# If $f$ be a uniformly continuous function on $(a,b)$ then $f$ is bounded there.

We define a continuous extension of $$f(x)$$ to the set $$[a,b]$$, by $$g(x)=f(x), x\in (a,b)$$ and $$g(a)= \lim_{x \to a^{+}}f(x)$$ and $$g(b)=\lim_{x \to b^{-}}f(x)$$. $$g(x)$$ being continuous on a compact set in $$\mathbb{R}$$, it is bounded. So, $$f$$ being a restriction of $$g$$, it must also be bounded.

Is this correct?

• I think you need to argue that the limits exist. If the limits exist, then this method works. – Clayton May 16 at 1:21
• @Clayton they must. Because $f$ is uniformly continuous. (Proof is manageable). And after showing that the limits exist, then I move on to my procedure. Will that be alright? – Subhasis Biswas May 16 at 1:27
• Correct! ${}{}{}$ – Clayton May 16 at 1:36
• You can show that the limits exist but you can also show that $f$ is bounded without first showing that the limits exist. – DanielWainfleet May 16 at 2:47