Uniformly Continuous on $C^1[a,b]$

Question

Stumped on a seemingly easy question, don't know why I can't think of a proof for this:

Show that if $f\in C^1[a,b]$ with $a,b\in\mathbb{R}$, then $f$ must be uniformly continuous on $[a,b]$.

Would a direct proof be sufficient for this question?

Any tips would be appreciated!

Have you proved earlier that continuous functions on compact sets are uniformly continuous? — Aniruddha Deshmukh, Sep 22 '19 at 07:21
While any continuous function $[a,b]\to\Bbb R$ is uniformly continuous, the assumption that it has a continuous derivative makes it easier to prove. Try proving that $f$ satisfies a Lipschitz condition. — Angina Seng, Sep 22 '19 at 07:36
@AlvinLepik This is not a duplicate. Here we have the stronger assumption that the map is continuously differentiable. — mathcounterexamples.net, Sep 22 '19 at 07:40

mathcounterexamples.net · Accepted Answer · 2019-09-22T09:07:07.310

3

Hint

Some cooking ingredients...

A continuous map on a bounded interval is bounded. I let you know to which map this should be applied.
The mean value theorem may be your friend.
A lischitz map is uniformly continuous.

I let you use the ingredients for a good recipe.

edited Sep 22 '19 at 09:07

answered Sep 22 '19 at 07:44

i think i know what you mean, but do you mind providing some more context? – Sep 22 '19 at 08:21

score 0 · Answer 2 · answered Sep 22 '19 at 10:56

0

Since $f'$ is continuous on $[a,b]$ then exists $M>0$ such that $|f'(x)| \leq M, \forall x \in [a,b]$

Let $x,y \in [a,b], x<y$

By M.V.T on the interval $[x,y]$ exists $c \in (a,b)$ such that $$|f(y)-f(x)|=|f'(c)||x-y| \leq M|x-y|$$

answered Sep 22 '19 at 10:56

Marios Gretsas

2 Answers2