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exam3solns

# exam3solns - Exam 3 Solutions 1 Using the bilinearity of...

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Exam 3 Solutions 1. Using the bilinearity of the inner product we have k x + y k 2 - k x - y k 2 = h x + y, x + y i - h x - y, x - y i = h x, x i + 2 h x, y i + h y, y i - ( h x, x i + 2 h x, - y i + h- y, - y i ) = h x, x i + 2 h x, y i + h y, y i - ( h x, x i - 2 h x, y i + h y, y i ) = 4 h x, y i . 2. For ( x, y ) 6 = (0 , 0) we have f = ( π 1 · π 2 ) · ( g ( π 2 1 + π 2 · π 2 + π 2 2 ) , where g ( x ) = 1 x . Since x 2 + xy + y 2 = ( x + y ) 2 - xy , it follows that ( π 2 1 + π 2 · π 2 + π 2 2 )( x, y ) 6 = 0 when ( x, y ) 6 = (0 , 0) and thus f is continuous at all such points. However, f is not continuous at (0 , 0) because f (0 , 1 k ) 0 and f ( 1 k , 1 k ) 1 3 as k → ∞ . 3. (a) Take d f ( S ). Then there is c S such that f ( c ) = d . Since c S there is ( x k ) in S such that x k c . Since f ( x k ) f ( c ) = d by the continuity of f and f ( x k ) f ( S ) for each k 1, we have d = f ( c ) f ( S ). (b) Let f : R R be given by f ( x ) = 1 x 2 +1 . Let S = Z + . Then S = S so f ( S ) = f ( S ) = 1 k 2 + 1 : k Z + . Since 0 / f ( S ) and 0 f ( S ), it follows that f ( S ) 6 = f ( S ). 4. C is compact so C is closed and since A is closed it follows that A C is closed. Since A C

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