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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.

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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 $y(x) = C e^{Mx}$.)

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  • $\begingroup$ Thanks @MartinR. I see now there were some tricks I didn't think of; namely looking at the function $h$. Very nice. $\endgroup$ – fxy Aug 3 '15 at 20:32
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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.

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  • $\begingroup$ You're right (of course!). I could only remember the method, not the name of the theorem :) $\endgroup$ – Martin R Aug 3 '15 at 20:52
  • $\begingroup$ Funny, in the german Wikipedia it is called "Gronwallsche Ungleichung", without umlaut. I would have expected it the other way around. $\endgroup$ – Martin R Aug 3 '15 at 20:56
  • $\begingroup$ @MartinR: The English Wikipedia page says he was Swedish, not German, and dropped the umlaut from his name after moving to the US. (Also, the integral form used here is apparently due to Bellman instead. Stigler's law strikes again...) $\endgroup$ – Nate Eldredge Aug 3 '15 at 20:57
  • $\begingroup$ Thanks for the insight @Nate Eldredge. I'll have to file that one away in my mind for when it inevitably arises again. $\endgroup$ – fxy Aug 3 '15 at 21:20

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