By Kehe Zhu (auth.)

Several typical Lp areas of analytic features were broadly studied long ago few many years, together with Hardy areas, Bergman areas, and Fock areas. The phrases “Hardy areas” and “Bergman areas” are by means of now common and good verified. however the time period “Fock areas” is a distinct story.

Numerous very good books now exist with reference to Hardy areas. a number of books approximately Bergman areas, together with a few of the author’s, have additionally seemed some time past few many years. yet there was no booklet out there in regards to the Fock areas. the aim of this booklet is to fill that void, particularly while many leads to the topic are entire by means of now. This e-book offers vital effects and strategies summarized in a single position, in order that new comers, particularly graduate scholars, have a handy connection with the subject.

This ebook includes proofs which are new and easier than the prevailing ones within the literature. particularly, the e-book avoids using the Heisenberg crew, the Fourier remodel, and the warmth equation. this assists in keeping the necessities to a minimal. a typical graduate path in each one of genuine research, complicated research, and sensible research may be adequate education for the reader.

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**Additional resources for Analysis on Fock Spaces**

**Sample text**

For α > 0 and p > 0, we use the notation Lαp to denote the space of Lebesgue 2 measurable functions f on C such that the function f (z)e−α |z| /2 is in L p (C, dA). p For f ∈ Lα , we write f p p,α = pα 2π α C f (z)e− 2 |z| 2 p dA(z). 1) Similarly, for α > 0 and p = ∞, we use the notation L∞ α to denote the space of Lebesgue measurable functions f on C such that f ∞,α = esssup | f (z)|e−α |z| 2 /2 : z ∈ C < ∞. 2) ∞ Obviously, we have Lαp = L p (C, dλ pα /2) for 0 < p < ∞. But L∞ α = L (C, dA). p When 1 ≤ p ≤ ∞, Lα is a Banach space with the norm f p,α .

Furthermore, equality is attained if and only if F is constant. This shows that the extremal functions are of the form f (w) = eα zw−(α |z| 2 /2)+iθ This proves the desired results for 0 < p < ∞. If p = ∞, it follows from the definition of f with f ∞,α ≤ 1. Therefore, sup{| f (z)| : f ∞,α ∞,α . that | f (z)| ≤ eα |z| 2 /2 ≤ 1} ≤ eα |z| 2 /2 . for all f 38 2 Fock Spaces On the other hand, the function f (w) = kz (w) is a unit vector in Fα∞ and kz (z) = 2 eα |z| /2 . Thus, we actually have sup{| f (z)| : f ≤ 1} = eα |z| 2 /2 ∞,α .

Once again, we consider functions of the form fx,k (z) = e−x|z| zk , 2 z ∈ C, where x > 0 and k is a positive integer. 2 Some Integral Operators 47 = α (β −α )|z|2 e β +x = α (β −α )|z|2 e β +x α β +x = 1+k C e(β +x)[α z/(β +x)]w¯ wk dλβ +x(w) αz β +x k e(β −α )|z| zk . 2 Suppose 1 < p ≤ 2 and 1/p + 1/q = 1. If the operator Pα is bounded on L p (C, dλβ ), then the operator Pα∗ is bounded on Lq (C, dλβ ). So there exists a positive constant C, independent of x and k, such that C |Pα∗ ( fx,k )|q dλβ ≤ C C | fx,k |q dλβ .