# Proof Uniformly Continous

• Sep 30th 2007, 07:12 AM
ml692787
Proof Uniformly Continous
Need help with this proof. Thanks so much!

Let j be an interval and f a differentiable function on j whose derivative is bounded. Show that f is uniformly continuous (use the mean value theorem).
Observe that boundedness of f ' is not necessary for uniform continuity of f. Indeed, the function x sqrtx (x > 0) is uniformly continuous and differentiable, but its derivative is not bounded.
• Sep 30th 2007, 08:31 AM
ThePerfectHacker
Quote:

Originally Posted by ml692787
Need help with this proof. Thanks so much!

Let j be an interval and f a differentiable function on j whose derivative is bounded. Show that f is uniformly continuous (use the mean value theorem).

Let $\displaystyle J$ be an open inverval. Say $\displaystyle f: J\mapsto \mathbb{R}$ is a differenciable function. And $\displaystyle |f'|\leq C$ is bounded by positive constant. Let $\displaystyle x,y\in J$ be two distinct points. Then $\displaystyle |f(x)-f(y)|\leq C|x-y|$ so if we choose $\displaystyle |x-y| < \delta$ where $\displaystyle \delta = \frac{\epsilon}{C}$ this will complete the proof.
• Sep 30th 2007, 08:36 AM
ThePerfectHacker
Quote:

Originally Posted by ml692787
Observe that boundedness of f ' is not necessary for uniform continuity of f. Indeed, the function x sqrtx (x > 0) is uniformly continuous and differentiable, but its derivative is not bounded.

For any $\displaystyle \epsilon >0$ choose $\displaystyle \delta = \epsilon^2$ then $\displaystyle |f(x)-f(y)| = |\sqrt{x}-\sqrt{y}|$ say $\displaystyle x\not = y$ because otherwise this difference is zero which is certainly less than $\displaystyle \epsilon$. In that case we get $\displaystyle |\sqrt{x}-\sqrt{y}| = \frac{|x-y|}{\sqrt{x}+\sqrt{y}} \leq \frac{|x-y|}{\sqrt{|x-y|} }= \sqrt{|x-y|} < \sqrt{\delta} = \epsilon$.