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Holomorphic on Cby fauboca
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#19
Feb2112, 04:39 PM

P: 159

We have a theorem we call our qtheorem.
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. $$ q(z) =\frac{f(z)f(z)}{za} $$ But this only for a finite number of points and I have an infinite number. 


#20
Feb2112, 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)}{ub}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
Feb2112, 05:13 PM

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#22
Feb2112, 05:17 PM

P: 159




#23
Feb2112, 05:24 PM

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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)}{wz} dw + \int_C \frac{f(w)}{wz} dw = \int_{\gamma_\epsilon} \frac{f(w)}{wz} 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]. 


#24
Feb2112, 05:28 PM

P: 159

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


#25
Feb2112, 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]. 


#26
Feb2112, 05:35 PM

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#27
Feb2112, 05:37 PM

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Try to think it over for a bit...



#28
Feb2112, 05:42 PM

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#29
Feb2212, 01:56 PM

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#30
Feb2712, 12:00 PM

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#31
Feb2712, 05:37 PM

P: 159

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? 


#32
Feb2712, 07:53 PM

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Thanks
PF Gold
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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".



#33
Feb2812, 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)}{za}[/itex] Step 1 is the CauchyGoursat 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 subrectangles so that each subrectangle has at most one [itex]\xi_j[/itex] inside it. By CauchyGoursat, [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 CauchyGoursat. 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 subrectangles 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 welldefined. 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? 


#34
Feb2912, 10:41 AM

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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 1ε 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. 


#35
Feb2912, 06:55 PM

P: 159




#36
Feb2912, 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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