Show that $f(x) = 0$ for all $x \in [a,b]$ given $|f'(x)| \leq C|f(x)| $

Suppose for real numbers $a<b$ one has a function with continuous derivative $f:[a,b] \to \mathbb{R}$ such that $f(a) = 0$ and there exists a real number $C$ with $$|f'(x)| \leq C|f(x)|$$ for all $x \in [a,b]$.

Show that $f(x) = 0$ for all $x \in [a,b]$.

Well, since $f(a) = 0$, we have that $|f'(a)| \leq C|0|$, so $|f'(a)| = 0$. Since $f$ has a continuous derivative, we also know that $f$ is continuous. Since $f$ is continuous on a compact interval, $f$ obtains a maximum, say at $\xi \in [a,b]$. So, $|f'(x)| \leq C|f(\xi)|$. Since the derivative is bounded, we obtain that $f$ is Lipschitz, so $f$ is also uniformly continuous.

Suppose to the contrary that $f(\xi) > 0$? The above is pretty much everything I could figure out about $f$, so I'm not sure what to try next. This one is also from an old qual and possibly uses methods from beyond our course.

I think maybe I should try to show that $|f'(x)| = 0$ for all $x$, but I don't know how.

Solution 1:

Assume that there exists $x_0$ such that $f(x_0)\ne 0.$ Since $f$ is continuous there exists a maximal interval $(c,d)$ such that $f(x)\ne 0$ on $(c,d)$ and $f(c)=0.$ (Note that it can be $c=a.$)

Now we have that $g(x)=\ln |f(x)|$ satisfies $|g'(x)|\le C.$ Thus, for any $t\in (0,1)$ and $\epsilon=x_0-c$

$$|g(x_0)-g(c+t\epsilon)|=\left|\int_{c+t\epsilon}^{x_0}g'(x)dx\right|\le C(x_0-c-t\epsilon).$$ Take the limit as $t\to 0$ to get a contradiction.