# Holomorphic on C

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.

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morphism
Homework Helper
One approach is to use Morera's theorem.

One approach is to use Morera's theorem.
I haven't learned that Theorem yet. Is there another approach.

morphism
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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?

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

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
Homework Helper
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].

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.

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morphism
Homework Helper
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].

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
Homework Helper
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].

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
Homework Helper
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.

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
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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]).

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
Homework Helper
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.

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.

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.

morphism
Homework Helper
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].

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
Homework Helper
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].

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 $\gamma$.