Condition to guarantee $f=0$ on $[a,b]$

I have been stuck for several days on this old Analysis problem (I am doing some study on my own). I have tried several things (which I'll indicate below), but I cannot seem to figure it out. Here is how the problem is presented:

Problem: "Let $f$ be a continuous real-valued function on $[a,b]$. Suppose there exists a constant $M \geq 0 $ such that

$$|f(x)| \leq M \int_a^x |f(t)| dt$$

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

My Thoughts: I have tried using the mean value theorem iteratively, but that seems to always lead me down a dead end road. I deduced that $f(a)=0$. If only $f$ were assumed to be differentiable, then maybe I could play with trying to get the derivative to be $0$, but unfortunately it's only continuous. My other thought was to (somehow) use the condition to show that $\int_a^b |f(x)| dx = 0$. I also played around a bit with contradiction, but to no avail. Even if one of these hair-brained thoughts is correct, I am not really sure what to do next.

If you have any ideas, suggestions, or solutions, I would really appreciate it if you are willing to share them. Thank you for your time.

Solution 1:

Define $g(x) := \int_a^x |f(t)| dt$ for $x \in [a, b]$. Then $g$ is differentiable, non-negative, $g(a) = 0$ and $$ g'(x) = |f(x)| \leq M \int_a^x |f(t)| dt = M g(x) \, . $$ Now let $h(x) := g(x)e^{-Mx}$. Then $h$ is non-negative, $h(a) = 0$ and $$ h'(x) = g'(x) e^{-Mx} - Mg(x) e^{-Mx} \le 0 \, . $$ So $h$ is decreasing on $[a, b]$ and therefore identical to zero.

It follows that $g$ is identically zero, and therefore $|f(x)| = g'(x)$ is also zero on the interval.

(The idea is that $g$ satisfies a "differential inequality" $g' \le Mg$, and to compare it with solutions of the corresponding differential equality $y' = My$, which of course are $y(x) = C e^{Mx}$.)

Solution 2:

This is a special case of Grönwall's inequality, an essential tool for any analyst. See the "Integral form for continuous functions" section at the above Wikipedia link, and apply it with $\alpha(t) = 0$ and $\beta(t) = M$. You conclude $f(x) \le 0$ everywhere. Then apply it again to $-f$.

The Wikipedia page also gives the proof, which is pretty much the "integrating factor" argument given by Martin R's answer.