Let $f : [0,\infty) \longrightarrow \Bbb{R}$ and suppose that $f''$ exists. Is it possible to have $f(x)f''(x) \leq -1$ for all $x$ ?

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    $\begingroup$ you mean $\forall x \in [0,\infty)$ $\endgroup$ – user44197 Dec 31 '13 at 6:52
  • $\begingroup$ As @user44197 asks, is this for all $x\in\left[0,\infty\right)$ or if there is at least one $x\in\left[0,\infty\right)$ such that this holds? I haven't found a solution for the first case, but the second case has a rather easy solution. $\endgroup$ – Brian Dec 31 '13 at 7:07
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    $\begingroup$ @pin2 First case is getting to be really hard! No obvious solution. Can you give some context for the problem? $\endgroup$ – user44197 Dec 31 '13 at 7:10
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    $\begingroup$ @BrianScholl Yes, it is for all $x$. $\endgroup$ – pin2 Dec 31 '13 at 7:11
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    $\begingroup$ @pin2 : you need to fix the question and the title, not just leave a comment. Also, please tell us what you've tried. $\endgroup$ – Stefan Smith Dec 31 '13 at 7:16

First of all note that if this inequality holds for all $x\ge 0$ then $f''(x)$ does not change sign and so does $f(x)$ for all $x\ge 0.$ Changing $f(x)$ to $-f(x),$ we may assume that $f''(x)<0$ and $f(x)>0$ for all $x\ge 0.$ Note, that $f'(x)$ does not change sign too, otherwise $f(x)$ would decrease at infinity and that would lead to $f(x)$ being negative at some point.

So we finally have the function with $f(x)>0,$ $f'(x)\ge 0,$ $f''(x)<0$ and $f(x)f''(x)\le -1.$ Consider the function $$g(x)=\left(f'(x)\right)^2+2\ln f.$$

Note, $$g'(x)=2f'(x)\frac{f''(x)f(x)+1}{f}\le 0.$$ So $g$ is a decreasing function and so is $f'(x)$ because $f''(x)<0.$

Since $g$ is decreasing, $f(x)$ is bounded and thus $\lim_{x\to\infty}f(x)=c<\infty$ and thus $f''(x)$ is bounded away from $0.$ In other words, we have $$f''(x)\le -c_0,$$ for some $c_0>0.$ But then, integrating the last inequality we get $$f'(t)-f'(0)\le \int_0^tf''(x)\le -c_0t$$ or $$f'(t)\le -c_0t+f'(0)$$ and we can make $f'(t)$ to get negative. This contradiction finishes the proof.

  • $\begingroup$ In line 3, how did you conclude that $f'≥0$? $\frac1x$, for example, is decreasing but always positive. $\endgroup$ – user112018 Jan 1 '14 at 3:55
  • $\begingroup$ but $(1/x)''>0$ and in our case it is not possible by assumption. $\endgroup$ – leshik Jan 1 '14 at 4:11

Here's another approach.

The answer is no. Suppose otherwise. Clearly $f$ and $f''$ can never vanish, so neither may change sign. Moreover, their signs are opposite; by replacing $f$ with $-f$ if necessary, we can assume that $f''(x)<0<f(x)$ for all $x\geq0$. The negativity of $f''$ means that $f$ is strictly concave, i.e., $f(x)< f(x_0) + f'(x_0)(x-x_0)$ for all $x,x_0\geq0$. On the other hand, $1<-f''(x)f(x)$. Combining these, we get \begin{align*} 1 & < -f''(x)f(x) < -f''(x)[f(x_0) +f'(x_0)(x-x_0)]. \end{align*} This implies \begin{align*} 0< {1\over f(x_0) + f'(x_0)(x-x_0)} < -f''(x). \end{align*} Integrating gives \begin{align*} 0<\int_0^x{1\over f(x_0) + f'(x_0)(t-x_0)}\,dt < -f'(x)+f'(0), \end{align*} and the middle quantity tends to infinity with $x$. This implies that $f'(x)<0$ for $x$ large enough. But this is a contradiction, since if $f'(x_0)<0$ then $f(x) < f(x_0) + f'(x_0)(x-x_0) <0$ for large $x$.

