This is baby Rudin exer 5.26, and the solution I have:
Exercise: 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 $[a,b].$
Solution: Suppose there is an interval $(c,d)\subset[a,b]$ such that $f(c)=0$ but $f(x)\ne0$ for $c<x<d.$ By passing to consideration of $-f(x)$ if necessary, we can assume $f(x)>0$ for $c<x<d$. The function $g(x)=\ln f(x)$ is then defined for $c<x<d$ and its derivative satisfies $$|g'(x)|=|\frac{f'(x)}{f(x)}|\le A.$$ The mean value theorem then implies $g(x)\ge g\left(\frac{c+d}{2}\right)-A\left(\frac{d-c}{2}\right)$ for all $x\in(c,d)$. But this is contradiction since, $g(x)\to-\infty$ as $x\to c$.
I can't understand how did we get $g(x)\ge g\left(\frac{c+d}{2}\right)-A\left(\frac{d-c}{2}\right)$.
I see that by MVT, $$g(x)-g\left(\frac{c+d}{2}\right)=g'(y)\left(x-\frac{c+d}{2}\right)\le g'(y)\left(d-\frac{c+d}{2}\right)=g'(y)\left(\frac{d-c}{2}\right)$$ for some $y\in (c,d)$. Hence, $$\left|\left|g(x)\right|-\left|g\left(\frac{c+d}{2}\right)\right|\right|\le A\left(\frac{d-c}{2}\right),$$ but, that's not desired inequality.
If $x\in(c,d)\setminus\left\{\frac{c+d}2\right\}$, then$$\frac{g(x)-g\left(\frac{c+d}2\right)}{x-\frac{c+d}2}\in[-A,A].$$Now, if $x\in\left(\frac{c+d}2,d\right)$, this implies that\begin{align}g(x)-g\left(\frac{c+d}2\right)&\geqslant(-A)\times\left(x-\frac{c+d}2\right)\\&\geqslant(-A)\times\left(d-\frac{c+d}2\right)\\&=-A\times\frac{d-c}2.\end{align}And if $x\in\left(c,\frac{c+d}2\right)$, this implies that\begin{align}g(x)-g\left(\frac{c+d}2\right)&\geqslant A\times\left(x-\frac{c+d}2\right)\\&\geqslant A\times\left(c-\frac{c+d}2\right)\\&=-A\times\frac{d-c}2.\end{align}