1. ## 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.

2. 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

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

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

If $d(f(x),f(y))\leq Md(x,y)$, then $f$ is said to be Lipschitz continuous, with the Lipschitz constant being the smallest such $M$ that satisfies that inequality. Lipschitz continuity always implies uniform continuity, but not necessarily the other way around.

3. 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 $\vert f(x)-f(y) \vert = \vert f(c)(x-y) \vert \leq M\vert x-y\vert$and so $f$ is Lipschitz and thus uniformly continous. Without continous derivative though...

4. Originally Posted by Jose27
If the derivative is continous you could use the MVT $\vert f(x)-f(y) \vert = \vert f(c)(x-y) \vert \leq M\vert x-y\vert$and so $f$ is Lipschitz and thus uniformly continous. Without continous derivative though...
MVT only requires the derivative to exist, not for it to be continuous.