# Uniformly continuous function on bounded open interval is bounded

Let $$f(x)$$ be uniformly continuous on a bounded open interval $$a. Show that $$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

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.

• And the contradiction will be that: as $f(x_{n_k})$ is Cauchy so it is bounded $\forall x_{n_k} \in (a,b)$? Feb 15, 2021 at 19:02
• 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.
– Mark
Feb 15, 2021 at 19:10
• Beautiful result! Thank you very much! Feb 15, 2021 at 19:22

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 $$[a,b].$$

• Oh, of course... I forgot it... Then as it is continious on closed inverval we can conclude that $f$ is bounded Feb 15, 2021 at 18:42

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 $$\{a_n\}$$.

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 $$2\delta. Then $$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.