If $f'(x)$ is bounded in $[a,b]$ then $f(x)$ is uniformly continuous in $[a,b]$ Help me please with this question.
Let $f(x)$ is a differentiable function in$ [a,b]$.
How using Mean value theorem to show that if $f'(x)$ is bounded in $[a,b]$ then $f(x)$ is uniformly continuous in $[a,b]$?
Thanks!
 A: Suppose $\left| f^\prime(x) \right| \leq B$ for all $x \in [a,b]$.
Fix an $\varepsilon > 0$.
We want to find a $\delta > 0$ that depends $\textit{only on }$ $\varepsilon$ such that, whenever $\left| x-y \right| < \delta$, then we have $\left| f(x) - f(y) \right| < \varepsilon$. [This is just the definition of uniform continuity, right? We have not actually done any math yet.]
The mean value theorem is very useful, provided that you apply it on the correct interval. In this problem, we have some points $x,y \in [a,b]$, so let's assume that $x<y$ and try the interval $[x,y]$, which will be a subset of $[a,b]$ [Can you figure out why this fact is important?]. So the mean value theorem states that there is a point $c \in (x,y)$ such that $$f^\prime(c)(y-x) = f(y)-f(x).$$
Now, we can use our assumption that $f^\prime$ is bounded when we take absolute values:
$$\left| f(x) - f(y) \right| = \left| f^\prime(c) \right| \left| x - y \right| \leq B \left| x- y \right|.$$
This is looking more like what we want, but we need to find a $\delta$ that does not depend on $x$ or $y$. This is why it is important to keep in mind that 'uniform continuity' is with respect to an interval, and not a single point, like 'continuity'. We need to determine how small we need to make $\left| x-y \right|$ (i.e., how close we need to make $x$ and $y$ to one another) to make $B \left| x-y \right| < \varepsilon$.
It looks like taking $\left| x-y \right| < \varepsilon /B$ is what we want, i.e., let $\delta = \varepsilon / B$. Then we have
$$\left| f(x) - f(y) \right| = \left| f^\prime(c) \right| \left| x - y \right| \leq B \left| x- y \right| < B \left(\frac{\varepsilon}{B}\right) = \varepsilon.$$
So we have shown: if $\left| x-y \right| < \delta = \varepsilon / B$, then $\left| f(x)-f(y) \right| < \varepsilon$, which is precisely the definition of uniform continuity.
There is actually one important possibility here that I neglected to consider: if $B=0$, then our choice of $\delta = \varepsilon / B$ will be undefined! But everything will work out in this case, too (although for a different reason than the one above). Can you figure out why?
A: You can show that $f$ is Lipschitz: for $x<y \in [a,b]$, there exists $z \in [x,y]$ such that $f(x)-f(y)=f'(z)(x-y)$, hence $|f(x)-f(y)| \leq K|x-y|$ with $|f'| \leq K$.
A: $f$ is uniformly continuous anyhow since $[a,b]$ is compact..
