# Holomorphic on C

1. Feb 21, 2012

### 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.

2. Feb 21, 2012

### morphism

One approach is to use Morera's theorem.

3. Feb 21, 2012

### fauboca

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

4. Feb 21, 2012

### 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?

5. Feb 21, 2012

### fauboca

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

6. Feb 21, 2012

### morphism

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

7. Feb 21, 2012

### fauboca

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

8. Feb 21, 2012

### 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].

9. Feb 21, 2012

### fauboca

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.

Last edited: Feb 21, 2012
10. Feb 21, 2012

### morphism

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.)

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. Feb 21, 2012

### fauboca

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. Feb 21, 2012

### morphism

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].

13. Feb 21, 2012

### fauboca

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. Feb 21, 2012

### 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.

15. Feb 21, 2012

### fauboca

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.

16. Feb 21, 2012

### morphism

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. Feb 21, 2012

### fauboca

Does it have to do with the winding number?

18. Feb 21, 2012

### morphism

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.

19. Feb 21, 2012

### 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.

20. Feb 21, 2012

### fauboca

Let $x\in[2,5]$. Then
$$g(b) = \frac{1}{2\pi i}\int_C\frac{f(u)}{u-b}du,$$
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.