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As the title states, I'm wondering why:

If $A$ is a bounded subset of $\mathbb{R}$ and $f:A\to \mathbb{R}$ is uniformly continuous on $A$, then $f$ must be bounded on $A$.

Proof:

Since it is uniformly continuous, the function is a Lipschitz function.

$|f(x)-f(y)| \leq L|x-y|$.

Since $A$ is bounded, $|x-y|$ does not get arbitrarily big and that too is bounded by a constant. Let $|x-y| \leq M$.

Then we have a Lipschitz condition where

$|f(x)-f(y)| \leq LM$.

The function is then bounded by the product of two constants, $LM$, which means that it is bounded. Can someone check this?

user108831
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    Not each uniformly continuous function is a Lipschitz function. Consider $x\mapsto\sqrt x$. However, this function is still bounded on bounded sets. – Stefan Hamcke Nov 15 '13 at 01:25
  • In order to be Lipschitz it must be $C^1$. Now consider that every bounded set of $\Bbb R$ has supremum and infimum$. – Haha Nov 15 '13 at 01:42
  • @Dimitris I don't quite get what you mean by $\mathbb{C^1}$ but since $A$ has a supremum and infimum, then in order for $f$ (on $A$) to be uniformly continuous, then $f$ has to also have a supremum and infimum. I can see it in my head but I don't see how to make that leap. – user108831 Nov 15 '13 at 01:49
  • @Dimitris Lipschitz implies $C^1$, you say? Hardly. – Did Nov 15 '13 at 02:05
  • since $A$ is bounded it's closure is bounded then you can extend $f$ to $\overline{A}$, which is compact, hence it's image is compact. – user40276 Nov 15 '13 at 02:33
  • @Did No,i say that if $f$ is uniformly continuous and $C^1$ then is Lipschitz. – Haha Nov 15 '13 at 10:54
  • @Dimitris: Your first comment reads: "In order to be Lipschitz it must be $C^1$". That means that Lipschitz implies $C^1$, which is not true. – robjohn Nov 25 '13 at 01:41
  • @robjohn,yes i know.i wrote it wrong. what i wanted to say is my second comment i say that if f is uniformly continuous and C1 then is Lipschitz. – Haha Nov 25 '13 at 10:04

2 Answers2

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Hint: Since $f$ is uniformly continuous, there is a constant $M$ so that $$ |x-y|\le1\implies|f(x)-f(y)|\le M $$ Thus, $f$ is bounded on each interval $[k,k+1]$. Since $A$ is bounded, you can cover it with a finite number of such intervals.

robjohn
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  • What is $A{}{}{}$ ? – GinKin Jun 17 '14 at 19:12
  • @GinKin: from the question: $A$ is a bounded subset of $\mathbb{R}$. – robjohn Jun 17 '14 at 19:14
  • Oh right. Is it also true if $A=\mathbb R$ or $A=(-\infty,\infty)$ ? (I guess here $\mathbb R=(-\infty,\infty)$) – GinKin Jun 17 '14 at 19:17
  • @GinKin: No. That is why it is specified that $A$ is bounded. – robjohn Jun 17 '14 at 19:19
  • Given $\epsilon$ we can find $\delta$. How are you doing it the other way round? – Not Euler Dec 03 '19 at 11:14
  • Take any $\delta,\epsilon$. The triangle inequality says we can let $M=\lceil1/\delta\rceil\epsilon$. – robjohn Dec 03 '19 at 11:51
  • will it count as valid proof if we use the theorem :- a function $ f$ is uniformly continuous on the interval $(a, b)$ if and only if it can be defined at the endpoints a and b such that the extended function is continuous on$ [a, b]$ and since a continuous function on closed bounded interval is bounded we conclude that the uniformly continuous function in open bounded interval is bounded – pie Jul 25 '23 at 02:40
  • @pie: You would need to address the fact that $A$ may not be an interval. – robjohn Jul 25 '23 at 10:02
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If not $\forall~n\in\mathbb Z^+,~\exists~a_n\in A$ such that $f(a_n)>n.$

$(a_n)_n$ being bounded it has a convergent subsequence $(a_{r_n})_n.$

Thus $(a_{r_n})_n$ is a Cauchy sequence in $A$ which maps to a non Cauchy sequence in $\mathbb R,$ a contradiction.

Sugata Adhya
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