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Given a well-behaved function $f(x,t)$ such that $\frac{\partial f}{\partial x}<0$ and $\frac{\partial^2f}{\partial x^2}>0$, is there a way to show \begin{equation} \left(\frac{\partial^2f}{\partial x^2}\right)^2 > \left(\frac{\partial^3f}{\partial x^3}\right)\left(\frac{\partial f}{\partial x}\right) \end{equation} or under what conditions this applies? I've tried to brute-force it with forward finite differences and it's equivalent to \begin{equation} (f(x+2h)-f(x+h))^2 > (f(x+3h)-f(x+2h))(f(x+h)-f(x)) \end{equation} but that's where I get stuck. Any thoughts?

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    $\begingroup$ You can hide the parameter $t$ that plays no role. You may also replace $f$ with $f(-x)$ so now $f'(x)>0$ and $f''(x)>0$. Then, the inequality now is $f'''f' > (f'')^2$, which exactly says that $f'$ is a logarithmically convex function. $\endgroup$ Jan 16, 2014 at 19:34
  • $\begingroup$ (so, for the old $f$, it is $-f'$ log.c.) $\endgroup$ Jan 16, 2014 at 19:45
  • $\begingroup$ Thanks, I'll see if I can find a way to prove log convexity for $-f'$. $\endgroup$
    – Em F
    Jan 16, 2014 at 20:13
  • $\begingroup$ What is $t$ doing in your question? $\endgroup$ Jan 16, 2014 at 20:27
  • $\begingroup$ @AlexandreEremenko, $t$ is there because the full problem is a PDE with a known terminal boundary condition $g(x,T)$. I think I can assume the above inequality holds for that boundary condition. Further, $g(0,t) = k>0$ and $g(\infty,t) = 0$. $\endgroup$
    – Em F
    Jan 16, 2014 at 23:14

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Forget about $f$; consider $g(x):=\ln(-f'_x)$: it is well defined. Then you say that $g'<0$ and ask if this implies that $g''>0$. Why would this be true?

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  • $\begingroup$ I might be misinterpreting your notation: $g(x) = \ln\left(-\frac{\partial f}{\partial x}\right)$ implies that $g'(x) = \frac{f_{xx}}{f_x}$ (where subscripts denote partial derivatives), so how does the original equation imply $g'<0$? $\endgroup$
    – Em F
    Jan 16, 2014 at 19:06
  • $\begingroup$ You claim that $f_x<0$ and $f_{xx}>0$, don't you? $\endgroup$ Jan 16, 2014 at 19:21
  • $\begingroup$ Right, sorry, I misread it. $\endgroup$
    – Em F
    Jan 16, 2014 at 20:14

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