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!

Answer 1

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.

Answer 2

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 $$|f(y)-f(x)|=|f'(c)||x-y| \leq M|x-y|$$