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
$$g(b) = \frac{1}{2\pi i}\int_C\frac{f(u)}{u-b}du,$$
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]

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.