On inequality of exponentially activated, unit transform of vector

Let $$x$$ be a vector and $$A$$ a matrix. Let $$y := \dfrac{Ax}{\lVert A \rVert}$$ bet a unit transform of vector $$x$$, and $$\widehat{y}:= y/\lVert x\rVert$$ the same transformation with normalization on $$x$$. Also define a vector $$z$$ point-wisely by $$z:=e^y$$ with $$\overline{z} := z/\lVert z \rVert$$.

Question. Suppose $$\widehat{y}_1>\widehat{y}_2$$ (or equivalently $$y_1 >y_2$$) where $$y_1$$ and $$y_2$$ are the first and second elements of $$y$$, respectively. Then $$\overline{z}_1 - \overline{z}_2 \geq \widehat{y}_1 - \widehat{y}_2?$$

The assumption that the transformation is a unit one might be removed; we might simply replace $$A/\lVert A \rVert$$ by $$M$$. Would it then be easier to prove or disprove?

Additional Assumption. Suppose $$x$$ is such that $$y_i >0$$ for all $$i$$ and $$y_1 = \max y_i$$.

• When you are comparing two vectors, are you comparing them element wise? – sudeep5221 May 21 at 13:05
• @sudeep5221 Sorry for unclear writing. I will clarify. – jachilles May 21 at 13:08

Let $$A$$ be identity matrix, $$x = (4a, a, 8a)$$. Then $$\hat y = (\frac{4}{9}, \frac{1}{9}, \frac{8}{9})$$ and $$\hat y_1 - \hat y_2 = \frac{1}{3}$$.
But $$\|z\| > e^{8a}$$ so $$\bar z_1 < e^{-4a}$$. As $$\bar z_2 > 0$$ we have $$\bar z_1 - \bar z_2 < e^{-4a}$$, so for large enough $$a$$ we have $$\bar z_1 - \bar z_2 < \frac{1}{3}$$.
• Thank you for your answer, sincerely. But since the derivative of $e^y$ is smaller than linear one when $y_i<0$, I think the above conjecture would not hold. And you provided a counter example. Sorry to bother but how about when $x$ is such that all the components $y_i>0$ – jachilles May 22 at 5:22
• In my example all $y_i > 0$. – mihaild May 22 at 8:23
• Take $x = (4 \varepsilon, 3 \varepsilon)$ for some small $\varepsilon$. Then $\hat y_1 - \hat y_2 = \frac{1}{5}$, but $\bar z_1 - \bar z_2 \approx \frac{\varepsilon}{2}$. – mihaild May 22 at 8:43
• If we omit $A$ then it probably is, but I can't came with formal sentence. I believe it's better to be asked as separate question - much more people will notice a question then comments buried in existent question. – mihaild May 22 at 9:11
• And of course if we don't omit $A$, lower bound on $x$ is irrelevant, as $\frac{A}{\|A\|}$ can decrease norm. – mihaild May 22 at 9:27