Logistic function as "difference of convex functions" (DC) is there a way to express the logistic function $$\frac{1}{1+\exp(-x)}$$ as the difference of two convex functions?
Thanks
 A: $$
\frac{d^2}{dx^2} (1+e^x)^{-1} = \frac{d}{dx} \left(-(1+e^x)^{-2}e^x\right)
= e^x\left( -(1+e^x)^{-2}+2(1+e^x)^{-3}e^x \right)
$$
$$
= \frac{e^x (-(1+e^x)+2e^x)}{(1+e^x)^3} = \frac{ e^x(e^x-1) }{(1+e^x)^3}.
$$
This is a bounded function because it is everywhere continuous and goes to $0$ as $x\to\pm\infty$.
So let $f(x) = Ax^2 + \dfrac{1}{1+e^x}$ with $A$ big enough so that the second derivative $f''$ is always positive.  Then the logistic function is the difference between the convex function $f$ and the convex function $x\mapsto Ax^2$.
A: A different approach.
If $f$ is any function with continuous second derivative let
$$
f''_+=\max(f'',0)\quad f''_-=-\min(f'',0).
$$
Then $f''_+$ and $f''_-$ are continuous, non-negative and $f''=f''_+-f''_-$. Now let $F_+$ and $F_-$ be such that $F_+''=f''_+$ and $F_-''=f''_-$. $F_+$ and $F_-$ are convex, and the constants of integration can be chosen so that $f=F_+-F_-$.
This is analogous to writing any $C^1$ function as the difference of two increasing functions.
A: A convex function has an increasing derivative and we can write any function of bounded variation as a difference of two increasing functions. So let us look at the derivative
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\frac1{1+e^{-x}}
&=\frac{e^{-x}}{(1+e^{-x})^2}\\
&=\frac1{(e^{x/2}+e^{-x/2})^2}\\
&=\tfrac14\,\mathrm{sech}^2(x/2)
\end{align}
$$
Thus, we can break this up into the difference of two increasing functions:
$$
f'(x)=\left\{\begin{array}{l}
\tfrac14&\text{when }x\ge0\\
\tfrac14\,\mathrm{sech}^2(x/2)\hphantom{-4}&\text{when }x\lt0
\end{array}\right.
$$
and
$$
g'(x)=\left\{\begin{array}{l}
\tfrac14-\tfrac14\,\mathrm{sech}^2(x/2)&\text{when }x\ge0\\
0&\text{when }x\lt0
\end{array}\right.
$$
Then if
$$
f(x)=\left\{\begin{array}{l}
\tfrac14x+\tfrac12&\text{when }x\ge0\\
\tfrac12\tanh(x/2)+\tfrac12&\text{when }x\lt0
\end{array}\right.
$$
and
$$
g(x)=\left\{\begin{array}{l}
\tfrac14x-\tfrac12\tanh(x/2)&\text{when }x\ge0\\
0&\text{when }x\lt0
\end{array}\right.
$$
$f$ and $g$ are convex and $f(x)-g(x)=\frac12+\frac12\tanh(x/2)=\dfrac1{1+e^{-x}}$
