Is it true that if $f$ is uniformly continuous on $[c,d] \subseteq (a,b)$ then it is uniformly continuous on $(a,b)$? Furthermore, is it true that if $f$ is uniformly continuous on $[a,b] \ \forall a,b \in \mathbb{R}$ then $f$ is uniformly continuous on $\mathbb{R}$?
How would I go about proving these two statements?
No: $$f:[1,2] \rightarrow \mathbb{R}: x \rightarrow \frac{1}{x}$$ is uniformly continuous. But $$f:]0,3[ \rightarrow \mathbb{R}: x \rightarrow \frac{1}{x}$$ is not uniformly continuous. If you meant for every $[c,d] \subset (a,b)$ then it is still not true: $$\forall [c,d] \subset (0,+\infty): f:[c,d] \rightarrow \mathbb{R}: x \rightarrow \frac{1}{x}$$ is uniformly continuous.
For the second: $$\forall [c,d] \subset (0,+\infty): g:[c,d] \rightarrow \mathbb{R}: x \rightarrow e^{x}$$ is uniformly continuous. But $$g:\mathbb{R} \rightarrow \mathbb{R}: x \rightarrow e^{x}$$ is not.
Let's see why this is a counterexample: Given: $\epsilon>0$, we search:
$$\delta>0 : \forall |x-y|<\delta \Rightarrow |e^{x}-e^{y}|<\epsilon$$ This can be rewritten: $$|e^{x}-e^{y}|<\epsilon$$ $$e^{y}-\epsilon<e^{x}<e^{y}+\epsilon$$ Taking the $ln$ (assuming $e^{y}-\epsilon>0$): $$ln(e^{y}-\epsilon)<x<ln(e^{y}+\epsilon)$$ Take $x = y+\frac{\delta}{2}$. Then: $$ln(e^{y}-\epsilon)<y+\frac{\delta}{2}<ln(e^{y}+\epsilon)$$ $$ln(e^{y}-\epsilon)-y<\frac{\delta}{2}<ln(e^{y}+\epsilon)-y$$ Taking the limit as $y \rightarrow \infty$: $$0\leq\frac{\delta}{2}\leq0$$ Thus no such $\delta$ exists. The problem why this doesn't work, is that $e^{x}$ increases more reapidly when $x$ gets bigger.