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Problem:

Let $f\colon \mathopen[0,1\mathclose] \to \mathbb{R}$ be defined as $$ f(x) = \frac{e^{\rho x}-1}{e^{\rho x}-1+e^{\rho (1-\gamma x)}-e^{\rho (1-\gamma) x}} $$ where $\rho$ and $\gamma$ are strictly positive real parameters. I am interested in showing that, for any fixed $\rho$, there exists a value $\gamma_{\rho}$ such that for any $\gamma\geq \gamma_{\rho}$, the function $f$ is first convex on an interval $[0,\bar{x}[$ and then concave on $[\bar{x},1]$, where $\bar{x}$ is the inflection point of $f$. This essentially boils down to showing that $f''$ is positive before $\bar{x}$ and negative afterwards.

What I have tried:

I observed that for all $x \in \mathopen]0,1\mathclose[$, the function $f$ satisfies the differential equation $$ f'(x) = \alpha(x) f(x) (1-f(x)) $$ where $$ \alpha(x) = \rho \left( \frac{1}{e^{\rho x}-1} + \frac{1}{e^{\rho (1-x)}-1} + (1+\gamma) \right). $$ From this, I deduced that $$ f(x) = \frac{f(x_{0})}{ f(x_{0}) + \big( 1-f(x_{0}) \big) \, e^{ -\int_{x_{0}}^{x} \alpha(s) \mathrm{d}s } } $$ for some $x_{0} \in \mathopen]0,1\mathclose[$. Consequently, $f''(x)$ has the same sign as $$ \frac{\sigma''(A(x))}{\sigma'(A(x))} + \frac{\alpha'(x)}{\alpha(x)^2} $$ where $$ \sigma(x) = \frac{f(x_{0})}{ f(x_{0}) + ( 1-f(x_{0}) ) e^{ -x} } $$ is the sigmoid function (which is a convex-concave function) and $$ A(x) = \int_{x_{0}}^{x} \alpha(s) \mathrm{d}s. $$ The challenge is that the signs of $\frac{\sigma''(A(x))}{\sigma'(A(x))}$ and $\frac{\alpha'(x)}{\alpha(x)^2}$ go in opposite directions, making it difficult to determine the sign of $f''$. A similar problem arises when you directly take the second derivative of $f$.

Question:

How can we determine the sign of $f''(x)$ and thus establish the regions of convexity and concavity of $f$? Are there any other techniques or insights that can help navigate this problem?

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  • $\begingroup$ When you say "for sufficiently large $\gamma$", do you mean "regardless of $\rho$" or "for fixed $\rho$ and sufficiently large in terms of $\rho$"? $\endgroup$
    – fedja
    Aug 12, 2023 at 14:10
  • $\begingroup$ Thank you for pointing out this imprecision and I apologize for my late reply. What I meant was that for any $\rho$, there exists a value $\gamma_{\rho}$ such that for any $\gamma\geq \gamma_{\rho}$ the function $f''$ is strictly positive and then strictly negative on the interval $\mathopen[0,1\mathclose]$. What my simulations suggest is that, for a fixed $\rho$, if $\gamma$ is less than a certain $\gamma_{\rho}$, then $f''$ is positive everywhere on $\mathopen[0,1\mathclose]$ so that $f$ is convex. I've corrected the original post. $\endgroup$
    – vico
    Aug 21, 2023 at 15:33
  • $\begingroup$ Do you have a response to the answer below? $\endgroup$ Aug 22, 2023 at 15:24

1 Answer 1

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This conjecture is true.

Indeed, letting $r:=\rho$, $a:=\gamma$, $t:=rx$, and $u:=e^r-1$, we see
\begin{equation*} f(x)=R(t):=\frac{e^t-1}{(1+u)e^{-a t}-e^{t-a t}+e^t-1}. \tag{10}\label{10} \end{equation*} So, the problem can be restated as follows:

Show that for each real $u>0$ there is some real $a_u>0$ such that for each $a\ge a_u$ there is some $t_{u,a}\in(0,\ln(1+u))$ such that $R''\ge0$ on $[0,t_{u,a}]$ and $R''\le0$ on $[t_{u,a},\ln(1+u)]$.

In what follows, it is assumed that $u>0$, $a$ is a large enough (depending on $u$) positive real number, and $t\in(0,\ln(1+u)]$ -- unless otherwise specified. Note that $(1+u)e^{-a t}-e^{t-a t}+e^t-1>e^{-a t}-e^{t-a t}+e^t-1=(e^t-1)(1-e^{-a t})>0$, so that \eqref{10} makes sense.

We have \begin{equation*} \begin{aligned} g(t)&:=R''(t)e^{-a t} \left(-e^{a t}+e^{a t+t}-e^t+u+1\right)^3 \\ &=-a^2 (u+1) e^{a t}+e^{a t+2 t} \left(-\left(a^2 (u+3)\right)-2 a u-u\right) \\ &+e^t (u+1) \left(a^2 (u+3)+2 a u+u\right)+e^{2 t} \left(-\left(a^2 (2 u+3)\right)-2 a u+u\right) \\ &+e^{a t+t} \left(a^2 (2 u+3)+2 a u-u\right)+a^2 e^{3 t}+a^2 e^{a t+3 t}-a^2 (u+1)^2. \end{aligned} \end{equation*} Note that $-e^{a t}+e^{a t+t}-e^t+u+1>-e^{a t}+e^{a t+t}-e^t+1=(e^t-1)(e^{a t}-1)>0$, so that $g(t)$ equals $R''(t)$ in sign. So, the problem can be further restated as follows:

Show that for each real $u>0$ there is some real $a_u>0$ such that for each $a\ge a_u$ there is some $t_{u,a}\in(0,\ln(1+u))$ such that $g\ge0$ on $[0,t_{u,a}]$ and $g\le0$ on $[t_{u,a},\ln(1+u)]$.

