# s7 - Math 8601 REAL ANALYSIS Fall 2010 Homework#7 Problems...

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Math 8601: REAL ANALYSIS. Fall 2010 Homework #7. Problems and Solutions. #1. Let f L 1 ( R 1 ). Show that S n ( x ) = 1 n n - 1 X j =0 f x + j n S ( x ) = Z x +1 x f ( t ) dt in L 1 ( R 1 ) as n → ∞ , i.e. Z | S n ( x ) - S ( x ) | dx 0 as n → ∞ . Proof. By Theorem 2.26, ε > 0 there are a constant A > 0 and a continuous function g ( x ) on R 1 , such that Z R 1 | f ( x ) - g ( x ) | dx < ε, and g ( x ) 0 for | x | ≥ A. We can write Z R 1 | g ( x + t ) - g ( x ) | dx = A +1 Z - A - 1 | g ( x + t ) - g ( x ) | dx for | t | ≤ 1 . By the dominated convergence theorem (Theorem 2.27(a)), Z R 1 | g ( x + t ) - g ( x ) | dx 0 as t 0 . Hence Z R 1 | f ( x + t ) - f ( x ) | dx Z R 1 | f ( x + t ) - g ( x + t ) | dx + Z R 1 | g ( x + t ) - g ( x ) | dx + Z R 1 | g ( x ) - f ( x ) | dx = Z R 1 | g ( x + t ) - g ( x ) | dx + 2 Z R 1 | g ( x ) - f ( x ) | dx < Z R 1 | g ( x + t ) - g ( x ) | dx + 2 ε. This implies lim sup t 0 Z R 1 | f ( x + t ) - f ( x ) | dx 2 ε, and since ε > 0 can be made arbitrarily small, ω ( h ) := sup | t |≤ h Z R 1 | f ( x + t ) - f ( x ) | dx 0 as h 0 + . 1

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Further, writing S n - S in the form S n ( x ) - S ( x ) = n - 1 X j =0 1 /n Z 0 f x + j n - f x + j n + t dt, and changing the order of integration, one can get Z R 1 | S n ( x ) - S ( x ) | dx n - 1 X j =0 1 /n Z 0 Z R 1 f x + j n - f x + j n + t dx dt = n
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