Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let be $\mathcal M(\partial\mathbb D)$ denote the set of all Borel complex probability measures on $\partial\mathbb D$ (unit circle in the complex plane). Define a mapping $\Phi:\mathcal M(\partial\mathbb D)\to \ell^{\infty}(\mathbb N)$ by $$ \Phi(\mu)=(c_0,c_1,\ldots) $$ where $c_k$, are the coefficients of the Taylor expansion (around $z=0$) of the function $$ f(z)=\int_{\partial\mathbb D}\frac{1}{1-wz} d\mu(w), $$ Question: Who is the set $\Phi(\mathcal M(\partial\mathbb D))$ ? Given a sequence is there a simple criterion to verify if it belongs to this space?

Motivation: Let be $A=(a_0,a_1,\ldots)\in\Phi(\mathcal M(\mathbb D))$, $g:U\subset\mathbb C\to \mathbb C$ analytic with $$g(z)=\sum_{k=0}^{\infty}b_kz^k$$ Define $A*g:U\subset\mathbb C\to \mathbb C$ by $$ A*h(z)=\sum_{k=0}^{\infty} a_kb_kz^k. $$ The above integral representation, can be used to give a short proof of $$ \|A*h\|_{U}\leq \|h\|_{U}, $$ where $\|g\|_{U}=\sup_{z\in U}|g(z)|$.

PS:This question it is a generalization of a question that arose in a discussion at Area 51.

Edition: I am correcting the question, because in the previous version the integrals could have no meaning. In fact, I was looking for the criterion by a line integral representation.

share|improve this question
    
Shouldn't the integral be over the whole disc, rather than the boundary? (Since the measure $\mu$ is not defined on the boundary). If so, the integral is a kind of Bergman-like projection of the measure. (The Bergman projection has $(1-\bar{w}z)^{-2}$ instead). It's a kind of weighted operator of Hankel type; so I would guess Hankel operators, Toeplitz operators, weighted Bergman spaces, maybe weighted Hardy spaces are probably useful phrases to search for. –  Zen Harper Aug 13 '10 at 9:21
    
Thank you Zen for the answer. I was think about the representation by a line integral. I edited the question. –  Leandro Aug 13 '10 at 14:22
    
What is a "complex probability" measure? I know "complex measure" and "probability measure". –  Gerald Edgar Aug 13 '10 at 16:25
add comment

2 Answers 2

up vote 3 down vote accepted

There's probably something I do not understand about your question, in case just forget my babbling. Anyway let me try: if you expand $$\frac{1}{1-wz}=\sum_{k\ge0}z^kw^k$$ and write $$f(z)=\sum_{k\ge0}z^k\int_{\partial D}w^k d\mu(w)= \sum_{k\ge0}z^k\int_0^{2\pi}e^{ik\theta}d\mu(\theta)$$ you see that the $c_k$ are the Fourier coefficients of positive order of the measure $\mu$. Now I would not go too far and say that any bounded sequence may apply, but this is well documented in several places (see e.g. the trigonometric moment problem).

EDIT: and maybe this is a good starting point.

share|improve this answer
    
Hi Piero, Thank you the answer and the reference. The question really reduce to find those Toeplitz positive semi-definite matriz. –  Leandro Aug 13 '10 at 18:53
    
Yes, so now the integral is on the boundary $\partial D$, the question becomes a bit easier (using Hardy spaces instead of Bergman spaces). Unfortunately I don't think the positive semi-definite matrix methods in the trigonometric moment problem can be generalised sensibly for general complex measures $\mu$; also it involves only finitely many Fourier coefficients. I think there's no simple characterisation of Fourier coefficients of general complex measures, only partial results (and I think it's generally believed that no good answer is possible). But the cauchy.pdf file is very good! –  Zen Harper Aug 16 '10 at 2:44
    
@Zen Harper: Yes, I'am enjoying a lot reading that paper. I think you right about the characterization of complex measures. In fact we are not restrict to positive measures as it is considered in wikipedia trigonometric moment problem link. –  Leandro Aug 16 '10 at 3:48
add comment

EDIT: this was for a different problem, but it has now been changed; so ignore this!

Too long for a comment...

If the integral is over $D$, then $f'(z)$ is (after changing variable $w$ to $\bar{w}$ in the measure $\mu$) \[ \frac{d}{dz} \int_D \frac{d\mu(w)}{1-\bar{w}z} = \int_D \frac{\bar{w} \, d\mu(w)}{(1-\bar{w}z)^2} = P_{L^2_a}(\bar{w} \mu), \] which is (formally) the $L^2_a(D)$ Bergman space projection operator applied to the measure $\bar{w} \mu$; but I don't know if this function of $z$ necessarily does lie in $L^2_a$.

So it's closely connected to Bergman space Toeplitz operators, which are known to be very tricky (much worse than on the Hardy space). Even worse, there is usually no nice way to characterise various function spaces which arise in terms of Taylor coefficients.

So your question really combines two interesting and highly non-trivial problems! I doubt there is any really simple answer.

share|improve this answer
    
Really nice the comment you made about the two connections. But in fact, I think what I would like to ask it was "less connective". Is there any hope that the things less hard replacing the integral on the disc for the line integral ? Thank you again ! –  Leandro Aug 13 '10 at 14:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.