Background: Let $S$ denote the so-called Schur class of complex analytic functions from the open unit disk $D$ in $\mathbb{C}$ to the closed unit disk $\overline{D}$. Given distinct points $z_1,\ldots,z_n\in D$ and points $w_1,\ldots,w_n\in\overline{D}$, G. Pick's Theorem (1916) says that there is an $f\in S$ such that $f(z_j)=w_j$ for each $j$ if and only if the matrix $\left(\frac{1-w_j\overline{w_k}}{1-z_j\overline{z_k}}\right)_{j,k=1}^n$ is positive semidefinite. Suppose that $g:D\to\mathbb{C}$ is a function. By Pick's Theorem it follows immediately that a necessary condition for $g$ to be in $S$ is that for each $n$ and each $n$-tuple of distinct points $z_1,\ldots,z_n\in D$, the matrix $\left(\frac{1-g(z_j)\overline{g(z_k)}}{1-z_j\overline{z_k}}\right)_{j,k=1}^n$ is positive semidefinite. I find it remarkable that this is also a sufficient condition (this too can be derived using Pick's Theorem, although there is a more general method, as indicated in Edit 2 and Yemon Choi's answer). However, I do not know whether Pick observed this fact, and I don't see it in I. Schur's related work from around that time (1917-1918).

Question: Who first observed that a function $g:D\to\overline{D}$ is analytic if for each $n$ and each $n$-tuple of distinct points $z_1,\ldots,z_n\in D$, the matrix $\left(\frac{1-g(z_j)\overline{g(z_k)}}{1-z_j\overline{z_k}}\right)_{j,k=1}^n$ is positive semidefinite?

To elaborate a little, the theory of reproducing kernel Hilbert spaces and their multipliers in which this characterization can now be found was developed well after people including Pick, Nevanlinna, and Schur had developed other aspects of the theory of bounded analytic functions on the unit disk. Part of the reason I would be interested in knowing the origin is that it would be interesting to see if completely different methods were used. If different methods weren't used, learning of the origin of this would help me to better understand the history of the uses of Hilbert space in function theory.

Edit 1: In an attempt to increase clarity, I changed the codomain from $\mathbb{C}$ to $\overline{D}$, so that the trivial condition of being bounded by 1 is no longer emphasized.

Edit 2: I'm still curious about the history of this, but I've realized that I was confused about the logical relationship between Pick's Theorem and the positivity criterion for analyticity. Yemon Choi's answer points out a better way to think about the latter; it is a simple criterion for being a multiplier of a reproducing kernel Hilbert space, and it works even for those RKHS for which the analogue of Pick's Theorem is false. The general criterion appears in section 2.3 of Agler and McCarthy's book Pick Interpolation and Hilbert Function Spaces, the identification of the multipliers of Hardy space appears in section 3.4, and Pick's Theorem is proved in section 5.2.

  • $\begingroup$ [Post edit] Ah, that was better. I deleted my correct but trivial answer to the wrong question. And no, I have no idea what the answer to the right question might be. $\endgroup$ Nov 24 '09 at 3:30

I don't know the answer to your question, but it might be of interest to observe that one can interpret the Pick condition for a function $g$ as saying that $g$ is a bounded multiplier for a certain reproducing kernel Hilbert space (see the book of Agler and McCarthy). The fact that said RKHS is the classical Hardy space, and consideration of $g\cdot 1$ where $1$ is the constant function, then tells us that $g$ is analytic.

However, as I don't have Agler and McCarthy's book to hand, I don't know what machinery gets used in proving the above. It could well be that they deduce it from something simpler and older which proves the automatic analyticity result more directly. In fact, thinking about this a little more, that must surely be the case. You probably end up showing that $g$ must agree on any finite subset of the disc with some bounded analytic function (a priori depending on the subset), and then e.g. Morera's theorem plus approximation ought to do the job.

  • $\begingroup$ Yes, that is interesting and precisely the context in which I was introduced to this condition. I don't have Agler and McCarthy on hand right now either, but I have seen this in there. If I'm not mistaken they prove it along the lines I vaguely indicated, by choosing a weak limit point of a bounded sequence of finite interpolating functions, where the finite sets of points are chosen so that their union is, e.g., dense in D. But as you point out there is something much more general going on, a characterization of multipliers of certain RKHS (those that have what A&M call the Pick property). $\endgroup$ Nov 24 '09 at 5:25
  • $\begingroup$ I was mistaken. The direct way you mentioned in your first paragraph is quite straightforward and doesn't take too much theory, and that is how A&G do it. $\endgroup$ Dec 7 '09 at 1:29
  • $\begingroup$ Of course I meant Agler and McCarthy (A&M, not A&G). $\endgroup$ Dec 19 '09 at 4:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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