Baby Rudin 5.26 Could someone help me out with the solution written in Roger Cooke's solutions for Baby Rudin 5.26?
https://minds.wisconsin.edu/bitstream/handle/1793/67009/rudin%20ch%205.pdf?sequence=7&isAllowed=y
The question is,

Suppose $f$ is differentiable on $[a,b]$, $f(a)=0$ and there is a real
  number $A$ such that $|f'(x)| \le A|f(x)|$ on $[a,b]$. Prove that
  $f(x)=0$ for all $x\in [a,b]$.

The solution says in its last paragraph, that 

But by definition of $M_0$, this implies $M_0 \le \frac{M_0}{2}$, so that
  $M_0\le 0$,  i.e. $M_0=0$.

I'm interpreting this as 
$M_0(x_o-a)A\ge M_0$, is assured if  $x_0 \equiv a+\frac{1}{2A}$.
But if $M_0$ is extremely large and $(x_0-a)$ is not so small, would this inequality still hold?
More formally,
Are $|f(x)| \le M_0$ and $|f(x)| \le M_0 (x_0-a) A$, enough conditions for deriving $M_0(x_0-a)A\ge M_0$ ?
 A: We're given a number $A$ such that $|f'(x)| \leq A|f(x)|$ on $[a,b]$, so we can define $x_0$ as $a + \frac{1}{2A}$, $M_0$ to be the supremum of $|f(x)|$ on $[a, x_0]$, and $M_1$ to be the supremum of $|f'(x)|$ on the same interval.
By the assumption on the function, we have $M_1 \leq A M_0$.
We also know $|f(x_0)| \leq M_1 (x_0 - a)$, by the mean value theorem; since $f(a) = 0$ and $f'(c) \leq M_1$ for any $c \in [a, x_0]$.
In fact, for any $x \in [a,x_0]$, we can say $$|f(x)| \leq M_1 (x - a).$$ And since $x \leq x_0$, we have
$$M_1 (x - a) \leq A M_0 (x_0 - a),$$ which by choice of $x_0$ is equal to $\frac{M_0}{2}$.
Hence $M_0$, the supremum of the values of $|f(x)|$ over the interval $[a,x_0]$, must be bounded above by $\frac{M_0}{2}$; thus it must be zero. So $f$ is identically zero on the interval $[a, x_0]$.

A comment brought up the good question of how we can be sure that our $x_0$ is within $[a, b]$. That is,
$$
a + \frac{1}{2A} \leq b
$$
or
$$
2A \geq \frac{1}{b-a}.
$$
Choose $c \in [a, b]$ so $|f(c)| = M_0$. Then
$$
|f(c)| \leq M_1(c-a) \leq A M_0 (c - a) \leq A M_0 (b - a).
$$
So $M_0 \leq A M_0 (b-a)$, so $\frac{1}{b-a} \leq A$.
A: I did not use the hint.
Let $x\in [a,b]$.


$\begin{align}
|f'(x)| & \le A|f(x)| \\
\Rightarrow|f(x)| & \ge \cfrac{|f'(x)|}{A} \\
\Rightarrow|f(x)| & \ge \cfrac{\left |\cfrac{f(x_0)-f(x)}{x_0-x}\right |}{A} \\
& = \cfrac{\left |f(x_0)-f(x)\right |}{A|x_0-x|} \\ \end{align}$.


Since $f(a)=0$ we have,


$\begin{align}
 |f(x)-f(a)| & \ge \cfrac{\left |f(x_0)-f(x)\right |}{A|x_0-x|} \\
\Rightarrow A|x_0-x||f(x)-f(a)| & \ge |f(x_0)-f(x)| \\
& = |f(x)-f(x_0)|.
\end{align}$


This implies that $A\ge0$. Then


$\begin{align}|A(x-x_0)(f(a)-f(x))| & \ge |f(x)-f(x_0)| \\
\Rightarrow  -|A(x-x_0)(f(a)-f(x))| & \le f(x)-f(x_0)\le |A(x-x_0)(f(a)-f(x))| \\ \end{align}$


Choose $x_0=a$ and note that this means $A>0$. Then this implies that $A(x-x_0)=1$. Thus,


$f(x)-f(a)\le|f(a)-f(x)|$


or


$\begin{align} f(x)-f(a) & =|f(a)-f(x)| \\
\Rightarrow f(x) & =|-f(x)| \text{ (since }f(a)=0 \text{)} \\  
\Rightarrow |f(x)| & =||-f(x)|| \\
& =|-f(x)| \\
\Rightarrow f(x) & =-f(x) \\
\Rightarrow 2f(x) & =0 \\
\Rightarrow f(x) & =0. \end{align}$
