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Topic: Do you solve this problem ? (Read 3203 times) |
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immanuel78
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Do you solve this problem ?
« on: Jul 27th, 2006, 8:39pm » |
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Here is a problem.
Let f : C-->C be continuous.
suppose that f is analytic off [-1,1].
Then f is entire.
I have thought this problem for 4 hours, but I can't solve this problem.
If someone can solve this problem, please give me hint or solution.
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| « Last Edit: Jul 27th, 2006, 8:40pm by immanuel78 » |
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Icarus
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Re: Do you solve this problem ?
« Reply #1 on: Jul 28th, 2006, 1:03pm » |
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Consider the circle Q of radius 2 about the origin, and the line segment L = [-1, 2]. We can traverse L in two directions: Let L+ traverse it from -1 to 2, and L- traverse it from 2 to -1. Finally, we construct the contour G = Q + L- + L+. f is analytic on the interior of G and continuous on G itself, so Cauchy's formula applies:
f(a) = (1/2i pi)  G f(z)dz/(z - a).
But the integrals on L+ and L- cancel, so f(a) = (1/2i pi) Q f(z)dz/(z - a) for all a inside Q and not on L.
But every a on L is the limit of values off L, and both f and the integral formula are continuous, so by taking the limit, we see that the integral formula must converge to f(a) for every a inside Q. The integral formula is analytic on all open sets on which it converges, hence f is analytic everywhere inside Q, and thus everywhere on C.
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Eigenray
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Re: Do you solve this problem ?
« Reply #2 on: Aug 8th, 2006, 3:56am » |
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This is a typical application of Morera's theorem, and the main step in the proof of the Schwarz reflection principle.
Let T be a triangle. If T doesn't intersect [-1,1], then the integral of f over T is 0 by Cauchy's (or Goursat's) theorem. If T intersects [-1,1] in just an edge or a vertex, then the result follows from the previous case by considering slightly shrunken triangles and using continuity of f. Otherwise, cut T up into smaller triangles to which the previous cases apply.
Now let U be a region symmetric wrt the real axis R, and let U+, U- be the intersections of U with the upper-, lower- half planes, respectively, and let I be the intersection of U with R. The Schwarz reflection principle states that if f is holomorphic on U+, and extends continuously to a real-valued function on I, then f may be extended to U- via f(z') = f(z)' (where ' denotes conjugation), and it will be holomorphic on all of U. To prove this, it is easily checked that the extended f is holomorphic on both U+ and U-, and is continuous on all of U. So we are in the situation of your problem: it remains only to show f is holomorphic on I.
Proof of Morera's theorem: Let z0 be any fixed point in the region D, and let U be a ball around z0 contained in D. For any z in U, let F(z) be the integral of f along a polygonal path within U from z0 to z. By hypothesis, this is independent of the path chosen. Now, for any fixed z in U, for h sufficiently small, F(z+h)-F(z) is (again using independence of paths) the integral of f along a straight line from z to z+h:
F(z+h)-F(z) = [int] (f(z) + [f(w)-f(z)]) dw = f(z)*h + [int] [f(w)-f(z)]dw.
By continuity of f, |f(w)-f(z)| < epsilon for |w-z|<delta, and we have
|(F(z+h)-F(z))/h - f(z)| < epsilon
for h sufficiently small, i.e., F is holomorphic and F' = f on U. It follows f is holomorphic on U, and in particular at z0, which was arbitrary.
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immanuel78
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Re: Do you solve this problem ?
« Reply #3 on: Aug 22nd, 2006, 3:36am » |
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Thank you for solving this problem.
But there are something that I can't understand in the proof of ICARUS.
In my book Cauchy's Integral formula is expressed as follows:
Let D be open in C(complex plane) and f : D->C be analytic.
Let L be a closed rectifiable curve in D s.t. n(L;w)=0 for all w in C - D.
For a in D - L, f(a)n(L;a)=1/2i(pi)[int]L f(z)dz/(z-a)
In the proof of ICARUS, I can't choose D.
Do you know the more generalized formula than what I know?
I will wait for your reply. Thank you.
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| « Last Edit: Aug 22nd, 2006, 6:48pm by immanuel78 » |
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Eigenray
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Re: Do you solve this problem ?
« Reply #4 on: Aug 22nd, 2006, 10:23am » |
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The usual workaround for that sort of thing is as follows:
Let D = C\[-1,1]. Let Q be the circle of radius 2 about the origin, traversed counter clockwise, and for n>0, let Rn be the rectangle joining the four points (-1-1/n +/- i/n), (2 +/- i/n), traversed clockwise. Let Ln be the concatenation of Q and Rn.
For any w in C\D = [-1,1],
n(Ln;w) = n(Q;w) + n(Rn;w) = 1 - 1 = 0,
so the theorem you quoted applies to any a not in Ln.
Pick a inside Q, outside [-1,2]. For n large enough, n(Rn;a) = 0, and n(Q;a)=1. Thus
(2pi i)f(a) = [int]Ln f(z)dz/(z-a) = [int]Q + [int]Rn.
Now g(z) = f(z)/(z-a) is continuous away from a, so for x in [-2,2],
g(x + iy) - g(x - iy) -> 0 as y->0,
and moreover the convergence is uniform in x by compactness of [-2,2]. Hence the integral of g over the top and bottom of Rn goes to 0 as n->infinity, and the integrals over the left and right sides go to 0 as well, since the length of these sides go to 0, and g is bounded on any compact subset of C\{a}.
Letting n->infinity, we have
f(a) = 1/(2pi i)[int]Q f(z)dz/(z-a)
for a inside Q, not in [-1,2], and proceed as before.
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| « Last Edit: Aug 22nd, 2006, 11:04am by Eigenray » |
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Icarus
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Re: Do you solve this problem ?
« Reply #5 on: Aug 22nd, 2006, 4:02pm » |
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There are a number of versions of Cauchy's Integral formula, all of which are extensions of the original. Indeed, the version you know is significantly more powerful that that which Cauchy proved.
The version I used can be expressed as:
Given a closed curve G of finite length enclosing an open region D, suppose that f is analytic in D and continuous on G. Then for all a in D,
f(a) = (n(G;a)/2 i) G f(z)dz/(z - a).
It can be proved from your version by approximating G with curves contained inside D. Your version applies to these curves, and by taking the limit as they approach G, you get the result above. This is essentially what Eigenray has done for the particular curve I used.
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| « Last Edit: Aug 27th, 2006, 6:14pm by Icarus » |
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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