Is it true, that there exists a number C>0 such that for all r>0 and
each function g which is holomorphic on the set D(r):={|z| <= r} (a
subset of the complex numbers) the inequality

(int_{D(r)} exp(-|z|^2) |g(z)| dz)^2 <= C int_{D(r)} exp(-2|z|^2)|g(z)|^2 dz 

holds. An equivalent inequality is

(int_{D(r)} exp(-|z|^2+f(z)) dz)^2 <= C int_{D(r)} exp(-2|z|^2+2f(z)) dz 

where the real valued function f is harmonic, i.e. 
Laplacian(f)= (d^2/dz_1^2 + d^2/dz_2^2)f=0.

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Clarification of Question by florian22651-ga on 16 Mar 2005 01:15 PST

I should have emphasized that the point in the question is that C does
NOT depend on r. Cauchy-Schwarz only gives C<= Kr^2 where K does not
depend on r.
Meanwhile I discovered that function g(z)=exp(z^2) provides a counterexample and
the conjecture reads now

(int_{D(r)} exp(-|z|^2) |g(z)| dz)^2 <= C*(1+r) int_{D(r)} exp(-2|z|^2)|g(z)|^2 dz 

where C does not depend on r.

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Clarification of Question by florian22651-ga on 11 Apr 2005 06:09 PDT

I think that limeape-ga is correct. It would be too much work to
construct an explicit counterexample, therefore I am going to cancel
the question now. Thank you all for your comments!

I might be oversimplifying this but as your integrand is real valued
you might aswell replace it with an ordinary real integral over the
region D x^2+y^2<=1.  The inequality then holds (with c=1) as ^2
defines a norm, and we have the triangle inequality for integrals.

I'm with donatio-ga in seeing it as a fact about real-valued
integrals, but I view the proof in terms of a Cauchy-Schwarz
inequality (rather than a triangle inequality, although the two
estimates are intimately related).

Let G(z) = exp(-|z|^2)*|g(z)|, a real-valued function on the closed(?) disk D(r).

Whether you view this as a subset of C or R^2 makes no difference to
the usual notion of integration over an open domain; the two
dimensional "measure" is the same.

Taking a square root of both sides, the original formulation asks
whether the 1-norm of G(z) over D(r) is bounded by a constant times
its 2-norm.

Since the 1-norm of G(z) is the inner product of G(z) with the
constant function 1, we can estimate by the Cauchy-Schwarz inequality:

  ||G(z)||_1 = <G(z),1> <= ||G(z)||_2 * || 1 ||_2

Cf. the write-up at PlanetMath:

[Cauchy-Schwarz inequality]
http://planetmath.org/encyclopedia/CauchySchwarzInequality.html

So the constant C needed for the inequality depends on the size of
D(r), since in effect C is (at least) the square of the area of D(r).

regards, mathtalk-ga

Correction, C is simply the area of D(r).  Should have checked the
details more carefully.  The two-norm of constant function 1 is
||1||_2 = square root of area.

regards, mathtalk-ga

I believe that you can't do better than Cauchy-Schwartz. There exist
carefully chosen polynomials approaching the constants you get in the
C-S bound.

The idea is that you would really like the function to satisfy
|g(z)|=e^|z^2|. You can't do this for a holomorphic function, but you
can get close enough.

I have a fuller answer, but to give full details would entail checking etc....