- #1

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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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- #1

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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

morphism

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One approach is to use Morera's theorem.

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I haven't learned that Theorem yet. Is there another approach.One approach is to use Morera's theorem.

- #4

morphism

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- #5

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This is just a guess but let g be a primitive of f.

- #6

morphism

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

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

- #8

morphism

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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].

- #9

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I am not sure why you center your circle at 3.5.

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].

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.

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- #10

morphism

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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.)I am not sure why you center your circle at 3.5.

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].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.

- #11

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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 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].

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

morphism

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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].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.

- #13

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For now, can we say by the Integral Transform Theorem, we know g and f agree on all the points out side of C?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].

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

- #14

morphism

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The problem with your corollary is that it requires that [itex]\gamma[/itex] be a circle, but really any simple closed curve works.

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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.

The problem with your corollary is that it requires that [itex]\gamma[/itex] be a circle, but really any simple closed curve works.

- #16

morphism

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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]).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.

- #17

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Does it have to do with the winding number?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]).

- #18

morphism

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Not really. It has to do with allowing more general [itex]\gamma[/itex] instead of just circles.Does it have to do with the winding number?

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

- #19

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Suppose q(z) is any function that is holomorphic on a disc U except at a finite number of points in U, and assume the [itex]\lim_{z\to\xi_j}(z-\xi_j)q(z) = 0[/itex]. Where [itex]\xi_j[/itex] are those finite points. Then q has a primitive and is analytic inside the disc.

$$

q(z) =\frac{f(z)-f(z)}{z-a}

$$

But this only for a finite number of points and I have an infinite number.

- #20

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[tex]

g(b) = \frac{1}{2\pi i}\int_C\frac{f(u)}{u-b}du,

[/tex]

i.e. [itex]g = f[/itex] for all [itex]b\neq x[/itex] where x is a removable singularity.

Now g is holomorphic everywhere inside C so g is the extension of f to the interior of C.

Then let g=f on [itex]\mathbb{C}[/itex] too.

- #21

morphism

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No - you still need to show that g(z)=f(z) for every z in C\[2,5].Now g is holomorphic everywhere inside C so g is the extension of f to the interior of C.

- #22

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I don't know what to do.No - you still need to show that g(z)=f(z) for every z in C\[2,5].

- #23

morphism

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You're better off just abandoning this approach and using Morera's theorem for a more elegant solution.I don't know what to do.

But for the sake of completeness, this is how we could've wrapped up. Let [itex]\gamma_\epsilon[/itex] be a small [itex]\epsilon[/itex]-band around [2,5] fully contained in C (here I'm assuming the radius of C is > 1.5). Orient [itex]\gamma_\epsilon[/itex] clockwise and C counterclockwise so that

[tex] f(z) = \int_{\gamma_\epsilon} \frac{f(w)}{w-z} dw + \int_C \frac{f(w)}{w-z} dw = \int_{\gamma_\epsilon} \frac{f(w)}{w-z} dw + g(z)[/tex] for all z in between [itex]\gamma_\epsilon[/itex] and C. This holds for all small [itex]\epsilon[/itex], so by letting [itex]\epsilon \to 0[/itex] and using the continuity of f, we see that the the integral over [itex]\gamma_\epsilon[/itex] in the above equations goes to zero, and consequently f(z)=g(z) for all z in C\[2,5].

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How is it done via Morera?You're better off just abandoning this approach and using Morera's theorem for a more elegant solution.

But for the sake of completeness, this is how we could've wrapped up. Let [itex]\gamma_\epsilon[/itex] be a small [itex]\epsilon[/itex]-band around [2,5] fully contained in C (here I'm assuming the radius of C is > 1.5). Orient [itex]\gamma_\epsilon[/itex] clockwise and C counterclockwise so that

[tex] f(z) = \int_{\gamma_\epsilon} \frac{f(w)}{w-z} dw + \int_C \frac{f(w)}{w-z} dw = \int_{\gamma_\epsilon} \frac{f(w)}{w-z} dw + g(z)[/tex] for all z in between [itex]\gamma_\epsilon[/itex] and C. This holds for all small [itex]\epsilon[/itex], so by letting [itex]\epsilon \to 0[/itex] and using the continuity of f, we see that the the integral over [itex]\gamma_\epsilon[/itex] in the above equations goes to zero, and consequently f(z)=g(z) for all z in C\[2,5].

Doesn't Morera only say we have a primitive and closed curves are 0. How does that make it an easier approach?

- #25

morphism

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So you just need to show that [itex]\int_\gamma f = 0[/itex] for certain curves [itex]\gamma[/itex].

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