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        general >> complex analysis >> double integral and essential singularity
        
(Message started by: immanuel78 on Dec 27th, 2006, 12:05am)
Title: double integral and essential singularity
Post by immanuel78 on Dec 27th, 2006, 12:05am
Let f be analytic in the region G={z:0<|z-a]<R}.
Show that if f has an essential singularity at z=a, then [double integral in G] |f(x+iy)|^2 dxdy = (infinity).
Title: Re: double integral and essential singularity
Post by Icarus on Jan 17th, 2007, 10:45am
This hasn't be responded to for awhile, and I want to try out towr's new script for inserting math symbols on a real problem, so I thought I'd give my current thoughts on it.

I would suggest assuming that http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifG |f(x+iy)|2 dxdy < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif. From this, you ought to be able to show that limzhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gifa (z - a)nf(z) = 0. By the Riemann condition, that implies f has a pole of order http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif n at a.
Title: Re: double integral and essential singularity
Post by hiyathere on Jan 18th, 2007, 10:29am

on 01/17/07 at 10:45:51, Icarus wrote:
This hasn't be responded to for awhile, and I want to try out towr's new script for inserting math symbols on a real problem, so I thought I'd give my current thoughts on it.


So how do you insert the math symbols?
Title: Re: double integral and essential singularity
Post by towr on Jan 18th, 2007, 11:08am

on 01/18/07 at 10:29:07, hiyathere wrote:
So how do you insert the math symbols?
To save another thread from going offtopic, I've summarized it here (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_suggestions;action=display;num=1035249563;start=25#49)
Title: Re: double integral and essential singularity
Post by Icarus on Jan 29th, 2007, 11:07am
Okay - here is an approach that seems to work - though I doubt it was the intended solution:

To ease the notation, we can assume wlog that a = 0. Assume the integral is finite. Express it in polar coordinates:  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif0http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/supcr.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif0http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/suppi.gif |f2(rehttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/supi.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/suptheta.gif)| rdhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gifdr.  For the outer integral to exist, the inner integral must be finite for almost all values of r. Again wlog, we can assume that it is finite for r = R. Now taking the circle of Radius R about 0 as the path, |http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif f2 dz | http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif |f2| |dz|, which is exactly the inside integral. Therefore http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif f2 dz is finite. Define F(w) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif zf2(z)/(z-w) dz. We see that F(w) is finite at all points inside the circle, including 0. F(w) is also holomorphic (a common lemma used in proving Cauchy's formula), and in fact, F(z) = zf2(z) + a for some constant a. This shows that f2(z) can have at most a pole of order 1 at 0. But it is easy to show that if f2 did have a pole at 0, the double integral would be infinite. Hence f2 is analytic at 0. So f is the square root of an analytic function, and hence f has a removable singularity at 0, or has 0 as a branch point. But the latter condition is impossible, since f is analytic everywhere in G. Therefore f must have a removable singularity.

Taking the contrapositive gives that if f has a non-removable singularity at a, the integral must be infinite.
Title: Re: double integral and essential singularity
Post by Eigenray on Feb 2nd, 2007, 4:54pm

on 01/29/07 at 11:07:50, Icarus wrote:
For the outer integral to exist, the inner integral must be finite for almost all values of r.

This can't be enough to show that f doesn't have an essential singularity.  Indeed, if f is any function continuous on {0<|z|<R}, then the inner integrals will all be finite, as the integral of a continuous function over a compact interval.

In fact, if f(z) = e1/z - 1/z, then http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif|z|=r f2(z) dz = 0 for all r>0.


Quote:
Define F(w) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif zf2(z)/(z-w) dz. We see that F(w) is finite at all points inside the circle, including 0. F(w) is also holomorphic (a common lemma used in proving Cauchy's formula), and in fact, F(z) = zf2(z) + a for some constant a.

F(z) is holomorphic, but only because it is the regular part (nonnegative powers in z) of the Laurent series for zf2(z).

The key to solving the problem is to consider the Laurent series of f:

f(z) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifn=-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/supinfty.gif an zn,

converging absolutely and uniformly on compact subsets of 0<|z|<R.  Then we have

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif|z|=r |f(z)|2 dhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif = 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif |an|2 r2n,

by orthogonality of the functions {ehttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/supi.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/supn.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/suptheta.gif : n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif}.  So

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif |f(z)|2 dxdy
= http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif0R r 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif |an|2r2n  dr
= 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subn.gif=-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/supinfty.gif |an|2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif0R r2n+1dr,

which can only be finite if an=0 whenever n<0, that is, if f actually has a removable singularity at 0.

Of course one needs to check that things converge appropriately to justify interchanging the order of integration and summation, but this boring.


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