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Math Help - Analyticity on a closed set.

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    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 f is analytic on U, if so then I think you can do the following proof. For each z \in D(a,R) there is \epsilon_z > 0 so that D(z,\epsilon_z) \subset U. Therefore, \mathcal{C} = \{ D(z,\epsilon_z) : |z-a| < R \} is an open covering of \overline{D(a,R)}. Since the closed disk is compact it follows by the Heine-Borel theorem that there is a finite subcovering of \overline{D(a,R)}. Let V be the union of all these sets in the subcovering. Each element of V is bounded therefore V is an open bounded set which contains \overline{D(a,R)}. Let C = \partial \overline{D(a,R)} and for each z\in C let \delta_z = \sup\{ m>0 | D(z,m)\subseteq V \}. The supremum exists because we made sure that V was bounded. The function \delta_z : C \to \mathbb{R} is a continous function on a compact set C. Since \delta_z does not assume zero on C it means that \inf \{ \delta_z : z\in C\} = \delta > 0. Thus, the annulus R - \delta < |z-a| < R + \delta is contained in V. The "hole" of the annulus is contained in V because the hole is contained in D(a,R) and this disk is contained in V. Thus, D(a,r) is contained in U (if it is contained in V it is surly contained in U) for r>R, here we let r= R+\delta.
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  3. #3
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    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
    purely topological question and your proof answers it in the affirmative.

    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
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    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
    purely topological question and your proof answers it in the affirmative.

    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 z\in C (the boundary of the disk) we have r_z>0 so that f is differenciable on D(z,r_z). And again as above define the continous function r_z: C\to \mathbb{R}, notice that C is a compact set and r_z is a positive function therefore r = \inf \{r_z : z\in C\} >0. Now argue that f is analytic on 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).
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  5. #5
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    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
    much for your kindness.
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  6. #6
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    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 u(x,t) be a solution to u_{xx} = u_{tt} on R=[0,1]\times [0,\infty). When I first seen this it used to bother me too because u is \mathcal{C}^2 on R, a closed set, while differenciability is defined only on open sets! Well, what we mean to say is that u = v|_{R} (notation here means that u is a restriction of another function) where v is a \mathcal{C}^2 function on D, and open set, so that 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
    much for your kindness.
    See this.
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