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 $M_0(x_0-a)A\ge M_0$ ?

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 $\frac{1}{b-a} \leq A$.