Let R be a connected open region in the plane, and let fn be a sequence of holomorphic functions on R such that {Re(fn)} converges uniformly on compact subsets to a harmonic function u. Suppose there exists z0 in R such that v(z0)=lim Im(fn(z0)) exists. I wish to prove that {fn} converges uniformly on compact subsets.
Why do I wish to prove such a thing? Because I'm preparing to take my Ph.D. qualifying exams in September, which make strong women weep and strong men befoul themselves, and the more of these problems I can do, the better my chances are. Also, I had trouble finding references to the Herglotz integral formula online...Maybe someone out there will find this useful. Also I think the name 'Herglotz' is just really keen. Apparently people liked him. I'm a big fan of his hair. Can't find a biography of him at the University of St Andrews site, which is a shame...I like them.
{fn} is normal on R iff it is normal on all disks D whose closures are contained in R. It is normal in such a disk, by the Arzela-Ascoli Theorem, if it is equicontinuous on compact subsets, and pointwise bounded. Let K be a compact subset of D, and let z and w be any points of K. We may map D linearly to the open unit disk U, and apply the Herglotz integral formula, which is just like the Poisson formula, only for analytic rather than harmonic functions:
Thus on each compact subset K of a disk D in R, there is a constant C with |fn(z)-fn(w)| less than C|z-w| for all n and for all z, w in K.
Suppose in some disk D there is a point z with {fn(z)} bounded. For all z in D, there is a compact curve g contained in D (a straight line segment, even) connecting z and z. By the above, |fn(z)-fn(z)| is less than C|z-z| for all n, so |fn(z)| is less than sup|fn(z)|+C|z-z| for all n: {fn} is pointwise bounded on D. If E is the set of points z in R for which {fn(z)} is bounded, this shows E is open (and nonempty, since z0 is in it). Now suppose w is in R-E; since R is open, there is a disk D containing w whose closure lies in R, which contains an even smaller disk whose closure is a compact subset of D containing w. For all z in this smaller disk, |fn(z)-fn(w)| is less than C|z-w| for all n; since {fn(w)} is not bounded, neither is {fn(z)}. Thus R-E is open, so E is closed. Since R is connected and E is nonempty, open and closed, E=R: {fn} is always pointwise bounded. Hence the family is normal on all disks D, and thus on all of R. It contains a subsequence converging normally to some analytic function f on R. Of necessity, Re(f)=u. Suppose g is any other normal subsequential limit of {fn}. Then Re(g)=u also, so f-g is an analytic, pure imaginary function: the Cauchy-Riemann equations imply f-g is an imaginary constant A. But since lim Im(fn(z0)) exists, the imaginary parts of f and g are both equal at z0, so A=0. Thus any normally convergent subsequence converges to f. Suppose {fn} as a whole did not converge to f uniformly on some compact subset K: then on this subset, there would be e greater than zero such that for all N, there exists an n greater than N with sup|fn-f| greater than e. Therefore we can construct a subsequence {fnk} with sup|fnk-f| greater than e for all k. But this subsequence in turn contains a subsequence converging normally, necessarily to f, since {fn} is a normal family, which is a contradiction. Therefore {fn} is uniformly convergent on compacta.
Posted by aloysius at August 02, 2003 04:46 PM | TrackBack |