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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 $\epsilon$.

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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 $y=\ln t\ge-\frac c2+f(t)$.


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

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  • $\begingroup$ Thank you very much for your answer! I guess I misrepresented my question. The conclusion I want to have is x is close to c/2 and y is close -c/2, which might be weaker than my original question. I will revise my question. $\endgroup$
    – luw
    Oct 3 '20 at 2:27
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    $\begingroup$ @luw : As you noted, your original question was answered. You should not change your question so as to invalidate a valid answer. Also, the question in your comment is, not weaker, but stronger than your original question, and hence is answered, negatively, by the same counterexample. Next, your revised question is different from the question in your comment (as well as from your original question). Anyhow, only a slight modification of the original answer is needed to answer, again negatively, your revised question -- which is now done in the response to your revised question. $\endgroup$ Oct 4 '20 at 0:15

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