HW201 - Analysis 2 Homework 01 1. Show that the following...

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Analysis 2 Homework 01 1. Show that the following subset K of 2 is convex, closed and bounded, but does not contain an element of greatest norm: K = { z = ( z n ) n =1 : X n =1 (1 + 1 n ) 2 | z n | 2 6 1 } . Solution 1. A ‘little’ thought suggests considering the form on 2 defined by [ x | y ] = X n =1 (1 + 1 n ) 2 x n y n . This form is plainly conjugated when its variables are switched; it is also plainly conjugate-linear in the first and linear in the second. Further, [ z | z ] > 0 and = 0 if and only if z = 0: indeed, k z k 2 6 [ z | z ] 6 4 k z k 2 . Thus, the form [ •|• ] is an inner product, whose associated IP norm ||| • ||| (say) satisfies k z k 6 ||| z ||| 6 2 k z k . ( * ) Notice that K (as defined in the question) is precisely the ‘closed unit ball’ for ||| • ||| . This implies at once that K is convex. The right inequality at (*) shows that the |||•||| -closed set is k•k -closed. The left inequality at (*) shows that K is bounded, being contained in B 2 (for k • k of course). So far, K is convex, closed and bounded, with sup {k z k : z K } 6 1. In fact, this supremum is precisely 1: by direct calculation, ||| n n +1 e n ||| = 1 and k n n +1 e n
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HW201 - Analysis 2 Homework 01 1. Show that the following...

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