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Let $f(x)$ be uniformly continuous on a bounded open interval $a<x<b$. Show that $f$ is bounded (i.e. $\exists M$ such that $|f(x)|\le M \ \forall x\in (a,b)$).

To be honest, I have no idea how to solve thus problem. I tried to pass directly by the definition of uniformly continuous function and extract something, and I passed by cases distinction ($f$ monotonic or not), but I still can't conclude. Intuitively I see why it is true but can't find a good approach to this problem. If someone could give a hint, I would appreciate it. Thanks in advance

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4 Answers 4

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Assume it is not bounded. Then there is a sequence $(x_n)_{n=1}^\infty\subseteq (a,b)$ such that $|f(x_n)|>n$ for all $n\in\mathbb{N}$. Since this sequence is bounded, it must have some Cauchy subsequence $(x_{n_k})$. Now use uniform continuity to show that $f(x_{n_k})$ is also Cauchy (this easily follows from the definition), and this will be a contradiction.

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  • $\begingroup$ And the contradiction will be that: as $f(x_{n_k})$ is Cauchy so it is bounded $\forall x_{n_k} \in (a,b)$? $\endgroup$
    – Daniil
    Feb 15, 2021 at 19:02
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    $\begingroup$ Yes, a Cauchy sequence is bounded. On the other hand our construction of $x_n$ clearly shows that $f(x_{n_k})$ cannot be bounded. So it is a contradiction. $\endgroup$
    – Mark
    Feb 15, 2021 at 19:10
  • $\begingroup$ Beautiful result! Thank you very much! $\endgroup$
    – Daniil
    Feb 15, 2021 at 19:22
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If $f$ is uniformly continous then $\lim_{x\to a_+}f(x) =A$ and $\lim_{x\to b_-} f(x) =B $ exist moreover if you define $ f(a) =A $ and $f(b)=B$ then you obtain a continous function on $[a,b]. $

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  • $\begingroup$ Oh, of course... I forgot it... Then as it is continious on closed inverval we can conclude that $f$ is bounded $\endgroup$
    – Daniil
    Feb 15, 2021 at 18:42
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If $f$ was not bounded, there would be a sequence $\{c_n\}$ converging to let say $c$ such that $\{\vert f(c_n) \vert\}$ is unbounded. Derive a contradiction to uniform continuity using a subsequence of $\{a_n\}$.

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This is a simple proof using the $\epsilon,\delta$ definition,

We can find $\delta>0$ such that $|f(x)-f(y)|<1$ for all $x,y\in(a,b)$ with $|x-y|<\delta$. WLOG assume $2\delta<b-a$. Then $f$ is continuous on $[a+\delta,b-\delta]$, so is bounded on $[a+\delta,b-\delta]$.

Now if $x\in[b-\delta,b)$, then $|x-(b-\delta)|<\delta$, so $|f(x)-f(b-\delta)|<1$, therefore $|f(x)|<1+|f(b-\delta)|$. Similarly $|f(x)|<1+|f(a+\delta)|$ for all $x\in(a,a+\delta]$. Therefore $f$ is bounded.

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