
#1
Feb2112, 09:45 AM

P: 159

Suppose [itex]f : \mathbb{C}\to \mathbb{C}[/itex] is continuous everywhere, and is holomorphic at every point except possibly the points in the interval [itex][2, 5][/itex] on the real axis. Prove that f must be holomorphic at every point of C.
How can I go from f being holomorphic every except that interval to showing it is holomorphic at that interval? I am assuming it has to be due to continuity. But there are continuous functions that aren't differentiable every where. 



#2
Feb2112, 12:25 PM

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One approach is to use Morera's theorem.




#3
Feb2112, 12:26 PM

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#4
Feb2112, 12:38 PM

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Holomorphic on C
Yes, but it's tricky. The key idea is that if you can find a holomorphic function [itex]g \colon \mathbb C \to \mathbb C[/itex] that agrees with f everywhere except possibly on [2,5], then g must in fact agree with f on [2,5] too (why?), so f must be holomorphic on all of C because g is. But now, how does one find such a g? Do you have any ideas?




#5
Feb2112, 12:42 PM

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#6
Feb2112, 12:44 PM

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That's not a bad idea. Could you spell it out a bit more?




#7
Feb2112, 12:47 PM

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#8
Feb2112, 12:53 PM

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But how is g defined on [2,5]?
I have to run so I'll leave you with a hint: Let [itex]C = \{ z \in \mathbb C \mid z3.5 < 1 \}[/itex] and define [itex]g(z) = \frac{1}{2\pi i} \int_C \frac{f(w)}{wz} dw[/itex] in C. Try to show that f and g agree off of [2,5]. 



#9
Feb2112, 03:19 PM

P: 159

With our definition of g, I have a theorem from class that show f and g agree but it was only stated for one point not a set of points. We call it the Integral Transform Theorem: Let [itex]\gamma[/itex] be any path and [itex]g:\gamma\to\mathbb{C}[/itex] be continuous. Define for all [itex]z \notin \gamma[/itex] [tex]G(z) = \int_{\gamma}\frac{g(u)}{uz}du[/tex]. Then [itex]G(z)[/itex] is analytic at every point [itex]z_0\notin\gamma[/itex]. Then our corollary to it If [itex]f:\gamma\to\mathbb{C}[/itex] is holomorphic and [itex]\gamma[/itex] is inside a disc on which f is holomorphic and which [itex]\gamma[/itex] is a circle, then for all z we get $$ f(z) = \frac{1}{2\pi i}\int_{\gamma}\frac{f(u)}{uz}du $$ and so f is analytic for all z inside gamma. 



#10
Feb2112, 03:47 PM

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#11
Feb2112, 03:51 PM

P: 159

Also, we were told today that we have proven Morera's Theorem but just didn't name it before. So we could use that as well then. 



#12
Feb2112, 03:52 PM

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#13
Feb2112, 04:07 PM

P: 159

I am still not sure then how to get the points inside C but not on [2,5] 



#14
Feb2112, 04:12 PM

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Let's not even concern ourselves with points outside of C. All that matters is stuff inside of C.
The problem with your corollary is that it requires that [itex]\gamma[/itex] be a circle, but really any simple closed curve works. 



#15
Feb2112, 04:18 PM

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#16
Feb2112, 04:22 PM

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#17
Feb2112, 04:26 PM

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#18
Feb2112, 04:35 PM

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Like I said in an earlier post, this approach is tricky. Morera's theorem is definitely the way to go here. 


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