Exponentials inequality: $|e^x + e^y| \leq |x - y|$ for $x,y<0$.

Question

I am trying to prove $|e^z - e^\omega| \leq |z - \omega|$ for $z,\omega\in\{z\in\mathbb{C}\;:\;\Re(z)<0\}$ and I get stuck in an inequality of the stated in the title : $|e^x + e^y| \leq |x - y|$ for $x,y<0$.

This is what I have tried: \begin{equation}\left|e^z - e^\omega\right| = \left|e^{i\Im(z)}\right|\left|e^{\Re(z)} - e^{\Re(\omega)}e^{i(\Im(\omega) - \Im(z))}\right| \leq \left|e^{\Re(z)} + e^{\Re(\omega)}\right| \end{equation}

Is it true that $|e^x + e^y| \leq |x - y|$ for $x,y<0$? If yes, any help will be appreciated.

Thank you!

Answer 1

To prove $|e^z-e^w|\le|z-w|$ for $z$ and $w$ of negative real part, consider $$|e^z-e^w|=\left|\int_w^z e^u\,du\right|.$$ Here the integral is over a straight line segment, of length $|w-z|$ from $w$ to $z$, and $|e^u|<1$ on it. Therefore $|e^z-e^w|\le|z-w|$.

Answer 2

Um, take $x=y=-1$;

$$|e^{-1}+e^{-1}|>|x-y|=0$$

seems like your theorem isn't true.

Answer 3

The inequality $|e^x + e^y| \leq |x - y|$ is not true for all $x,y <0$.

Suppose that we have $|e^x + e^y| \leq |x - y|$ for all $x,y <0$.

With $x \to 0$ and $y \to 0$ we would get $2 \le0$. !