Functional Analysis

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By John Garnett

The publication, which covers a variety of attractive themes in research, is very good equipped and good written, with based, distinctive proofs. The e-book has informed an entire iteration of mathematicians with backgrounds in complicated research and serve as algebras. It has had a very good influence at the early careers of many prime analysts and has been commonly followed as a textbook for graduate classes and studying seminars in either the USA and abroad.

- From the quotation for the 2003 Leroy P. Steele Prize for Exposition

The writer has now not tried to provide a compendium. particularly, he has chosen quite a number subject matters in a many-faceted conception and, inside that variety, penetrated to enormous depth...the writer has succeeded in bringing out the great thing about a thought which, regardless of its particularly complicated age---now forthcoming eighty years---continues to shock and to thrill its practitioners. the writer has left his mark at the subject.

- Donald Sarason, Mathematical Reviews

Garnett's Bounded Analytic Functions is to operate thought as Zygmund's Trigonometric Series is to Fourier research. Bounded Analytic Functions is extensively considered as a vintage textbook used all over the world to coach modern-day practioners within the box, and is the first resource for the specialists. it truly is superbly written, yet deliberately can't be learn as a singular. relatively it supplies simply the ideal point of element in order that the inspired scholar develops the considered necessary abilities of the alternate within the strategy of studying the wonderful thing about the combo of actual, complicated and sensible analysis.

- Donald E. Marshall, collage of Washington

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1. 2. If u(z) is harmonic on H and if p > 1, then sup |u(x + i y)| p d x ≤ B p sup y y |u(x + i y)| p d x. 2 is false at p = 1. Take u(x, y) = Py ∗ f (x), f ∈ L 1 , f > 0. Then sup y |u(x, y)| ≥ M f (x) and M f ∈ L 1 . 3 (Fatou). assume Let u(z) be harmonic on H , let 1 ≤ p ≤ ∞ and sup u(x + i y) L p (d x) y < ∞. Then for almost all t the nontangential limit lim α (t) z→t u(z) = f (t) exists. If p > 1, u(z) is the Poisson integral of the boundary value function f (t), and if 1 < p < ∞, u(x + i y) − f (x) p →0 (y → 0).

The condition f ∈ L 1 (dt/(1 + t 2 )) merely guarantees that converges. Py (s) f (x − s) ds 22 Chap. I preliminaries Proof. We may assume t = 0. Let us first consider the points (0, y) on the axis of the cone α (0). Then u(0, y) = Py (s) f (s) ds, and the kernel Py (s) is a positive even function which is decreasing for positive s. That means Py (s) is a convex combination of the box kernels (1/2h)χ(−h,h) (s) that arise in the definition of Mf. Take step functions h n (s), which are also nonnegative, even, and decreasing on s > 0, such that h n (s) increases with n to Py (s).

If 1 2π v(z 0 + r eiθ ) dθ = −∞, then since v is bounded above and since Poisson kernels are bounded and positive, we have 1 2π Pz (θ)v(z 0 + r eiθ ) dθ = −∞, Consequently Un (z) → −∞ for each z ∈ −∞ on (z 0 , r ). The nonempty set {z ∈ |z| < 1. 3 v = : v(z) ≡ −∞ on a neighborhood of z} is then open and closed, and we again have a contradiction. 5. Let v(z) be a subharmonic function in the unit disc D. Assume v(z) ≡ −∞. For 0 < r < 1, let ⎧ |z| ≤ r, ⎨v(z), vr (z) = 1 ⎩ Pz/r (θ)v(r eiθ ) dθ, |z| < r.

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