# Decreasing functions

Let $u:[a,b]\to \mathbb{R}$ be a continuous and a. e. differentiable function (with respect to the Lebesgue measure).

Is it true that $u' < 0$ a. e. in $[a,b]$ implies $u$ strictly decreasing everywhere in $[a,b]$?

I know the answer is negative (thanks to Jonas and Cameron).

But, what appens if $u$ is absolutely continuous in $[a,b]$?

In other words, is it true that $u^\prime \leq 0$ implies $u$ decreasing in $[a,b]$ when $u$ lies in the Sobolev space $W^{1,1}(a,b)$?

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Note: Cameron Buie's answer was posted before Pacciu said that $u$ is supposed to be continuous. –  Jonas Meyer Jul 6 '12 at 14:20

Hint: Cantor-Lebesgue function minus $x$.

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+1: Nice one, sir! –  Cameron Buie Jul 6 '12 at 14:28
+1. Thank you Jonas. It's funny how I came up with the same function just 5 minutes after editing the question. XD –  Pacciu Jul 6 '12 at 14:37

Nope. Consider $u:[-1,1]\to\Bbb R$ given by $$u(x)=\begin{cases}-x & x\leq 0\\ 1-x & x>0\end{cases}.$$ Clearly, it's differentiable a.e. in $[-1,1]$ and with a negative derivative a.e., but not everywhere strictly decreasing.

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Good save. I was going to add the same function with worse formatting. –  Jonas Meyer Jul 6 '12 at 14:18
Thanks, Jonas. I'm glad I caught the error quickly. –  Cameron Buie Jul 6 '12 at 14:19
Thanks Cameron. But I was thinking about continuous functions... I've just corrected my question's text. –  Pacciu Jul 6 '12 at 14:21
Ah! Well, I'll have to think about that one, but it seems likely to be true in that case. –  Cameron Buie Jul 6 '12 at 14:24