Suppose f is differentiable on an interval I. Prove that f' is bounded on I if and only if exists a constant M such that $|f(x) - f(y)| \le M|x - y|$ Been working on my homework for the past 7 hours and I'm completely lost on these particular ones :(
The chapter on I'm on is about differentiation and the mean value theorem.
1) Suppose f is differentiable on an interval I. Prove that f' is bounded on I if and only if there exists a constant M such that $|f(x) - f(y)| \le M|x - y|\,\forall \,x,y \in I$
2) Prove that $|\sqrt(x) - \sqrt(y)|\le \frac{1}{2\sqrt(a)}|x - y|\,\forall\, x, y \in [a, \infty), a > 0$
 A: Hint $\lim_{x\to y}-M\le f'(y)=\lim_{x\to y}{f(x)-f(y)\over  x-y}\le\lim_{x\to y}M$
A: Hints: For the first problem's $(\Rightarrow)$ direction, isolate $M$ on one side. For the $(\Leftarrow)$ direction, use the mean value theorem. For the second problem, use the first problem.
A: 1) $( => )$ 
Let $x,y \in I$, by the mean value theorem exist $z$ between $x$ and $y$ such that $|f(x) - f(y)| = |f^{'}(z)| |x-y|$. exist $M>0$ such that $|f^{'}(u)| \leq M$ $\forall u$.
Then $|f(x) - f(y)| \leq M |x-y|$
$(<=)$
Fix $a \in I$. by hipothesis we have 
$$ \displaystyle\frac{f(x) - f(a)}{x-a}  \leq M \ \ and  -M \leq \displaystyle\frac{f(x) - f(a)}{x-a}$$
then
$$\displaystyle\lim_{x \rightarrow a} \displaystyle\frac{f(x) - f(a)}{x-a}  \leq M \ \ and  -M \leq \displaystyle\lim_{x \rightarrow a} \displaystyle\frac{f(x) - f(a)}{x-a}$$
We have $a$ arbitrary, then $f^{'}$ is bounded.
2) We have $f^{'}(x) = 1 /2 \sqrt{x}$. Take $x,y \in [a, + \infty)$. Again by the mean value theorem exists $z $ between $x$ and $y$ such that 
$$ |\sqrt{x} - \sqrt{y}|  = \Bigl| \displaystyle\frac{1}{2 \sqrt{z}} \Bigl| |x-y|$$
Note that $ z > a =>  \Bigl| \displaystyle\frac{1}{2 \sqrt{z}} \Bigl| \leq \Bigl| \displaystyle\frac{1}{2 \sqrt{a}} \Bigl| $ . Then
$$|\sqrt{x} - \sqrt{y}|  \leq \Bigl| \displaystyle\frac{1}{2 \sqrt{a}} \Bigl| |x-y|$$
3) The statement is false .take $f(x) = |x|$ and $x_0 = 0$. We have
$$ \displaystyle\lim_{h \rightarrow 0}  \displaystyle\frac{f(0+h) - f(0-h)}{h} = \lim_{h \rightarrow 0}  \displaystyle\frac{|h| - |-h|}{h} = \lim_{h \rightarrow 0}  \displaystyle\frac{0}{h}  = 0$$.
But we know that the function $f$ given by $f(x) = |x|$ is not diferentiable at $x = 0$.
