Prove Holomorphic on C: Continuity and Differentiability

  • Thread starter fauboca
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In summary, the key idea is to find a holomorphic function g \colon \mathbb C \to \mathbb C that agrees with f everywhere except possibly on [2,5], and then use the Integral Transform Theorem to show that f and g agree on all points outside of C. However, we run into problems when trying to apply this to points inside C but not on [2,5], as the corollary to the theorem requires f to be holomorphic inside the circle. Therefore, alternative approaches such as Morera's theorem may need to be used to show that f is holomorphic at all points of C.
  • #1
fauboca
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0
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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  • #2
One approach is to use Morera's theorem.
 
  • #3
morphism said:
One approach is to use Morera's theorem.

I haven't learned that Theorem yet. Is there another approach.
 
  • #4
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
morphism said:
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.
 
  • #6
That's not a bad idea. Could you spell it out a bit more?
 
  • #7
morphism said:
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.
 
  • #8
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].
 
  • #9
morphism said:
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.
 
Last edited:
  • #10
fauboca said:
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].
 
  • #11
morphism said:
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.
 
  • #12
fauboca said:
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].
 
  • #13
morphism said:
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]
 
  • #14
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
morphism said:
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.
 
  • #16
fauboca said:
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]).
 
  • #17
morphism said:
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?
 
  • #18
fauboca said:
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.
 
  • #19
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.
 
  • #20
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.
 
  • #21
fauboca said:
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].
 
  • #22
morphism said:
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.
 
  • #23
fauboca said:
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].
 
  • #24
morphism said:
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?
 
  • #25
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
morphism said:
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?
 
  • #27
Try to think it over for a bit...
 
  • #28
morphism said:
Try to think it over for a bit...

By Goursat's rectangle method?
 
  • #29
fauboca said:
By Goursat's rectangle method?
Yes, something like that would work.
 
  • #30
morphism said:
Yes, something like that would work.

I am still lost on how to do this though.
 
  • #31
morphism said:
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?
 
  • #32
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
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?
 
  • #34
HallsofIvy said:
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.
 
  • #35
morphism said:
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)?
 
<h2>1. What does it mean for a function to be holomorphic on C?</h2><p>A function is considered holomorphic on C if it is complex differentiable at every point in the complex plane. This means that the function has a well-defined derivative at every point in the complex plane, which is a stronger condition than just being differentiable on the real line.</p><h2>2. How can I prove that a function is holomorphic on C?</h2><p>To prove that a function is holomorphic on C, you must show that it satisfies the Cauchy-Riemann equations. These equations relate the partial derivatives of the function's real and imaginary parts, and if they are satisfied, the function is complex differentiable and therefore holomorphic on C.</p><h2>3. Is a holomorphic function on C always continuous?</h2><p>Yes, a holomorphic function on C is always continuous. This is a consequence of the Cauchy-Riemann equations, which imply that the function's real and imaginary parts are continuously differentiable. Therefore, the function is continuous on the complex plane.</p><h2>4. Can a function be holomorphic on C but not differentiable at a specific point?</h2><p>No, a function cannot be holomorphic on C if it is not differentiable at a specific point. Holomorphicity requires complex differentiability at every point in the complex plane, so if a function is not differentiable at a specific point, it cannot be holomorphic on C.</p><h2>5. Are all analytic functions on C holomorphic?</h2><p>Yes, all analytic functions on C are holomorphic. This is because an analytic function is one that can be expressed as a convergent power series, and any function that can be represented as a power series is complex differentiable and therefore holomorphic on C.</p>

1. What does it mean for a function to be holomorphic on C?

A function is considered holomorphic on C if it is complex differentiable at every point in the complex plane. This means that the function has a well-defined derivative at every point in the complex plane, which is a stronger condition than just being differentiable on the real line.

2. How can I prove that a function is holomorphic on C?

To prove that a function is holomorphic on C, you must show that it satisfies the Cauchy-Riemann equations. These equations relate the partial derivatives of the function's real and imaginary parts, and if they are satisfied, the function is complex differentiable and therefore holomorphic on C.

3. Is a holomorphic function on C always continuous?

Yes, a holomorphic function on C is always continuous. This is a consequence of the Cauchy-Riemann equations, which imply that the function's real and imaginary parts are continuously differentiable. Therefore, the function is continuous on the complex plane.

4. Can a function be holomorphic on C but not differentiable at a specific point?

No, a function cannot be holomorphic on C if it is not differentiable at a specific point. Holomorphicity requires complex differentiability at every point in the complex plane, so if a function is not differentiable at a specific point, it cannot be holomorphic on C.

5. Are all analytic functions on C holomorphic?

Yes, all analytic functions on C are holomorphic. This is because an analytic function is one that can be expressed as a convergent power series, and any function that can be represented as a power series is complex differentiable and therefore holomorphic on C.

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