By Ronald Larsen

**Read Online or Download Functional analysis;: An introduction (Pure and applied mathematics, v. 15) PDF**

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**Extra resources for Functional analysis;: An introduction (Pure and applied mathematics, v. 15)**

**Sample text**

21. that converges to V V is normable -- that is, there exists a norm on topology and the product topology number of spaces V Letting *22. (xn) is converges to 0. Prove that every be a family of normable topological linear Let letting and (an) is bounded. Prove that the topological product spaces. 0 (anxn) be a topological linear space. (V,T) Prove 4. is bounded if and only if, whenever is a sequence of scalars in Cauchy sequence in is bounded. be a topological linear space over (V,T) that a subset B In particular, is bounded.

All seminormed linear spaces are examples of these topological linear spaces, but the converse is not true. 2. There it will be seen that a topological linear space is finite dimensional if and only if its topology is locally compact. 3. They will be seen to be precisely those topological linear spaces whose topology has, at the origin, a neighborhood base consisting of convex open sets. 4 some properties of gauges in topological linear spaces will be given, and these results will then be used to characterize those topological linear spaces that are normable.

Because is bounded, there exists some B and since if ab < 1, then ab(x/ab) = x such that b E abU d U, Hence U. as and this holds for b > I/a, By the definition of x E bB. ab > 1, is balanced, U as B C all, such that b > 0 But this implies that x/ab E U. contradicting the choice of any for which a > 0 is absorbing, there also exists some B Clearly then x E bB. is also balanced. U it then follows q q(x) > 1/a > 0. that Therefore it then x f 0, q(x) 7 0, and so is a norm. q This completes the proof of part (vi) and the theorem.