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Let $$f\in L^{\infty }(T)$$ f ∈ L ∞ ( T ) with $$\Vert f\Vert _{\infty }\le 1$$ ‖ f ‖ ∞ ≤ 1 . If $$f(0)\ne 0,$$ f ( 0 ) ≠ 0 , $$n,k\in {\mathbb {Z}} $$ n , k ∈ Z , and $$b_{n,n-k}=\int _{E}f(x)^{n}e^{-2\pi i(n-k)x}dx$$ b n , n - k = ∫ E f ( x ) n e - 2 π i ( n - k ) x d x , $$E=\{x\in T:|f(x)|=1\}$$ E = { x ∈ T : | f ( x ) | = 1 } , we prove that the arithmetic means $$\frac{1}{N} \sum _{n=M}^{M+N}|b_{n,n-k}|^{2}$$ 1 N ∑ n = M M + N | b n , n - k | 2 decay like $$\{\log N\log _{2}N\cdot \cdot \cdot \log _{q}N\}^{-1}$$ { log N log 2 N · · · log q N } - 1 as $$N\rightarrow \infty $$ N → ∞ , uniformly in $$k\in {\mathbb {Z}} $$ k ∈ Z .
Analysis and Mathematical Physics – Springer Journals
Published: Dec 6, 2018
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