Norm of the Backward Shift and Related Operators in Hardy and Bergman Spaces

Transcription

1 Norm of the Backward Shift and Related Operators in Hardy and Bergman Spaces Tim Ferguson University of Alabama SEAM, University of Tennessee, 2017 Ferguson (UA) Backward Shift SEAM / 16

2 Suppose f is a function defined on the unit disc. Define B(f ) = f f (0). Define B(f ) = [f f (0)]/z. The norm of each of these operators can be at most 2 on any Hardy space H p (or real harmonic Hardy space h p ). But is it actually this large? Ferguson (UA) Backward Shift SEAM / 16

3 The H case 1 ɛ 1 2δ 1 Consider the conformal mapping f of the unit disc onto the region shown, that maps 0 to 1 ɛ and 1 to 1 if extended to the boundary. Then f = 1 and f = 2 ɛ. Thus B H = 2. Ferguson (UA) Backward Shift SEAM / 16

4 The h 1 case For harmonic functions in h 1, B(f ) will be large when most of the mass of f is concentrated in a small set on which f is large and has nearly constant sign Ferguson (UA) Backward Shift SEAM / 16

5 The H 1 case For h 1 (the space of harmonic functions with bounded M 1 integral means), the Poisson kernel P has BP h 1 = 2. However, the analytic completion of P is not in H 1, so perhaps B H 1 < 2. In fact, this is true. It is related to the fact that H 1 functions that have mass concentrated on a small set must have signs that oscillate a lot on this set. Ferguson (UA) Backward Shift SEAM / 16

6 First H 1 concentration theorem and a Bound Theorem Suppose that f H 1 = 1 and that for some set A T, we have Re f dt/2π 1 ɛ for ɛ < 1/4. Then A Theorem log(γ + ɛ) log(1 2ɛ) m(a) max 0<γ<1 log γ The norm of B on the Hardy space H 1 is at most Ferguson (UA) Backward Shift SEAM / 16

7 Second H 1 concentration theorem and a Bound Theorem Suppose that f H 1 and that f H 1 = 1. Furthermore, suppose that f L 1 (E) 1 ɛ for some set E T where m(e) δ and 0 < ɛ, δ < 1/2. Then Theorem ( ) 1 ɛ δ ( ) ɛ 1 δ f (0). δ 1 δ The norm of the backward shift operator on H 1 is at most In the bound above, the first term is at most 1.4 if ɛ = δ, the second is small for small ɛ and δ. Also, the bound decreases as ɛ and δ decrease. Ferguson (UA) Backward Shift SEAM / 16

8 Bergman spaces Theorem Suppose that the norm of the operator B is equal to K on H p. Let µ be a radial weight such that µ(d) <. Then the norm of B is at most K on the Bergman space A p (µ). Theorem Suppose that the norm of the operator B is equal to K on H 1. Let µ be a finite radial weight that is increasing. Then the norm of B is at most 2K on the Bergman space A 1 (µ). Lemma Let µ be an increasing radial measure. Then zf A 1 (µ) (1/2) f A 1 (µ). Ferguson (UA) Backward Shift SEAM / 16

9 The real harmonic Bergman space Theorem The norm of the backward shift on the real harmonic Bergman space ar 1 is at most In fact, the same estimate holds on any subspace X of L 1 with the property that u X implies that u(re iθ ) (1 r) 2 and the property that the average value of any u X on circles centered at the origin is constant. To prove the theorem, we basically use the fact that the mass of u on the circle of radius r cannot be too concentrated on a set of small angular measure, due to the bound on the size of u. Ferguson (UA) Backward Shift SEAM / 16

10 The operator B r. We define the operator B r : h 1 R h1 R to be the operator f (Bf ) r, where (Bf ) r (z) = (Bf )(rz). Equivalently B r can be thought of as the operator obtained by applying B and then restricting the function obtained to the circle centered at the origin with radius r. Suppose that f and g are functions defined on the interval [a, b]. By the convolution of f and g, we mean the function f g(x) = 1 b a b a f (y) g(x y) dy, where f and g are the periodic extensions of f and g to the real line. Ferguson (UA) Backward Shift SEAM / 16

11 A useful lemma Lemma Let f be a real function with average µ and let ν be a finite measure. Then f µ dν = 2 f µ dν. {x:f >µ} Ferguson (UA) Backward Shift SEAM / 16

12 Theorem Suppose we are given an m n matrix A. Let µ be a fixed number, and let C j = m k=1 a kj. Define D j = max(c j µ, 0), and let D = n j=1 D j. Suppose that D j D k but a ij a ik. Then we do not decrease D by interchanging a ij and a ik. The continuous form of the above theorem yields the following results: Lemma Suppose that P and f are nonnegative integrable functions on [a, b] and that f 1 = 1, where 1 denotes the L 1 norm with normalized Lebesgue measure. Then P f P f 1 1 P P 1 1. Ferguson (UA) Backward Shift SEAM / 16

13 Corollary Suppose that u is a nonnegative harmonic function in the unit disc and that 0 r < 1. Then 1 2π 2π 0 u(re iθ ) u(0) dθ u(0) 1 2π where P r is a Poisson kernel. = u(0) 2π P r (e iθ ) 1 dθ 0 (2 4π ) arccos(r). Note that this corollary holds for r = 1 if we replace 2π 0 u(e iθ ) u(0) dθ by u u(0) h 1. R 1 2π Ferguson (UA) Backward Shift SEAM / 16

14 We must now deal with functions that are allowed to be negative. The following is a continuous form of a previous theorem. Lemma Suppose that P is a nonnegative integrable function on [α, β] and f is an integrable function such that f 1 = 1, where 1 denotes the L 1 norm with normalized Lebesgue measure. Let P denote the decreasing rearrangement of P, so that P (t) = inf{x : m({y [α, β] : P(y) > x}) t}. Then P f P f 1 1 Q Q 1 1 where Q is some function of the form ap (x) bp (α + β x), where a + b = 1. Ferguson (UA) Backward Shift SEAM / 16

15 We need to have a condition under which we can conclude that Q(x) = P (x). The following theorem provides such a condition. Theorem Suppose that P is a continuous function on [α, β] and that P is nonnegative with average 1 and decreasing. Let Q(x) = ap(x) bp(β + α x) for some nonnegative real numbers a and b such that a + b = 1. Then there is a c between α and β such that ap(c) bp(β + α c) = a b. If for some such c, α+c α β P dx + P dx 2c, β c then P P 1 1 Q Q 1 1, where 1 denotes the L 1 norm with normalized Lebesgue measure. Ferguson (UA) Backward Shift SEAM / 16

16 Theorem Suppose that 0 r < 1 and u h 1 R. Then 1 2π u(re iθ ) u(0) dθ inf 2π u a 0 a R h 1 R where P r is a Poisson kernel. = inf a R u a h 1 R 2π 1 P r (e iθ ) 1 dθ 2π 0 (2 4π ) arccos(r) Note that the theorem holds for r = 1 if we replace 2π 0 u(e iθ ) u(0) dθ by u u(0) h 1. R 1 2π Corollary The value of B h 1 R a 1 R = 1. Ferguson (UA) Backward Shift SEAM / 16