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Let's say that we have function $u:\mathbb R_0\to \mathbb R$ with $u'(x)>0$, $u''(x)<0$, $u'''(x)>0$, $\lim_{x\to 0} u'(x) = \infty, \lim_{x\to 0}u'(x) = 0$.

Take $x_1 < x_2$. Does $$\frac{u'''(x_1)}{u'''(x_2)}\leq \frac{u'(x_1)}{u'(x_2)}$$ always hold?

Thanks, I'm lost here.

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What do you mean by R_0? –  Robert Israel Nov 14 '12 at 6:57
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... and how can $\lim_{x \to 0} u'(x)$ be both "inf" (I assume $\infty$ and $0$? –  Robert Israel Nov 14 '12 at 7:06
    
I would guess that this is from the positive reals to the reals and the second limit is as $x \to \infty$ –  Henry Nov 14 '12 at 7:39
    
For some basic information about writing math at this site see e.g. here, here, here and here. –  Julian Kuelshammer Nov 14 '12 at 8:08

1 Answer 1

No, the ratio $u'''/u'$ need not be monotone. Since there is no restriction on the fourth derivative, nothing prevents us from adding a spike to $u'''$ that will destroy its monotonicity without violating any other constraints.

More concretely, let $u$ be a function as in the statement. Let $$v(x)=u(x)+ \delta \epsilon^3 \sin (2\pi \epsilon^{-1}x)\chi_{[1\le x\le 1+\epsilon]} \tag1$$ Ignoring insufficient smoothness* at $1$ and $1+\epsilon$, we see that $v$, $v'$ and $v''$ are within $\delta\epsilon $ of the corresponding $u$ derivatives. Also, $v'''-u''' $ is of order $\delta$. Fix $\delta$ small enough so that $v'''$ is positive. Now make $\epsilon$ very small. The added term in (1) has huge 4th derivative (first positive, then negative), and this makes $v'''$ non-monotone.

(*) Instead of one period of sine function, one should use a smoother compactly supported function, such as convolution of one period of sine with a smooth bump function.

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