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Holomorphic on C

by fauboca
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fauboca
#1
Feb21-12, 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.
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morphism
#2
Feb21-12, 12:25 PM
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One approach is to use Morera's theorem.
fauboca
#3
Feb21-12, 12:26 PM
P: 159
Quote Quote by morphism View Post
One approach is to use Morera's theorem.
I haven't learned that Theorem yet. Is there another approach.

morphism
#4
Feb21-12, 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?
fauboca
#5
Feb21-12, 12:42 PM
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Quote Quote by morphism View Post
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?
This is just a guess but let g be a primitive of f.
morphism
#6
Feb21-12, 12:44 PM
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That's not a bad idea. Could you spell it out a bit more?
fauboca
#7
Feb21-12, 12:47 PM
P: 159
Quote Quote by morphism View Post
That's not a bad idea. Could you spell it out a bit more?
If g is a primitive of f, then [itex]g' = f[/itex]. As long as f is on an open set which is the case here.
morphism
#8
Feb21-12, 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 |z-3.5| < 1 \}[/itex] and define [itex]g(z) = \frac{1}{2\pi i} \int_C \frac{f(w)}{w-z} dw[/itex] in C. Try to show that f and g agree off of [2,5].
fauboca
#9
Feb21-12, 03:19 PM
P: 159
Quote Quote by morphism View Post
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 |z-3.5| < 1 \}[/itex] and define [itex]g(z) = \frac{1}{2\pi i} \int_C \frac{f(w)}{w-z} dw[/itex] in C. Try to show that f and g agree off of [2,5].
I am not sure why you center your circle at 3.5.

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)}{u-z}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
$$
f(z) = \frac{1}{2\pi i}\int_{\gamma}\frac{f(u)}{u-z}du
$$
and so f is analytic for all z inside gamma.
morphism
#10
Feb21-12, 03:47 PM
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Quote Quote by fauboca View Post
I am not sure why you center your circle at 3.5.
The radius should have been 1.5 and not 1. I basically wanted a circle centered on the real axis with [2,5] as a diameter. (Now that I think about it a bit more, the radius should probably be 1.5+0.01 (the small increment is to ensure that [2,5] is contained in the interior of C). But this doesn't matter much.)

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)}{u-z}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
$$
f(z) = \frac{1}{2\pi i}\int_{\gamma}\frac{f(u)}{u-z}du
$$
and so f is analytic for all z inside gamma.
You will need to modify the corollary a bit to conclude that f and g agree for all points inside C but not on [2,5].
fauboca
#11
Feb21-12, 03:51 PM
P: 159
Quote Quote by morphism View Post
The radius should have been 1.5 and not 1. I basically wanted a circle centered on the real axis with with [2,5] as a diameter. (Now that I think about it a bit more, the radius should probably be 1.5+0.01 (the small increment is to ensure that [2,5] is contained in the interior of C).)


You will need to modify the corollary a bit to conclude that f and g agree for all points inside C but not on [2,5].
I am not sure how to alter the corollary besides saying we can run this process for all real numbers in [2,5] where each f* agrees with f.

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.
morphism
#12
Feb21-12, 03:52 PM
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Quote Quote by fauboca View Post
I am not sure how to alter the corollary besides saying we can run this process for all real numbers in [2,5] where each f* agrees with f.
The corollary requires f to be holomorphic inside [itex]\gamma[/itex]. But if [itex]\gamma=C[/itex], we run into problems, because we don't know if f is holomorphic on [2,5].
fauboca
#13
Feb21-12, 04:07 PM
P: 159
Quote Quote by morphism View Post
The corollary requires f to be holomorphic inside [itex]\gamma[/itex]. But if [itex]\gamma=C[/itex], we run into problems, because we don't know if f is holomorphic on [2,5].
For now, can we say by the Integral Transform Theorem, we know g and f agree on all the points out side of C?

I am still not sure then how to get the points inside C but not on [2,5]
morphism
#14
Feb21-12, 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.
fauboca
#15
Feb21-12, 04:18 PM
P: 159
Quote Quote by morphism View Post
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.
The definition of C you provided is a circle. Since [itex]\mathbb{C}[/itex] is open, there is an open disc around C. So using that definition, we would have inside C is analytic.
morphism
#16
Feb21-12, 04:22 PM
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Quote Quote by fauboca View Post
The definition of C you provided is a circle. Since [itex]\mathbb{C}[/itex] is open, there is an open disc around C. So using that definition, we would have inside C is analytic.
Yes, g is analytic inside C. But we don't know that f is. And your corollary doesn't show that f=g inside C (but off of [2,5]), which is really what we want to show. So the corollary has to be tweaked to show that f=g inside C (but off of [2,5]).
fauboca
#17
Feb21-12, 04:26 PM
P: 159
Quote Quote by morphism View Post
Yes, g is analytic inside C. But we don't know that f is. And your corollary doesn't show that f=g inside C (but off of [2,5]), which is really what we want to show. So the corollary has to be tweaked to show that f=g inside C (but off of [2,5]).
Does it have to do with the winding number?
morphism
#18
Feb21-12, 04:35 PM
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Quote Quote by fauboca View Post
Does it have to do with the winding number?
Not really. It has to do with allowing more general [itex]\gamma[/itex] instead of just circles.

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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