uniformly continuous in $(a,b)$ can I say that its image is bounded?

($a$ and $b$ being finite numbers).

I tried proving and disproving it. Couldn't find an example for a non-bounded image.

Is there any basic proof or counter example for any of the cases?

Thanks a million!

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4 Answers 4

up vote 6 down vote accepted

There is a $\delta >0$ such that for all $x,y\in (a,b)$ with $|x-y|\leq \delta$ we have $|f(x)-f(y)|\leq 1$. Let $p=\min (\delta,\frac{b-a}{3})$. Then $f$ is continuous on $I=[a+p,b-p]$. Hence $f$ is bounded on $I$. In addition, $f$ is bounded by $|f(a+p)|+|f(b-p)|+1$ on $(a-b)-I$. This means $f$ is bounded on $(a,b)$.

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Hint: Prove first that a uniformly continuous function on an open interval can be extended to a continuous function on the closure of the interval.

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1  
I remember figuring this out with my friend a few days before the final in calculus II. That was a hoot. For some reason they forgot to mention this theorem in class, but when we asked the professor he said it would be fine to use it... –  Asaf Karagila Feb 12 '13 at 0:38

Let $\delta$ be so that $|x-y|< \delta \Rightarrow |f(x)-f(y)|<1$. Now prove that there exists a finite set $F$ so that

$$(a,b) \subset \cup_{z \in F} (z-\frac{\delta}{2}, z+\frac{\delta}{2} ) \,.$$

Now, what can you say about $|f(x)|$ and $\max\{f(z)|z \in F \}+1\,.$?

P.S. The existence of the finite set is really the fact that $(0,1)$ is precompact in the topology of $\mathbb R$. So this solution is not really different than Ittay's...

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If, for instance, $\limsup\limits_{x\rightarrow a^+} f(x)=\infty$, then given any $\delta>0$, one could find $x$ and $y$ with $a<x<\delta$ and $a<y<\delta$ such that $|f(x)-f(y)|>1$. Could $f$ then be uniformly continuous on $(a,b)$?

Note if $f$ is continuous on $(a,b)$ and unbounded, then either $f$ is "unbounded at $a$" or "unbounded at $b$" (aside from the above, there are three other cases to consider).

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