During my research, I come across this question.
Let $f \in C^2(\mathbb R, \mathbb R_+^*)$ with $\forall x \in\mathbb R, f'(x) \geq |f''(x)+f(x)|$.
Is it true that $\forall x \in \mathbb R, f''(x) \geq 0$ ?
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2$\begingroup$ What is $\mathbb R_+^*$? $\endgroup$– Iosif PinelisCommented Jan 25 at 13:24
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$\begingroup$ $\mathbb R_+^*=] 0,+\infty[$ $\endgroup$– DattierCommented Jan 25 at 13:27
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1$\begingroup$ I don't think it can be true, because for instance one may modify exp(2x) locally at 0, making $f''<0$ without affecting the inequality $\endgroup$– Pietro MajerCommented Jan 25 at 14:20
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$\begingroup$ @PietroMajer: It's not purely a local question though (near $x=0$, $f(x)=\sin x$ also works fine as a counterexample), and in any event, $f(x)=e^{2x}$ doesn't satisfy the inequality. $\endgroup$– Christian RemlingCommented Jan 25 at 15:40
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1$\begingroup$ @PietroMajer: I think the question has more content than you gave it credit for. Let's say we indeed try $f(x)=\sin x$, lazily until the maximum $x=\pi/2$. But now we're doomed, we can't continue past this point because we can no longer afford $f''<0$ (this would make $f'$ negative), so must make $f'\ge 1$ and thus can't keep the function $C^2$. $\endgroup$– Christian RemlingCommented Jan 25 at 15:50
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1 Answer
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There is no such function.
In terms of $g=f'/f$, the inequality becomes $g\ge |1+g^2+g'|$ or $|g'/g+g+1/g|\le 1$, at least when $g>0$. This shows that $g'/g\le -1$ or $g'\le -g$, and this last conclusion is clearly also correct when $g=0$.
Gronwall's inequality now shows that $g(x)\le ae^{-x}$. Since $g=(\log f)'$ and this bound is integrable on $x>0$, it follows that $f$ is bounded.
Since $f$ is also increasing, $L=\lim_{x\to\infty} f(x)$ exists, and $f'(x)\to 0$. However, then the inequality forces $f''(x)\to -L$, which will make $f'$ negative eventually, leading to a contradiction.
answered Jan 25 at 17:05
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$\begingroup$ @IosifPinelis: I was indeed a bit sloppy at the end, but everything is fine as written anyway: $g=f'/f\le ae^{-x}$, so $f'$ does go to zero. Even if it didn't, we still have $f'\in L^1$ from $f\to L$, so $f''+f\in L^1$, and then integrating we obtain $f'(x)=-Lx + o(x)$, which is good enough. $\endgroup$ Commented Jan 25 at 18:00
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$\begingroup$ In fact, it would have been slightly easier to not go back to $f$ at all: $g(x)\to 0$, so $g'\to -1$, which would make $g$ negative eventually, but clearly $g\ge 0$ under our assumptions on $f$. $\endgroup$ Commented Jan 26 at 0:26