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Holomorphic on C

by fauboca
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fauboca
#19
Feb21-12, 04:39 PM
P: 159
We have a theorem we call our q-theorem.

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.
fauboca
#20
Feb21-12, 05:06 PM
P: 159
Let [itex]x\in[2,5][/itex]. Then
[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.
morphism
#21
Feb21-12, 05:13 PM
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Quote Quote by fauboca View Post
Now g is holomorphic everywhere inside C so g is the extension of f to the interior of C.
No - you still need to show that g(z)=f(z) for every z in C\[2,5].
fauboca
#22
Feb21-12, 05:17 PM
P: 159
Quote Quote by morphism View Post
No - you still need to show that g(z)=f(z) for every z in C\[2,5].
I don't know what to do.
morphism
#23
Feb21-12, 05:24 PM
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Quote Quote by fauboca View Post
I don't know what to do.
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].
fauboca
#24
Feb21-12, 05:28 PM
P: 159
Quote Quote by morphism View Post
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].
How is it done via Morera?

Doesn't Morera only say we have a primitive and closed curves are 0. How does that make it an easier approach?
morphism
#25
Feb21-12, 05:30 PM
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Morera tells you that f must be holomorphic in a region G if [itex]\int_\gamma f = 0[/itex] for all closed curves [itex]\gamma[/itex] in G. The converse is of course Cauchy's theorem.

So you just need to show that [itex]\int_\gamma f = 0[/itex] for certain curves [itex]\gamma[/itex].
fauboca
#26
Feb21-12, 05:35 PM
P: 159
Quote Quote by morphism View Post
Morera tells you that f must be holomorphic in a region G if [itex]\int_\gamma f = 0[/itex] for all closed curves [itex]\gamma[/itex] in G. The converse is of course Cauchy's theorem.

So you just need to show that [itex]\int_\gamma f = 0[/itex] for certain curves [itex]\gamma[/itex].
How is that shown?
morphism
#27
Feb21-12, 05:37 PM
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Try to think it over for a bit...
fauboca
#28
Feb21-12, 05:42 PM
P: 159
Quote Quote by morphism View Post
Try to think it over for a bit...
By Goursat's rectangle method?
morphism
#29
Feb22-12, 01:56 PM
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Quote Quote by fauboca View Post
By Goursat's rectangle method?
Yes, something like that would work.
fauboca
#30
Feb27-12, 12:00 PM
P: 159
Quote Quote by morphism View Post
Yes, something like that would work.
I am still lost on how to do this though.
fauboca
#31
Feb27-12, 05:37 PM
P: 159
Quote Quote by morphism View Post
Yes, something like that would work.
So I am trying to use Morera's Theorem:
Let U be an open set in C and let f be continuous on U. Assume that the integral of f along the boundary of every closed rectangle in U is 0. Then f is holomorphic.

So let [itex]U = \mathbb{C} - [2,5][/itex] Let R be rectangles in U which are parallel to the coordinate axes. So [itex]\int_{\partial R}f=0[/itex].

Now how can I use this?
HallsofIvy
#32
Feb27-12, 07:53 PM
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Thanks
PF Gold
P: 39,682
By the way, "holomorphic at every point" doesn't make sense- "holomorphic" is not a point property. You may mean "analytic at every point". "Holomorphic" means "analytic at every point in the complex plane".
fauboca
#33
Feb28-12, 09:15 PM
P: 159
I have this proof for finite points but how would I modify it for infinite many points between [2,5]?

Assume [itex]q(z)[/itex] is any function that is holomorphic on a disc U except at a finite number of points [itex]\xi_1,\ldots, \xi_n\in U[/itex], and assume [itex]\lim_{z\to\xi_j}(z-\xi_j)q(z)=0[/itex] for [itex]1\leq j\leq n[/itex]. Let [itex]U'=U-\{\xi_1,\ldots\xi_n\}[/itex]. Then q is holomorphic on U'.
Note [itex]F(z)=q(z)-f'(a)[/itex] so [itex]q(z)=\frac{f(z)-f(a)}{z-a}[/itex]

Step 1 is the Cauchy-Goursat argument:
[itex]\int_{\partial R}q(z)dz=0[/itex] for all rectangles R in U such that [itex]\xi_j\notin\partial R[/itex] for j.

proof:

Let R be a rectangle with the boundary of R in U'. First subdivide R into sub-rectangles so that each sub-rectangle has at most one [itex]\xi_j[/itex] inside it. By Cauchy-Goursat, [itex]\int_{\partial R}=\sum_i\int_{\partial R_i}[/itex]. So it suffices to show [itex]\int_{\partial R}q=0[/itex] if R contains at most one [itex]\xi_j[/itex].
If R contains no [itex]\xi_j[/itex], then we are done by Cauchy-Goursat. Assume [itex]\xi=\xi_j[/itex] is inside R. Let [itex]\epsilon>0[/itex] be given. Put [itex]\xi[/itex] in a square of size x at the center of this where x is chosen small enough so that [itex]|(z-\xi)q(z)|<\epsilon[/itex] for z in and on this square. Subdivide this rectangle so the square which is x by x is its own rectangle around [itex]\xi[/itex]. As before, the integrals of the sub-rectangles that don't contain [itex]\xi[/itex] are 0. So [itex]\int_{\partial R}q=\int_{\text{square with xi}}q[/itex].
$$
\left|\int_{\text{square with xi}}q\right|\leq \underbrace{||q||_{\text{square with xi}}}_{\frac{\epsilon}{\min\{|z-\xi|\}}}\times \underbrace{(\text{length of square with xi}}_{4x}\leq\frac{\epsilon}{x/2}4x=8\epsilon
$$
So [itex]\left|\int_{\text{square with xi}}q\right|=0[/itex].

