# Does a Holomorphic Function Extend Continuously to the Unit Circle?

• symbol0
In summary: The maximum modulus principle says that |f| \leq 1 on D. The minimum modulus principle says that 1 \leq |f|. This means that |f|=1. There is not much to it.
symbol0
Show that there is no holomorphic function f in the unit disc D that extends continuously to |z|=1 such that f(z) =1/z for |z|=1

Some thoughts that might not be relevant:
If such f existed then, I can see that f maps the unit circle to the unit circle and the unit disc onto the unit disc.
On |z|=1 f would be equal to the conjugate function which is not differentiable anywhere.

I'm kind of stuck. I appreciate any suggestions.

Edit: wow it's way too late. Think integration

Do you mean that if I integrate f on circles with radii approaching 1, I get
0 = int f approaches int 1/z =2ipi ?

but does the continuity of f imply the continuity of the integral like that?

Office_Shredder said:
Think integration

Does integration work with this problem? I assume you had an argument like the following in mind:
If $f$ had an analytic extension to $\partial D$, then take the curve $\gamma:[0,1] \rightarrow \mathbb{C}$ given by $\gamma(t)=\exp(2 \pi i t)$. Notice that $\int_{\gamma} f = 2\pi i$ by calculation but that $\int_{\gamma} f = 0$ since $f$ is analytic. This is a contradiction so $f$ cannot have such an analytic extension.
But since the extension of $f$ to $\partial D$ is only continuous, none of the integration theorems for analytic functions will work here. The notion of what it means for $f$ to be analytic on $D \cup \partial D$ is ill-defined as well, since $D \cup \partial D$ is not open. Maybe I am just missing something here and you found a way to do this with integration (which would be pretty neat).

I am pretty sure that you can do this without integration though. Notice that the maximum/minimum modulus principle apply since $f$ is holomorphic on $D$ and continuous on $\partial D$. This means $|f|$ assumes its maximum and minimum values on $\partial D$; in particular, it follows that $|f|=1$ on $D$. From here you should be able to derive a contradiction.

Thank you jgens.
About your approach, you say f doesn't have a min in D.
How do we know 0 is not a min ?

symbol0 said:
About your approach, you say f doesn't have a min in D.

I said $|f|=1$ on $D$ which means that every point of $D$ is a minimum for $|f|$.

I want to understand how do you conclude |f| = 1 on D.
I know f cannot have a maximum in D. So |f|<1 on D.
...what else?

symbol0 said:
I want to understand how do you conclude |f| = 1 on D.
I know f cannot have a maximum in D. So |f|<1 on D.
...what else?

The maximum modulus principle says that $|f| \leq 1$ on $D$. The minimum modulus principle says that $1 \leq |f|$. This means that $|f|=1$. There is not much to it.

To get the minimum you can take 1/f as long as f us non-zero.

If you integrate around circles of radius r and let r approach one you can get a contradiction can't you?

To jgens: you say 1 leq |f|. How do we know f is not a function like f(z)=z which has a minimum in D ?

To office shredder: to get that contradiction don't we need to know that integrals are continuous ? is that a fact?

wouldn't the winding number around the origin be positive? whereas it seems to be -1.

If the curve is going counterclockwise, the winding number is greater or equal to 0.
And how is that related to the question?

## 1. What does it mean when a function doesn't exist?

A function not existing means that there is no mathematical rule or equation that can be used to describe the relationship between the input and output values. In other words, the function does not have a defined domain and range.

## 2. How can you tell if a function doesn't exist?

A function can be determined to not exist if there is a discontinuity or a break in the graph, or if the function produces undefined or imaginary output values for certain input values.

## 3. Can a function not exist at a single point?

Yes, a function can fail to exist at a single point, such as when there is a hole or a vertical asymptote in the graph. In this case, the function still exists for all other input values.

## 4. What is the difference between a function not existing and a function being undefined?

A function not existing means that there is no mathematical rule or equation to describe the relationship between the input and output values. On the other hand, a function being undefined means that the function is not defined for certain input values, but it may still have a defined domain and range.

## 5. Can a function not exist for all input values?

Yes, it is possible for a function to not exist for any input values. This can happen when the function has an infinite discontinuity or when the function approaches infinity as the input values get larger or smaller.

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