Saturday, February 9, 2019

real analysis - Prove that a function whose derivative is bounded is uniformly continuous.



Suppose that $f$ is a real-valued function on $\Bbb R$ whose derivative exists at each point and is bounded. Prove that $f$ is uniformly continuous.


Answer



Since $f'$ is bounded then there's $M>0$ s.t.
$$|f'(x)|\leq M\quad \forall x\in\mathbb{R}$$
hence by mean value theorem we find
$$|f(x)-f(y)|\leq M|x-y|\quad \forall x,y\in\mathbb{R}$$
so $f$ is a lipschitzian function on $\mathbb{R}$ and therefore it's s uniformly continuous on $\mathbb{R}$.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...