# $|e^a-e^b| \leq |a-b|$ for complex numbers with non-positive real parts

Came across this problem on an old qualifying exam: Let $a$ and $b$ be complex numbers whose real parts are negative or 0. Prove the inequality $|e^a-e^b| \leq |a-b|$.

If $f(z)=e^z$ and $z=x+iy$, then $|f'(z)|=e^x\leq 1$ given that $x \leq 0$. I played around with the limit definition of the derivative, but wasn't able to get anywhere. Not sure what else to try; a hint would be very helpful!

• can you please mention in which Qualifying exam it has appeared,would be helpful for practice @dls – BAYMAX May 20 '17 at 6:07
• @BAYMAX Idk if it's still useful to you, but it came in the TIFR PhD entrance test in 2017. – MathIsNice1729 Mar 6 at 16:52

Consider integrating $f'(z) dz$ along the line segment from $a$ to $b$

• Neat. But could anyone explain this idea further? – yoyostein Dec 18 '16 at 14:51
• This takes another approach to the mean-value theorem -- the integral approach. It works neatly with holomorphic functions (and the MV theorem itself isn't very elegant), but I still would've liked to see the connection drawn. – Jonathan Y. Aug 3 '17 at 10:23

Prove and then use the following fact:

Let $D \subseteq \mathbb C$ be a convex region and let $f: D \to \mathbb C$ be holomorphic with $|f'|\le 1$ on $D$. Then for $a,b\in D$ we have

$$|f(b) - f(a)| \le |b-a|$$

• What I'd also like to see here is a reference to the mean value theorem, and why although it fails in general with complex functions the norm can still be bounded (similarly as with vector-valued real functions). – Jonathan Y. Aug 3 '17 at 10:21

An interesting related article A norm inequality for Hermitian operators by Ritsuo Nakamoto

The American Mathematical Monthly; Mar 2003; 110, 3;

If you put $\exp(z)$ in here and turn what you see about how two complex numbers $z$, $w$ in the left half plane $\Re(z)\le 0$ suffer under the map $\exp$ into a statement, that statement would be exactly the wanted inequality.