# Inequality involving sigmoid function

Let $$\sigma$$ denote the sigmoid function $$\sigma(x) = \frac{1}{1+e^{-x}}$$, let $$x,y \in \mathbb{R}$$. Given the following two conditions: $$|\sigma(-x) - \sigma(y)| < \epsilon$$ and $$x - y > c > 0,$$ where $$\epsilon$$ can be regarded as a small positive number and $$c$$ as a large positive number.

Revised question: can we show that $$x > \alpha c + f(\epsilon)$$ and $$y < - \alpha c + g(\epsilon)$$, where $$\alpha$$ is some positive constant, $$f$$ and $$g$$ are some functions. (Intuitively I want to show $$x$$ is bounded above from 0 and $$y$$ is bounded below from 0)

Original question (which has been answered by losif Pinelis): Can we draw the conclusion that $$x > \frac{c}{2} - f(\epsilon), y <-\frac{c}{2} + f(\epsilon)$$, where $$f(\epsilon)$$ is some function of $$\epsilon$$.

• Does not this hold trivially? Tak $f(\epsilon)\equiv-\infty$ and $g(\epsilon)\equiv+\infty$ (well, almost). Oct 3 '20 at 3:27

Such a function $$f$$ does not exist.
Suppose the contrary. Let $$t:=\epsilon\downarrow0$$ and let $$c$$ go to $$\infty$$ fast enough so that $$c\ge\ln\frac1t$$ and $$c\ge2f(t)+2\ln\frac1t$$. Let then $$x=c$$ and $$y=\ln t$$. Then eventually $$y\ge-x$$, $$\sigma(y)\ge\sigma(-x)$$, $$|\sigma(-x)-\sigma(y)|=\sigma(y)-\sigma(-x)<\sigma(y)=\frac t{1+t} and $$x-y=c+\ln\frac1t>c>0$$. However, the imposed condition $$c\ge2f(t)+2\ln\frac1t$$ means that $$y=\ln t\ge-\frac c2+f(t)$$.
In response to the revised question: The answer is still no: Use the same $$t:=\epsilon$$, $$x=c$$, and $$y=\ln t$$ as before, but now choose $$c$$ so that $$c\ge\ln\frac1t$$ and $$c\ge(g(t)+\ln\frac1t)/\alpha$$, which latter is equivalent to $$y\ge-\alpha c+g(t)$$, which latter is the negation of your desired condition $$y<-\alpha c+g(t)$$.