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$|e^a-e^b| \leq |a-b|$

Could someone help me through this problem? Let a, b be two complex numbers in the left half-plane. Prove that $|e^{a}-e^{b}|<|a-b|$

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marked as duplicate by Ross Millikan, Adrián Barquero, Zev Chonoles Apr 24 '12 at 3:01

This question was marked as an exact duplicate of an existing question.

By no mean this is true... – Lierre Apr 24 '12 at 2:32
I think that being in the complex plane if this inequality can be fulfilled – Chalie Her Apr 24 '12 at 2:48
Oh... Sorry I read a little bit too fast, there is a condition with the half-plane. But that's funny, nobody seems to see it, two wrong answers already ;) – Lierre Apr 24 '12 at 2:51
The old question only asks for a non-strict inequality, but a minimally careful bound on the integral suggested by GEdrgar there will show the stricy inequality as asked here. – Mariano Suárez-Alvarez Apr 24 '12 at 3:03
up vote 0 down vote accepted

By mean value theorem, $$ |e^a - e^b| \leqslant |a-b|\max_{x\in [a,b]} e^x $$ But $a$ and $b$ have a negative real part, and then all $x$ in $[a,b]$ also have a negative real part. Hence the $\max$ is less than one. And thus $$ |e^a - e^b| <|a-b|. $$

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Mean value theorem does not hold for complex-valued functions. – dls Apr 24 '12 at 3:03
@dls — I think I gave a wrong name, what is the english name for the theorem I used ? – Lierre Apr 24 '12 at 3:04
the mean value theorem holds for complex functions – Chalie Her Apr 24 '12 at 3:18
The mean value theorem for derivatives does not hold for complex-valued functions. The standard example is $f(x)=e^{ix}$ on the interval $[0,2\pi]$. Then $f(2\pi)-f(0)=0$ but $2\pi|f'(x)|=2\pi$ for every $x$. – dls Apr 25 '12 at 4:20
@dls — But still, the $|f(x)-f(y)| \leq \max_{z\in[x,y]} |\nabla f(z)|$, by integrating. What is the name of this inequality ? – Lierre Apr 25 '12 at 11:21

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