# Uniform continuity and Bounded Derivative

1. Jul 16, 2011

### Bacle

Hi, All:

Let f R-->R be differentiable. If |f'(x)|<M< oo, then f is uniformly continuous, e.g.,

by the MVTheorem. Is this conditions necessary too, i.e., if f:R-->R is differentiable

and uniformly continuous, does it follow that |f'(x)|<M<oo ?

Thanks.

2. Jul 16, 2011

### micromass

Hi Bacle!

This is not true. Consider the function

$$f:]0,1[\rightarrow\mathbb{R}: x\rightarrow \sqrt{x}$$

then this is uniform continuous (continuous on compact interval). But the derivative $f^\prime(x)$ grows very large if x gets closer to 0.

In fact, the condition you mention is equivalent to the Lipschitz-condition (for differentiable functions). That is, if $f:[a,b]\rightarrow \mathbb{R}$ is continuous and differentiable on ]a,b[, then the following are equivalent
• $|f~{}^\prime(x)|\leq k~$ for all x in ]a,b[
• f is k-Lipschitz, that is $|f(x)-f(y)|\leq k|x-y|~~$ for all x,y in [a,b]

The proof uses the mid-value theorem.

It is easy to see that being Lipschitz is stronger than uniform continuity.

3. Jul 16, 2011

### gb7nash

Unless I'm reading something wrong, the statement in the original post requires differentiability on all R. This example isn't applicable.

4. Jul 16, 2011

### micromass

Fine, consider

$$f:[0,1]\rightarrow \mathbb{R}:x\rightarrow x^{3/2}\sin(1/x)$$

this extends to an uniform continuous, differentiable function on $\mathbb{R}$, but it's derivative is not bounded.

5. Jul 16, 2011

### Bacle

Thanks, micromass; I agree, because continuous+compact implies uniformly-continuous;

I was thinking more of what if the extension outside of

[a,b] is non-trivial (e.g., a constant function, or without exhausting , by covering

the remainder by compact sets and pasting/smoothing at the endpoints), if the condition on

the derivative must be satisfied in (-00,a)\/(b,oo), i.e., if the function is not defined

piecewise in countably-many compact intervals.