Cauchy's mean value theorem to solve this problem, it is appropriate to apply Cauchy's theorem?
Let $ h\colon [a,b​​] \to\mathbb R$ a continuous function $ f $ and a differentiable function of $(a,b​)$ such that $ f(a) = 0$. Prove that if there is $ L \neq0$ such that for every $x \in [a, b]$
$$ | L f '(x) + h (x) f (x) | \le | f (x )|,$$
then $f(x)\equiv 0 $ for every $x \in [a, b]$.
 A: I have tried to resolve the question this way:
the function $h$ is continuous on $[a, b​​]$ then, for the theorem
Bolzano (Weierstrass) it's limited, that there exists $A$ such that for every $x\in[a,b​​]$ we have $$|h(x)|\le A.$$
From assumption
$$ | L f '(x) + h (x) f (x) | \le | f (x )|,$$
from which follows
$$ |f '(x)| \le \frac{1+A}{| L|}.$$
Let $[c,d]\subset[a,b]$ of length less than
$$\frac{1}{2}\cdot\frac{1+A}{| L|}= \frac{B}{2}$$
and such that
$$f(c)=0$$
We know that in a range with these properties exists (in fact just take for example $c = a$).
Now, if $x_0\in[c, d]$ we can write,
$$|f(x_0)-f(c)|=|f'(x_1)||x_0-c|\le\frac{B}{2}\cdot\frac{|f'(x_1)|}{B}=\frac{|f(x_1)|}{2} \frac{B}{2}$$
Repeating this reasoning we thus find a sequence $(x_n)$ is strictly
decreasing and such that
$$f(x_0)\le \frac{|f(x_1)|}{2}\le \frac{|f(x_2)|}{2^2}\le\cdots\le \frac{|f(x_n)|}{2^n}$$
Obviously, this last inequality (here we use the fact that $|f(x_n)|$ is limited) implies
$$f(x_0)=0$$
To complete the solution is sufficient to cover $[a, b​​]$ with finite number of
subintervals of length less than $\frac{B}{2}$ and use the fact that $f$ is zero on each subinterval.
We note that the same conclusion holds if we assume $h$ limited and not necessarily
continuous on [a, b​​].
A: I don't know how to use Cauchy's mean value theorem, but here is a solution. We assume $a=0$ to simplify; we have for all $0\leq x\leq b$
$$|Lf'(x)|\leq |Lf'(x)+h(x)f(x)|+|h(x)f(x)|\leq |(1+h(x))|\cdot|f(x)|,$$
and putting $M:=\sup_{0\leq x\leq b}\left|\frac{1+h(x)}L\right|$, we get $|f'(x)|\leq M|f(x)|$. Now we show by induction that for all $k\geq 1$ and $0\leq x\leq b$:
$$\tag{1}|f(x)|\leq \frac{M^k}{(k-1)!}\int_0^x(x-t)^{k-1}|f(t)|dt.$$ 
For $k=1$, we have, since $f(0)=0$: 
$$|f(x)|=\left|\int_0^xf'(t)dt\right|\leq M\int_0^x|f(t)|dt,$$
and if it's true for $k$, then 
\begin{align*}
|f(x)|&\leq \frac{M^k}{(k-1)!}\int_0^x(x-t)^{k-1}|f(t)|dt\\
&\leq \frac{M^k}{(k-1)!}\int_0^x(x-t)^{k-1}\int_0^t|f'(s)|dsdt\\
&\leq \frac{M^k}{(k-1)!}\int_0^x(x-t)^{k-1}\int_0^tM|f(s)|dsdt\\
&\leq\frac{M^{k+1}}{(k-1)!}\left(\left[\frac{(x-t)^k}k\int_0^t|f(s)|ds\right]_{t=0}^{t=x}+\int_0^x\frac{(x-t)^k}k|f(t)|dt\right)\\
&=\frac{M^{k+1}}{k!}\int_0^x(x-t)^k|f(t)|dt.
\end{align*}
Put $M':=\sup_{0\leq x\leq b}|f(x)|$. Then thanks to (1)
$$|f(x)|\leq M'\frac{M^kx^k}{k!}\quad \forall k\geq 1$$
so taking the limit $k\to\infty$, we get $f(x)=0$.
