By Weber M.

This e-book provides in a concise and obtainable means, in addition to in a standard environment, a number of instruments and strategies coming up from spectral concept, ergodic concept and stochastic methods conception, which shape the foundation of and give a contribution interactively greatly to the present study on almost-everywhere convergence difficulties. Researchers operating in dynamical structures and on the crossroads of spectral concept, ergodic thought and stochastic approaches will locate the instruments, equipment, and effects offered during this e-book of significant curiosity. it truly is written in a method available to graduate scholars.

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Example text

D. ’s. If we write more simply An = AU n (f ), then A n − Am 2 = 1 1 − , n m for any integers n < m. Let 0 < ε < 1 be fixed. Thus An ≤ ε, if n ≥ ε12 . For each 1 ≤ n ≤ 2ε , we cover An with one ball of radius ε. Finally, if 2ε < n < ε12 , let mk = kε12 , 1 ≤ k ≤ 1ε . Notice that x ≥ x/2 if x ≥ 1, and x − y ≤ 3(x − y) if x − y ≥ 1/2. Thus 1ε ≤ m 1 ≤ 2ε , and ε 1 1 1 1 1 1 1 1 − = 2 ≥ 21 1 ≥ . = kε2 (k + 1)ε 2 ε k(k + 1) ε ε ( ε + 1) 1+ε 2 So 1 kε 2 − 1 (k+1)ε 2 1 1 − = mk mk+1 ≤ 3 1 , ε2 k(k+1) 1 kε 2 1 kε 2 − which implies 1 (k+1)ε 2 1 (k+1)ε 2 ≤ 4k(k + 1)ε4 1 1 − kε2 (k + 1)ε 2 ≤ 4k(k + 1)ε4 3 1 2 ε k(k + 1) = 12ε2 .

Let also t0 < t1 < · · · < tr be an ordered sequence of positive integers such that ATti f − ATtj f ≥ ε, ∀0 ≤ i < j ≤ r. Apply the previous theorem to the subsequence N = {t0 , t1 , . . , tr , tr+1 , tr+2 , . . } , where tr+j = tr + j , if j = 1, 2, . . Then ε2 r ≤ 4(2π + 1), and consequently, N(AT (f ), . , ε) ≤ 1 + This establishes the claimed inequality. 4(2π + 1) 30 ≤ 1+ 2. 6 Remarks. (1) The above estimate is also optimal. This can be seen by considering rotations. Take X = [−π, π ) provided with the normalized Lebesgue measure λ.

We also assume that each Tn is positively homogeneous, which means that Tn (cf ) = cTn f for nonnegative c and f ∈ L1 (μ). For instance, Tn can be the absolute value of any linear operator from L1 (μ) to L1 (μ). ∞ 2 1/2 . Then Let Sf (x) = n=1 Tn f (x) ∞ μ{|Tn f | ≥ λ} ≤ C sup sup λ λ≥0 f 1 ≤1 n=1 ⇒ sup sup λμ{Sf ≥ λ} ≤ 10C. 15) λ≥0 f 1 ≤1 38 1 The von Neumann theorem and spectral regularization The proof is rather elementary. As Sf ≤ S1 f + S2 f , where ∞ S1 f (x) = 1/2 (Tn f (x))2 1{Tn f ≤1} (x) , n=1 ∞ S2 f (x) = (Tn f (x))2 1{Tn f >1} (x) 1/2 , n=1 we get μ{Sf ≥ 2} ≤ μ{S1 f ≥ 1} + μ{S2 f ≥ 1} ∞ ≤ μ{S1 f ≥ 1} + μ (Tn f )2 1{Tn f >1} ≥ 1 n=1 ∞ ≤ μ{S1 f ≥ 1} + μ{Tn f > 1} ≤ μ{(S1 f )2 ≥ 1} + C f 1 n=1 ∞ =μ ∞ 1{2−k−1 ≤Tn f ≤2−k } ≥ 1 + C f (Tn f (x))2 .

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