To prove this, kill the exponentials of the form $e^{ct}$ for constant $c$'s one by one, by letting
\begin{equation*} D(t):=\frac{g'(t)}{e^t},\quad D_2(t):=\frac{D'(t)}{e^t},\quad D_3(t):=\frac{D_2'(t)}{e^t}, \end{equation*} \begin{equation*} D_4(t):=\frac{D_3'(t)}{e^{(a-3)t}},\quad D_5(t):=\frac{D_4'(t)}{e^t},\quad D_6(t):=\frac{D_5'(t)}{e^t}. \end{equation*} Then \begin{equation*} D_6(t)=6 a^3 (a+1) (a+2) (a+3) e^t \\ -2 a (a+1) \left(a^2+a-2\right) \left(a^2 (u+3)+2 a u+u\right), \end{equation*} which is clearly increasing in $t$, from $D_6(0)=-2 a (a+1) (a+2) \left(a^3 u+a^2 (u-12)-a u-u\right)<0$ eventually (that is, for all large enough $a$, depending on $u>0$) and $D_6(\ln(1+u))=2 a (a+1) (a+2) \left(2 a^3 u+4 a^2 (2 u+3)+a u+u\right)>0$ eventually (and, in fact, for all real $a>0$ and $u>0$).

So, eventually $D_6$ is $-+\,$: that is, for some $t_{6;a,u}\in(0,\ln(1+u))$ we have $D_6<0$ on $[0,t_{6;u,a})$ and $D_6>0$ on $(t_{6;u,a},\ln(1+u)]$.

So, eventually $D_5$ is down-up -- that is, for some $t_{6;a,u}\in(0,\ln(1+u))$ we have $D_5$ is decreasing on $[0,t_{6;u,a}]$ and increasing on $[t_{6;u,a},\ln(1+u)]$. Also, eventually $D_5(0)=-a (a+1) \left(10 a^3 u+a^2 (5 u-36)-13 a u-2 u\right)<0$ and $D_5(\ln(1+u))=a (a+1) \left(a^4 u^2+a^3 u (9 u+14)+a^2 \left(16 u^2+43 u+36\right)+a u (6 u+13)+2 u (2 u+1)\right)>0$. So, eventually $D_5$ is $-+$.

So, eventually $D_4$ is down-up. Also, eventually $D_4(0)=2 a^2 \left(-11 a^2 u+6 a (u+2)+5 u\right)<0$ and $D_4(\ln(1+u))=a (u+1) \left(2 a^4 u^2+7 a^3 u (u+2)+8 a^2 \left(u^2+3 u+3\right)+5 a u (u+2)+2 u^2\right)>0$. So, eventually $D_4$ is $-+$.

So, eventually $D_3$ is down-up. Also, eventually $D_3(0)=-a \left(14 a^2 u+3 a (3 u-4)+u\right)<0$ and $D_3(\ln(1+u))\frac{(1+u)^2}{a}=(u+1)^a\left(a^3 u^2+2 a^2 u (u+2)+a \left(-u^2+3 u+6\right)-u (2 u+1)\right) +6 a (u+1)^2>0$. So, eventually $D_3$ is $-+$.

So, eventually $D_2$ is down-up. Also, eventually $D_2(0)=-12 a (a+1) u<0$ and $D_2(\ln(1+u))(1+u)^2 =(u+1)^a\left(-a^2 u (u+1) (3 u+2)-a u (u+1) (7 u+8)-2 u (u+1)^2\right) +2 a^2 u (u+1)^2-4 a u (u+1)^2+2 u (u+1)^2<0$. So, eventually $D_2<0$.

So, eventually $D$ is decreasing. Also, eventually $D(0)=u \left(a^2 u+2 a (u-3)+u\right)>0$ and $D(\ln(1+u))(1+u)^2 =(u+1)^a\left(-a^2 u^2 (u+1)^2-a u (5 u+4) (u+1)^2-u (2 u+3) (u+1)^2\right) -2 a u (u+1)^3+3 u (u+1)^3<0$. So, eventually $D$ is $+-\,$: that is, for some $t_{1;a,u}\in(0,\ln(1+u))$ we have $D>0$ on $[0,t_{1;u,a})$ and $D<0$ on $(t_{1;u,a},\ln(1+u)]$.

So, eventually $g$ is up-down -- that is, for some $t_{1;a,u}\in(0,\ln(1+u))$ the function $g$ is increasing on $[0,t_{1;u,a})$ and decreasing on $(t_{1;u,a},\ln(1+u)]$. Also, eventually $g(0)=(2 a+1) u^2>0$ and $g(\ln(1+u))(1+u)^2 =(u+1)^a\left(-2 a u^2 (u+1)^3-u (u+2) (u+1)^3\right) +2 u (u+1)^4<0$.

So, eventually $g$ is $+-$. $\quad\Box$

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  • $\begingroup$ Thanks a lot for the answer! $\endgroup$
    – vico
    Aug 22, 2023 at 18:55

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