# uniform continuity

• Oct 13th 2009, 07:17 PM
friday616
uniform continuity
Suppose that f:R--> R is differentiable on R and that the derivative of f is bounded on R. Prove that f is uniformly continuous on R.
• Oct 13th 2009, 07:24 PM
redsoxfan325
Quote:

Originally Posted by friday616
Suppose that f:R--> R is differentiable on R and that the derivative of f is bounded on R. Prove that f is uniformly continuous on R.

Three words: MEAN VALUE THEOREM

$\displaystyle \frac{|f(x)-f(y)|}{|x-y|}=|f'(t)|\leq M\implies |f(x)-f(y)|\leq M|x-y|$.

Let $\displaystyle \delta=\frac{\epsilon}{M}$ and you're done.

If $\displaystyle d(f(x),f(y))\leq Md(x,y)$, then $\displaystyle f$ is said to be Lipschitz continuous, with the Lipschitz constant being the smallest such $\displaystyle M$ that satisfies that inequality. Lipschitz continuity always implies uniform continuity, but not necessarily the other way around.
• Oct 13th 2009, 07:29 PM
Jose27
Quote:

Originally Posted by friday616
Suppose that f:R--> R is differentiable on R and that the derivative of f is bounded on R. Prove that f is uniformly continuous on R.

If the derivative is continous you could use the MVT $\displaystyle \vert f(x)-f(y) \vert = \vert f(c)(x-y) \vert \leq M\vert x-y\vert$and so $\displaystyle f$ is Lipschitz and thus uniformly continous. Without continous derivative though...
• Oct 14th 2009, 05:51 AM
putnam120
Quote:

Originally Posted by Jose27
If the derivative is continous you could use the MVT $\displaystyle \vert f(x)-f(y) \vert = \vert f(c)(x-y) \vert \leq M\vert x-y\vert$and so $\displaystyle f$ is Lipschitz and thus uniformly continous. Without continous derivative though...

MVT only requires the derivative to exist, not for it to be continuous.