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I am wondering if the following assertion is true:

Let $f,g:\mathbb{R}_+\rightarrow [0,1]$ be completely monotone functions on $\mathbb{R}_+^*$, that is, $(-1)^n f^{(n)}(x)\geq 0$ and $(-1)^n g^{(n)}(x)\geq 0$ for any $x>0$ and any $n\in \mathbb{N}$. Assume that $f(0)=g(0)$ and $\lim_{x\rightarrow \infty} f(x) = \lim_{x\rightarrow \infty} g(x)=0$, and that there exist $0<a<A$ such that $f\leq g$ on $[0,a]$ and $f\leq g$ on $[A,\infty)$.

Then $f\leq g$ on $\mathbb{R}_+$ (?).

If not, does anyone have a counter-example?

N.B.: 1) one can think of $f$ and $g$ as two Laplace transforms of positive measures.

2) Removing any of the assumptions seems to lead to a counter-example.

Thank you for your help!

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  • $\begingroup$ Thanks Christian, do they cross? Btw I should add that $lim_{x\rightarrow \infty} f(x)=lim_{x\rightarrow \infty} g(x)=0$, $f$ and $g$ being truly Laplace transforms. $\endgroup$
    – Alphonse
    Commented Sep 26, 2015 at 16:23
  • $\begingroup$ Your conditions are contradictive: there are no such functions, if $f(0)=0$ and $\lim_\infty f=0$, how can $f$ be monotone? $\endgroup$
    – Alexandre Eremenko
    Commented Sep 26, 2015 at 17:11
  • $\begingroup$ Hi, i did not say that $f(0)=0$, just that $f(0)=g(0)$. $\endgroup$
    – Alphonse
    Commented Sep 26, 2015 at 17:15
  • $\begingroup$ yes, unless I'm missing something, this solution is $t=1$, which corresponds to $e^{-x/2}=1$, hence $x=0$. But anyway, I originally forgot to mention that $f$ and $g$ must have the same final value (which I edited), for which I apologize. $\endgroup$
    – Alphonse
    Commented Sep 26, 2015 at 18:02

1 Answer 1

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No. Let's take $g(x)=\epsilon e^{-\epsilon x}+(1-\epsilon)e^{-\alpha x}$, $$ f(x) = \int_{\epsilon}^{1+\epsilon} e^{-tx}\, dt = \frac{1}{x}e^{-\epsilon x}(1-e^{-x}) . $$ Then clearly $f\le g$ near infinity, and near zero, $f(x)\simeq 1-(1/2+\epsilon)x$, $g(x)\simeq 1-(\epsilon^2+(1-\epsilon)\alpha)x$. I also want $f\le g$ near zero, and this gives a condition on $\alpha$ (given $\epsilon$). For small $\epsilon$, I can take $\alpha\approx 1/2$.

On the other hand, as $\epsilon\to 0$, $\alpha\to 1/2$, we have that $f(1)\to 1-e^{-1}$, $g(1)\to e^{-1/2}$, and now a calculator will tell us that $f(1)>g(1)$ for sufficiently small $\epsilon$. (There might be easier counterexamples.)

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  • $\begingroup$ Thanks Christian, seems like a nice counterexample. Do you think it would still fail if I assume further that $f(x) = \int_0^\infty e^{-tx} \phi_f(t) dt$ and $g(x) = \int_0^\infty e^{-tx} \phi_g(t) dt$, where $\phi_f,\ \phi_g >0$ on $\mathbb{R}_+$? $\endgroup$
    – Alphonse
    Commented Sep 27, 2015 at 8:26
  • $\begingroup$ @Alphonse: I think essentially the same counterexample will still work, if we just take $\phi_{f,g}$ that approximate my choices $\phi_f=\chi_{(\epsilon, 1+\epsilon)}$, $\phi_g=\epsilon\delta_{\epsilon} + (1-\epsilon)\delta_{\alpha}$. $\endgroup$
    – Christian Remling
    Commented Sep 27, 2015 at 13:12
  • $\begingroup$ Yes that is what I thought at first, but then I realized that you really needed a compact support for your example to work. If you make the support of $\phi_f$ 'bigger and bigger', then you lose the $1/x$ part which makes it work. $\endgroup$
    – Alphonse
    Commented Sep 27, 2015 at 13:20
  • $\begingroup$ @Alphonse: I can always add the same positive function to both $\phi_f$, $\phi_g$, so a compactly supported counterexample would suffice. (There is the issue of not being allowed singular measures now, but I don't think this can be a serious problem.) $\endgroup$
    – Christian Remling
    Commented Sep 27, 2015 at 13:27

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