I got the following questions:
Let $f$ be a real valued function of a real variable:
(1) If $f$ is continuous and bounded on the interval (a,b) (meaning there exist $M,L\in \mathbb{R}$ such that $\forall x \in (a,b), M \leq f(x) \leq L$), then $f$ is uniformly continuous on (a,b)?
(2) If $f$ is bounded on $\mathbb{R}$ and uniformly continuous on every finite interval (meaning that the interval is of the form $[a,b]$ or $(a,b)$ or $[a,b)$ or $(a,b]$ where $a,b \in \mathbb{R}$), then $f$ is uniformly continuous on $ \mathbb{R}$?
(3) If $f$ is uniformly continuous on (0,1), then $f$ is bounded on (0,1)?
(4) If $f$ is bounded, continuous and monotonic on (0,1), then $f$ is uniformly continuous on (0,1)?
If the statement is true prove it and if not give a counter example:
I tried to prove and to find a counter example without so much success. Thanks.
1: Consider $f(x) = \sin\frac{1}{x}$ on $(0,1)$.
2: Consider $f(x) = \sin (x^2)$.
3: Uniform continuity means that you can pick an $\delta$ and always find a $\epsilon$ such that $|f(x) - f(y)| < \delta$ if $|x -y| < \epsilon$. Now pick any $\delta$, and find the corresponding $\epsilon$. Then you can cover $(0,1)$ with finitely many intervals of length $\epsilon$. Use that to put a global bound on $|f(x) - f(y)|$ for $x,y \in (0,1)$, using the local bound that uniform continuity gives you
4: Extend $f$ to $[0,1]$ by using the boundedness and monotonicity. Then use that $[0,1]$ is compact, in particular that every covering with open intervals contains a finite subcover. This allows you to convert the local bound on $|f(x)-f(y)|$ that you get from $f$'s continutity to a global bound that shows it's uniform continuity.