Re: Maximum modulus question 2.

On 23-10-2007 18:15, Cooper wrote:

In addtion to the prior problem, the another problem does also seem
invincible.
That is,
"Let h be harmonic funtion in open disk. Then there is a sequence
{z_n} in U s.t.
|z_n| -> 1 and {h(z_n)} is bounded."
Any comments or hints are great welcome.
Hint: minimum modulus theorem.
Sorry. I cannot find light with only this hint.
Could you give me more hints please?
Say v is a harmonic conjugate of u, so that f = u + iv
is analytic. Let g = something(f). Then g is analytic,
g has no zero, and |g| = somethingelse(u). Apply the
minimum modulus theorem to g.
A further hint: What function do you know that guarantees no zeros? Use that for the function something() in David's hint. If you choose
something() well, then the choice of somethingelse() should be obvious.
- -
Good!
Thank you very much^^
I'm glad to see that you solved your problem. Unfortunately, I wasn't
able to solve it, even with the hints.
I suppose that David Ullrich's idea was this: take g = exp(f); then _g_
has no zeros and |g| = e^u. Since _g_ has no zeros and it is not
constant, |g| has no minimum. And... ? I thought that I could take, for
each natural number _n_, a number z_n such that |z_n| = 1 - 1/n and that
|g(z_n)| is the minimum of the restriction of _f_ to the closed disk
centered at 0 with redius 1 - 1/n. Then the sequence (|g(z_n)|)_n is a
decreasing one and, since each term is positive, it converges. We can
assume that (z_n)_n converges to some _z_ such that |z| = 1.
Now if lim_n|g(z_n)| > 0, we're done; lim_n u(z_n) will be a real
number. But if lim_n|g(z_n)| = 0 then no conclusion is reached, because
then lim_n u(z_n) will be -oo.
What did I miss?
Hmm. What _I_ missed is that the original function was h, not u.
Anyway:
There exists a point on the circle |z| = 1-1/n where u is
no larger than u(0). There also exists a point on the
same circle where u is no smaller than u(0). Hence...
I was so focused on the minimum modulus theorem (because that was the
suggestion of The World Wide Wade) that it never occurred that I should
also use the maximum modulus theorem (in spite of the name of the
thread). :-(

Indeed, I did not solve this problem as David suggested since I am not
yet covered the concept of harmonic conjugate. It is just exercise in
Rudin RCA in chap 13 which concerns mainly Maximum Modulus Thm along
with minimum modulus thm.

However, I solved this problem with another method keeping in mind the
original hint The World wide wade suggested.

I divided the case with three, that is,
(1)h has no zero in U
(2)h has finitely many zero in U
(3)h has infinitely zero in U

In case (1), we can use minimum principle.(|z_n|=1-1/n)

Do you mean that you use both the minimum principle and the maximum
principle like David C. Ullrich suggested? Actually, you can do it
always (I mean, without assuming any hypotheses concerning the zeros
of _h_). For harmonic functions the maximum principle and the minimum
principle just say that _h_ has no local maximum and no local minimum.

In case (2), if h has m zeros(counting multiplcity),

What is the multiplicity of a zero of a real-valued function of a
complex variable?

then h(z)=prod_i(z-a_i) *g(z) (here g(z) has no zero)
Then, |h(z)| <= 2^m*|g(z)| and by (1), we can choose such sequence z_n
keeping |h(z_n)| bounded.

But then _g_ will not even be real-valued function. So, you cannot use
the previous case.

In case (3), by identity thm, there exist such sequence.

The identity theorem doesn't apply here. If h(x,y) = x^2 - y^2, then
_h_ is harmonic, but h(1/n,1/n) = 0 for each natural _n_, in spite of
the fact that _h_ is not the null function.

Best regards,

Jose Carlos Santos
.

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