Step 2 is to use step one to create a primitive for q on all of U.
Define [itex]g(a)=\int_{z_0}^{z_1}q(z)dz[/itex] where the path is from [itex]z_0[/itex] horizontal and then vertical. So [itex]g(a)[/itex] is well-defined. If a point is not unreachable, then [itex]\lim_{h\to 0}\frac{g(a+h)-g(a)}{h}=q(a)[/itex] by exactly the same means as before. Unreachable are the points vertical above [itex]\xi[/itex].
When computing [itex]\frac{g(a+h)-g(a)}{h}[/itex], only consider h in C with [itex]|h|<\frac{\delta}{2}[/itex]. The path for computing g(a+h) also misses [itex]\xi[/itex]. Then for these h [itex]g(a+h)-g(a)=\int_a^{a+h}q[/itex] so the same reason as before show [itex]\frac{g(a+h)-g(a)}{h}-g(a)\to 0[/itex] as [itex]h\to 0[/itex].

To handle all the bad exception.
Pick [itex]\epsilon\geq 0[/itex] such that the point [itex]z_1=z_0+\epsilon(1+i)[/itex] is not on any of the same vertical or horizontal lines as any [itex]\xi_j[/itex]. Of course epsilon is really small compared to the radius of U.
Define [itex]g_1(a)=\int_{z_1}^aq(z)dz[/itex]. This defines a primitive for q(z) on all of the disc except on the unreachable regions.
Define [itex]g_2(a)=\int_{z_1}^aq(z)dz[/itex] but this time move vertical and then horizontal. So [itex]g_2(a)[/itex] is also a primitive for q on U except at the its unreachable points (but g_1 reaches those points and vice versa). On the overlap, [itex]g_1[/itex] and [itex]g_2[/itex] are primitives for the same q. They differ only by a constant. But [itex]g_1(z_1)=0=g_2(z_1)[/itex] so [itex]g_1=g_2[/itex] on the overlap. For a small enough epsilon,
$$
g(z)=\begin{cases}g_1(z)\\g_2(z)\end{cases}
$$ is defined on all of U' and is a primitive for q.

How do I extend this to my problem?
morphism
#34
Feb29-12, 10:41 AM
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Quote Quote by HallsofIvy View Post
By the way, "holomorphic at every point" doesn't make sense- "holomorphic" is not a point property. You may mean "analytic at every point". "Holomorphic" means "analytic at every point in the complex plane".
Not true; what you're thinking of is "entire".

re: fauboca

Here's what you do. Let ##\gamma## be a closed curve in C. If ##\gamma## doesn't contain any point of [2,5] in its interior, then ##\int_\gamma f =0## since f is holomorphic away from [2,5]. So now suppose that ##\gamma## contains points of [2,5] in its interior. For the sake of illustration, suppose that ##\gamma## is a square with vertices (3,0), (3,1), (4,1) and (4,0). Cut the square horizontally into two rectangles of length 1-ε and ε, so that the smaller rectangle has its base on [3,4]. This breaks up your contour into two contours, and the integral of f over one of them is zero. Now use a limit argument.

The general case is similar.
fauboca
#35
Feb29-12, 06:55 PM
P: 159
Quote Quote by morphism View Post
Not true; what you're thinking of is "entire".

re: fauboca

Here's what you do. Let ##\gamma## be a closed curve in C. If ##\gamma## doesn't contain any point of [2,5] in its interior, then ##\int_\gamma f =0## since f is holomorphic away from [2,5]. So now suppose that ##\gamma## contains points of [2,5] in its interior. For the sake of illustration, suppose that ##\gamma## is a square with vertices (3,0), (3,1), (4,1) and (4,0). Cut the square horizontally into two rectangles of length 1-ε and ε, so that the smaller rectangle has its base on [3,4]. This breaks up your contour into two contours, and the integral of f over one of them is zero. Now use a limit argument.

The general case is similar.
Shouldn't the square contain the interval? So it would be (1.9,-.1), (1.9, .1), (5.1,-.1),(5.1,.1)?
fauboca
#36
Feb29-12, 09:13 PM
P: 159
Let ##\gamma## be a closed curve in ##\mathbb{C}##. If ##\gamma## doesn't contain any point from [2,5] in its interior, then ##\int_{\gamma}f=0## since f is holomorphic away from [2,5]. Suppose that ##\gamma## contains [2,5] in its interior. Let R be a rectangle oriented with the coordinate axes in ##\gamma## such that [2,5] is in R such that ##[2,5]\notin\partial R##. Divide the rectangle into two sub rectangles of length ##1-\epsilon## and ##\epsilon## such that [2,5] is contained in one the sub rectangles. WLOG suppose [2,5] is the rectangle of length ##\epsilon##. Then the over f of the rectangle of length ##1-\epsilon## is zero.

I am not sure how you mean to use the limit argument.


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