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 $\epsilon$.



Inequality involving sigmoid function
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 $y\ge-x$, $\sigma(y)\ge\sigma(-x)$, $|\sigma(-x)-\sigma(y)|=\sigma(y)-\sigma(-x)<\sigma(y)=\frac t{1+t}<t$ and $x-y=c+\ln\frac1t>c>0$. However, the imposed condition $c\ge2f(t)+2\ln\frac1t$ means that $y=\ln t\ge-\frac c2+f(t)$.




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**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 $y<-\alpha c+g(t)$.


