I used the change of variables t sy applying fubini

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(I used the change of variables t = sy .) Applying Fubini to h ( x, y ) gives the claim. 59). Let f ( x ) = x - 1 sin x . a). Show that 0 | f ( x ) | dx = : If n is an integer, then (2 n +1) π 2 sin x dx = cos(2 n + 1) π - cos 2 = 2. Therefore, 0 | f ( x ) | dx n =1 (2 n +1) π 2 sin x (2 n + 1) π dx = 2 (2 n + 1) π = . Professor Minicozzi, Fall 2002. 1
2 MATH 605, HW 5 SOLUTIONS b). Show that lim b →∞ b 0 f ( x ) dx = π/ 2: Apply Fubini on the square [0 , b ] × [0 , b ] to h ( x, y ) = e - xy sin x (when we integrate first in y we get the integral that we’re interested in; get the other way as in 57).
MATH 605, HW 6 SOLUTIONS Folland’s Real Analysis; Chapter 5: 1.) Let X be a normed v.s. over R . (a) Show that vector addition and scalar multiplication are continuous from X × X and X × R to X : I’ll just do this for addition. Define T : X × X X by T ( x, y ) = x + y . Given > 0, we have that if max {| x - x 1 | , | y - y 1 |} < / 2, then the triangle inequality gives | T ( x, y ) - T ( x 1 , y 1 ) | = | x + y - x 1 - y 1 | ≤ | x - x 1 | + | y - y 1 | < . (b) Show that || x | - | y || ≤ | x - y | (so that the norm is continuous from X to R ): This follows immediately from the triangle inequality since | x | = | x - y + y | ≤ | x - y | + | y | and | y | = | y - x + x | ≤ | x - y | + | x | . 5.) Let X be a normed v.s. over R and Y X a subspace. Show that the closure ¯ Y of Y is also a subspace: Suppose that x n , y n Y with x n x and y n y (so x and y are in ¯ Y ). We have to show that ax ¯ Y and x + y ¯ Y . This follows immediately from exercise 1 but I will prove it without using that by instead showing that ( x n + y n ) ( x + y ) and ax n ax . The triangle inequality gives the first: | ( x n + y n ) - ( x + y ) | ≤ | x n - x | + | y n - y | → 0 . The second is even easier since | ax n - ax | = | a ( x n - x ) | = | a | | x n - x | → 0. 7.) Let X be a Banach space. (a) Suppose that T L ( X, X ) and | I - T | < 1 where I is the identity operator. Show that 0 ( I - T ) n converges to T - 1 : First of all, note that | ( I - T ) n | = | I - T | n so that 0 | ( I - T ) n | < since it is a geometric series (here we use that | I - T | < 1). Since X is complete, so is L ( X, X ); therefore 0 ( I - T ) n converges to some limit, let’s call it S , in L ( X, X ) (by theorem 5.1). We now show that S is really T - 1 . To see this, note that ( I - T ) S = lim N →∞ ( I - T ) N 0 ( I - T ) n = lim N →∞ N +1 1 ( I - T ) n = S - I . In other words, S - TS = S - I so that TS = I (it’s easy to see that ST = TS ). (b) Suppose that T L ( X, X ) is invertible and and | S - T | < | T - 1 | - 1 . Show that S is invertible: We have that | ST - 1 - I | = | ( S - T ) T - 1 | ≤ | S - T | | T - 1 | < | T - 1 | - 1 | T - 1 | = 1 . Therefore, (a) gives [ ST - 1 ] - 1 . Now just check that multiplying this by T - 1 gives S - 1 . 11.) This is standard, see almost any analysis book. 12.) Ditto. 17.) Show that a linear functional f on a normed v.s. X is bounded iff f - 1 (0) is closed: By 5.2, f is bounded iff it is continuous iff it is continuous at 0. Therefore, f bounded Professor Minicozzi, Fall 2002. 1
2 MATH 605, HW 6 SOLUTIONS implies f - 1 (0) is closed. To go the other way, suppose f is not bounded. Then there must exist x n X with | x n | → 0 but f ( x n ) = 1. Choose y with f ( y ) = 1. Then f ( y - x n ) = 0 so ( y - x n ) f - 1 (0) but ( y - x n ) y / f - 1 (0).

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