  • $\begingroup$ Why the middle integral tends to infinity with $x$? $\endgroup$ – pin2 Dec 31 '13 at 19:44
  • $\begingroup$ @pin2 The integrand grows like $1/t$ (it is bounded below by $C/(1+t)$ for some constant $C>0$, for instance), so the integral grows like $\int_1^x{dt\over t} = \log{x}$. $\endgroup$ – Nick Strehlke Dec 31 '13 at 21:14

This answer is an expanded version of leshik's correct answer rephrased in the language of Newtonian mechanics. Let $m>0$ be a positive constant. Let $x:[0,\infty[\to \mathbb{R} $ be a twice differentiable function.

Prove that $$\tag{1} \forall t\in [0,\infty[:~~ m~ x(t) ~\ddot{x}(t) ~\leq~ -1$$ is impossible.

The proof is by contradiction, and evolves in several steps:

  1. Assume that eq. (1) holds.

  2. The acceleration $\ddot{x}$ does not change sign. Proof by contradiction: Assume that the pair $(\ddot{x}(t_1),\ddot{x}(t_2))$ has opposite sign. Define midpoint $t_3:=\frac{t_1+t_2}{2}$. Either $(\ddot{x}(t_1),\ddot{x}(t_3))$ or $(\ddot{x}(t_3),\ddot{x}(t_2))$ have opposite sign. Define $t_4$ to be the midpoint of the corresponding interval. Continuing this way, we get a converging sequence $(t_n)_{n\in\mathbb{N}}\to t_{*}\in [t_1,t_2]$. Both sequences $(\ddot{x}(t_n))_{n\in\mathbb{N}}$ and $(x(t_n))_{n\in\mathbb{N}}$ must have alternating signs. Since $x$ is continuous, we conclude that $x(t_{*})=0$. A contradiction.

  3. By reflection $x\to -x$ if necessary, we may from now on assume that $x>0>\ddot{x}$.

  4. The velocity $\dot{x}>0$ is positive. Proof by contradiction: Assume that $\dot{x}(t_1)\leq 0$. Since $\ddot{x}<0$, there exists $t_2>t_1$ such that $\dot{x}(t_2)< 0$, and then in turn, $$\tag{2}\forall t\in [t_2,\infty[:~~\dot{x}(t)~\leq~\dot{x}(t_2)~<~0.$$ Such a negative velocity will eventually lead to a negative position in the future no matter what. Contradiction since $x>0$.

  5. Next define the mechanical energy $$\tag{3}E(t)~:=~\frac{m}{2} \dot{x}(t)^2+\ln x(t).$$

  6. The energy function (3) is a weakly decreasing function $$\tag{4}\forall t_1,t_2\in [0,\infty[:~~ t_1 ~<~ t_2 ~~\Rightarrow~~ E(t_1)~\geq~E(t_2).$$ Proof: Multiply (1) with $\frac{\dot{x}(t)}{x(t)}>0$ and integrate from $t_1$ to $t_2$.

  7. The position function $x(t)\leq e^{E(0)}$ is bounded from above. Proof: Use eqs. (3) and (4).

  8. The acceleration function $\ddot{x}(t)\leq - \frac{e^{-E(0)}}{m}$ is bounded from above. Proof: Use eq. (1).

  9. Such a negative acceleration will eventually lead to a negative position in the future no matter what. Contradiction since $x>0$.

  • $\begingroup$ how does $\log x(t)$ find it's way to a mechanical energy function? $\endgroup$ – nbubis Jan 2 '14 at 22:50
  • $\begingroup$ The logarithmic potential $V=\ln x$ (and the corresponding force $F=−1/x$) turn the original inequality (1) into a one-sided Newton's second law $m\ddot{x}\leq F$. To imitate Newton's second law $m\ddot{x}=\sum_i F_i$, one could introduce non-conservative forces. $\endgroup$ – Qmechanic Jan 3 '14 at 0:33

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