hmwk13sol_a

# hmwk13sol_a - Math 202A Homework 13 Roman Vaisberg November...

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Unformatted text preview: Math 202A Homework 13 Roman Vaisberg November 28, 2007 Problem 17. Suppose f is defined on R 2 as follows: f ( x, y ) = a n if n x &lt; n + 1 and n y &lt; n + 1, ( n 0); f ( x, y ) =- a n if n x &lt; n + 1 and n + 1 y &lt; n + 2, ( n 0); while f ( x, y ) = 0 elsewhere. Here a n = k b k , with { b k } a positive sequence such that k =0 b k = s &lt; . (a) Verify that each slice f y and f x is integrable. Also for all x , R f x ( y ) dy = 0, and hence R ( R f ( x, y ) dy ) dx = 0. Proof. If y &lt; 0 then f y ( x ) = 0 If 0 y &lt; 1 then f y ( x ) = a x &lt; 1 elsewhere If n y &lt; n + 1 then f y ( x ) = - a n- 1 n- 1 x &lt; n a n n x &lt; n + 1 elsewhere Clearly f y is integrable for every y . If x &lt; 0 then f x ( y ) = 0 If n x &lt; n + 1 then f x ( y ) = a n n y &lt; n + 1- a n n + 1 y &lt; n + 2 elsewhere It is also clear that f x is integrable for every x and R f x ( y ) dy = 0. Thus R ( R f ( x, y ) dy ) dx = 0. (b) However, R f y ( x ) dx = a if 0 y &lt; 1, and R f y ( x ) dx = a n- a n- 1 if n y &lt; n + 1 with n 1. Hence y 7 R f y ( x ) dx is integrable on (0 , ) and Z Z f ( x, y ) dx dy = s. Proof. R (R f ( x, y ) dx ) dy = a + ( a 1- a ) + ( a 2- a 1 ) + = b k = s. (c) Note that R R R | f ( x, y ) | dxdy = . Proof. In the extended sense we have R R R | f ( x, y ) | dxdy = 2 a k and since { a n } is a positive increasing sequence we have the desired conclusion....
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## hmwk13sol_a - Math 202A Homework 13 Roman Vaisberg November...

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