# Thread: Analyticity on a closed set.

1. ## Analyticity on a closed set.

Hi:
Let C be the field of complex numbers and let D(z0,m) denote the open
disk with center z0 and radius m. Let U be an open subset of C,
closure of D(a,R) contained in U and f:U->C. Suppose further that
f is analytic on the closure of D(a,R). Proposition: for no r > R is f
analytic on D(a,r). Can this be true? Thank you for reading.

2. Originally Posted by ENRIQUESTEFANINI
Hi:
Let C be the field of complex numbers and let D(z0,m) denote the open
disk with center z0 and radius m. Let U be an open subset of C,
closure of D(a,R) contained in U and f:U->C. Suppose further that
f is analytic on the closure of D(a,R). Proposition: for no r > R is f
analytic on D(a,r). Can this be true? Thank you for reading.
Are you told that $\displaystyle f$ is analytic on $\displaystyle U$, if so then I think you can do the following proof. For each $\displaystyle z \in D(a,R)$ there is $\displaystyle \epsilon_z > 0$ so that $\displaystyle D(z,\epsilon_z) \subset U$. Therefore, $\displaystyle \mathcal{C} = \{ D(z,\epsilon_z) : |z-a| < R \}$ is an open covering of $\displaystyle \overline{D(a,R)}$. Since the closed disk is compact it follows by the Heine-Borel theorem that there is a finite subcovering of $\displaystyle \overline{D(a,R)}$. Let $\displaystyle V$ be the union of all these sets in the subcovering. Each element of $\displaystyle V$ is bounded therefore $\displaystyle V$ is an open bounded set which contains $\displaystyle \overline{D(a,R)}$. Let $\displaystyle C = \partial \overline{D(a,R)}$ and for each $\displaystyle z\in C$ let $\displaystyle \delta_z = \sup\{ m>0 | D(z,m)\subseteq V \}$. The supremum exists because we made sure that $\displaystyle V$ was bounded. The function $\displaystyle \delta_z : C \to \mathbb{R}$ is a continous function on a compact set $\displaystyle C$. Since $\displaystyle \delta_z$ does not assume zero on $\displaystyle C$ it means that $\displaystyle \inf \{ \delta_z : z\in C\} = \delta > 0$. Thus, the annulus $\displaystyle R - \delta < |z-a| < R + \delta$ is contained in $\displaystyle V$. The "hole" of the annulus is contained in $\displaystyle V$ because the hole is contained in $\displaystyle D(a,R)$ and this disk is contained in $\displaystyle V$. Thus, $\displaystyle D(a,r)$ is contained in $\displaystyle U$ (if it is contained in $\displaystyle V$ it is surly contained in $\displaystyle U$) for $\displaystyle r>R$, here we let $\displaystyle r= R+\delta$.

3. ## Analycity on a closed set.

I'm very sorry I didn't explain myself clearly. No. I'm only told f is analytic
on the closure of the disk. However, first I should have wondered if there
is such D(a,r) in S (I forgot to establish that D(a,r) must be contained in
S, but reading your proof I see you understood it that way). This is a

So the question remains. Is there a D(a,r) in S such that r > R and f
analytic on D(a,r)? Best regards and thank you for your time.

Enrique

4. Originally Posted by ENRIQUESTEFANINI
I'm very sorry I didn't explain myself clearly. No. I'm only told f is analytic
on the closure of the disk. However, first I should have wondered if there
is such D(a,r) in S (I forgot to establish that D(a,r) must be contained in
S, but reading your proof I see you understood it that way). This is a

So the question remains. Is there a D(a,r) in S such that r > R and f
analytic on D(a,r)? Best regards and thank you for your time.

Enrique
In this case I think you procede the exact same way i.e. this problem is more topological than analytic. The typical definition of analytic at a point means the function is differenciable in the neighborhood of this point*. Therefore for each $\displaystyle z\in C$ (the boundary of the disk) we have $\displaystyle r_z>0$ so that $\displaystyle f$ is differenciable on $\displaystyle D(z,r_z)$. And again as above define the continous function $\displaystyle r_z: C\to \mathbb{R}$, notice that $\displaystyle C$ is a compact set and $\displaystyle r_z$ is a positive function therefore $\displaystyle r = \inf \{r_z : z\in C\} >0$. Now argue that $\displaystyle f$ is analytic on $\displaystyle D(a,R+r)$. Does this work?

*)John Conway actually defined "analytic" to mean a function is differenciable on a neighborhood with a continous derivative (but then he later proves that differenciability on a neighborhood implies continuity of the derivative).

5. ## Analycity on a closed set.

Perfectly well. I have now revised the definition of analytic function as
given in the book I am studying and see it has a note saying: if T is an
arbitrary subset of the complex (not necessarily open), the terminology
"f is analytic on T" is used to mean that f is analytic on some open set
containing T. This would have saved me effort if only I had read more
carefully. By the way, it's like being in heaven to have such a nice people
at one's disposal to help in math problems.

P.S.: I could have used the note as is or I could have taken T to be
a point in the complex and used your proof. By the way again: whom
could I contact to get help on using the text editor. Thank you very

6. Originally Posted by ENRIQUESTEFANINI
if T is an
arbitrary subset of the complex (not necessarily open), the terminology
"f is analytic on T" is used to mean that f is analytic on some open set
containing T. This would have saved me effort if only I had read more
carefully. By the way, it's like being in heaven to have such a nice people
at one's disposal to help in math problems.
This definition is not only intrinsic for complex functions. In PDE's sometimes you will say let $\displaystyle u(x,t)$ be a solution to $\displaystyle u_{xx} = u_{tt}$ on $\displaystyle R=[0,1]\times [0,\infty)$. When I first seen this it used to bother me too because $\displaystyle u$ is $\displaystyle \mathcal{C}^2$ on $\displaystyle R$, a closed set, while differenciability is defined only on open sets! Well, what we mean to say is that $\displaystyle u = v|_{R}$ (notation here means that $\displaystyle u$ is a restriction of another function) where $\displaystyle v$ is a $\displaystyle \mathcal{C}^2$ function on $\displaystyle D$, and open set, so that $\displaystyle R\subset D$.

P.S.: I could have used the note as is or I could have taken T to be
a point in the complex and used your proof. By the way again: whom
could I contact to get help on using the text editor. Thank you very