By Robert E. Megginson

Getting ready scholars for extra research of either the classical works and present examine, this can be an obtainable textual content for college students who've had a path in actual and intricate research and comprehend the elemental houses of L p areas. it's sprinkled liberally with examples, old notes, citations, and unique assets, and over 450 routines offer perform within the use of the consequences built within the textual content via supplementary examples and counterexamples.

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It is enough to show that D includes a neighborhood of the origin. If A is a nonempty subset of D, then 1 1 1 1 1 1 0 E -A+ -(-A) c -D +-(-D)= -D + -D = D 2 2 -2 2 2 2 ' where the last equality comes from the convexity of D. Thus, it is enough to prove that Do =I 0, for then the neighborhood ~Do+ 4{ -D 0 ) of the origin must be included in D. Suppose that D has empty interior. For each positive integer n, the set nD is closed and has empty interior, and so X \ nD is an open set that is dense in X.

By the theorem, each such linear operator is a normed space isomorphism. 17 Corollary. Every finite-dimensional vector space has exactly one norm topology. PROOF. Let X be a vector space of finite dimension n and let T be a linear operator from X onto Euclidean n-space. Txll defines a norm on X. If ll·llo is any other norm on X, then the identity operator on X, viewed as a linear operator from (X, II·IIT) onto (X, ll·llo), is an isomorphism, so the two norms induce the same topology. 18 Corollary.

Then the following are equivalent. --t Y be a (a) The operator Tis continuous. (b) The operator T is continuous at 0. (c) The operator T is uniformly continuous on X. (d) The operator Tis bounded. (e) For some neighborhood U of 0 in X, the set T(U) is bounded in Y. (f) There is a nonnegative real number M such that liTxll 5 Mllxll for each x in X. (g) The quantity sup{ liTxll : x E Bx} is finite. PROOF. Suppose that Tis continuous at 0. x1-x2l1 < 6. Therefore Tis uniformly continuous on X, which proves that (b)=> (c).

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