Prove Holomorphic on C: Continuity and Differentiability

[fauboca]

Suppose f : \mathbb{C}\to \mathbb{C} is continuous everywhere, and is holomorphic at every point except possibly the points in the interval [2, 5] 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.

 

[morphism]

One approach is to use Morera's theorem.

 

[fauboca]

One approach is to use Morera's theorem.

I haven't learned that Theorem yet. Is there another approach.

 

[morphism]

Yes, but it's tricky. The key idea is that if you can find a holomorphic function g \colon \mathbb C \to \mathbb C 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]

Yes, but it's tricky. The key idea is that if you can find a holomorphic function g \colon \mathbb C \to \mathbb C 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]

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

 

[fauboca]

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

If g is a primitive of f, then g' = f. As long as f is on an open set which is the case here.

 

[morphism]

But how is g defined on [2,5]?

I have to run so I'll leave you with a hint:

Let C = \{ z \in \mathbb C \mid |z-3.5| < 1 \} and define g(z) = \frac{1}{2\pi i} \int_C \frac{f(w)}{w-z} dw in C. Try to show that f and g agree off of [2,5].

 

[fauboca]

But how is g defined on [2,5]?

I have to run so I'll leave you with a hint:

Let C = \{ z \in \mathbb C \mid |z-3.5| < 1 \} and define g(z) = \frac{1}{2\pi i} \int_C \frac{f(w)}{w-z} dw 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 \gamma be any path and g:\gamma\to\mathbb{C} be continuous. Define for all z \notin \gamma
G(z) = \int_{\gamma}\frac{g(u)}{u-z}du.
Then G(z) is analytic at every point z_0\notin\gamma.

Then our corollary to it
If f:\gamma\to\mathbb{C} is holomorphic and \gamma is inside a disc on which f is holomorphic and which \gamma 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]

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 \gamma be any path and g:\gamma\to\mathbb{C} be continuous. Define for all z \notin \gamma
G(z) = \int_{\gamma}\frac{g(u)}{u-z}du.
Then G(z) is analytic at every point z_0\notin\gamma.

Then our corollary to it
If f:\gamma\to\mathbb{C} is holomorphic and \gamma is inside a disc on which f is holomorphic and which \gamma 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]

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]

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 \gamma. But if \gamma=C, we run into problems, because we don't know if f is holomorphic on [2,5].

 

[fauboca]

The corollary requires f to be holomorphic inside \gamma. But if \gamma=C, 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]

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 \gamma be a circle, but really any simple closed curve works.

 

[fauboca]

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 \gamma be a circle, but really any simple closed curve works.

The definition of C you provided is a circle. Since \mathbb{C} is open, there is an open disc around C. So using that definition, we would have inside C is analytic.

 

[morphism]

The definition of C you provided is a circle. Since \mathbb{C} 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]

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]

Does it have to do with the winding number?

Not really. It has to do with allowing more general \gamma 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.

 

[fauboca]

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 \lim_{z\to\xi_j}(z-\xi_j)q(z) = 0. Where \xi_j 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]

Let x\in[2,5]. Then
<br /> g(b) = \frac{1}{2\pi i}\int_C\frac{f(u)}{u-b}du,<br />
i.e. g = f for all b\neq x 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 \mathbb{C} too.

 

[morphism]

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]

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]

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 \gamma_\epsilon be a small \epsilon-band around [2,5] fully contained in C (here I'm assuming the radius of C is > 1.5). Orient \gamma_\epsilon clockwise and C counterclockwise so that
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) for all z in between \gamma_\epsilon and C. This holds for all small \epsilon, so by letting \epsilon \to 0 and using the continuity of f, we see that the the integral over \gamma_\epsilon in the above equations goes to zero, and consequently f(z)=g(z) for all z in C\[2,5].

 

[fauboca]

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 \gamma_\epsilon be a small \epsilon-band around [2,5] fully contained in C (here I'm assuming the radius of C is > 1.5). Orient \gamma_\epsilon clockwise and C counterclockwise so that
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) for all z in between \gamma_\epsilon and C. This holds for all small \epsilon, so by letting \epsilon \to 0 and using the continuity of f, we see that the the integral over \gamma_\epsilon 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]

Morera tells you that f must be holomorphic in a region G if \int_\gamma f = 0 for all closed curves \gamma in G. The converse is of course Cauchy's theorem.

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

 

[fauboca]

Morera tells you that f must be holomorphic in a region G if \int_\gamma f = 0 for all closed curves \gamma in G. The converse is of course Cauchy's theorem.

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

How is that shown?

 

[morphism]

Try to think it over for a bit...

 

[fauboca]

Try to think it over for a bit...

By Goursat's rectangle method?

 

[morphism]

By Goursat's rectangle method?

Yes, something like that would work.

 

[fauboca]

Yes, something like that would work.

I am still lost on how to do this though.

 

[fauboca]

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

Now how can I use this?

 

[HallsofIvy]

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]

I have this proof for finite points but how would I modify it for infinite many points between [2,5]?

Assume q(z) is any function that is holomorphic on a disc U except at a finite number of points \xi_1,\ldots, \xi_n\in U, and assume \lim_{z\to\xi_j}(z-\xi_j)q(z)=0 for 1\leq j\leq n. Let U&#039;=U-\{\xi_1,\ldots\xi_n\}. Then q is holomorphic on U'.
Note F(z)=q(z)-f&#039;(a) so q(z)=\frac{f(z)-f(a)}{z-a}

Step 1 is the Cauchy-Goursat argument:
\int_{\partial R}q(z)dz=0 for all rectangles R in U such that \xi_j\notin\partial R 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 \xi_j inside it. By Cauchy-Goursat, \int_{\partial R}=\sum_i\int_{\partial R_i}. So it suffices to show \int_{\partial R}q=0 if R contains at most one \xi_j.
If R contains no \xi_j, then we are done by Cauchy-Goursat. Assume \xi=\xi_j is inside R. Let \epsilon&gt;0 be given. Put \xi in a square of size x at the center of this where x is chosen small enough so that |(z-\xi)q(z)|&lt;\epsilon for z in and on this square. Subdivide this rectangle so the square which is x by x is its own rectangle around \xi. As before, the integrals of the sub-rectangles that don't contain \xi are 0. So \int_{\partial R}q=\int_{\text{square with xi}}q.
$$
\left|\int_{\text{square with xi}}q\right|\leq \underbrace{||q||_{\text{square with xi}}}_{\frac{\epsilon}{\min\{|z-\xi|\}}}\times \underbrace{(\text{length of square with xi}}_{4x}\leq\frac{\epsilon}{x/2}4x=8\epsilon
$$
So \left|\int_{\text{square with xi}}q\right|=0.

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

To handle all the bad exception.
Pick \epsilon\geq 0 such that the point z_1=z_0+\epsilon(1+i) is not on any of the same vertical or horizontal lines as any \xi_j. Of course epsilon is really small compared to the radius of U.
Define g_1(a)=\int_{z_1}^aq(z)dz. This defines a primitive for q(z) on all of the disc except on the unreachable regions.
Define g_2(a)=\int_{z_1}^aq(z)dz but this time move vertical and then horizontal. So g_2(a) 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, g_1 and g_2 are primitives for the same q. They differ only by a constant. But g_1(z_1)=0=g_2(z_1) so g_1=g_2 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]

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]

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]

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.