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Download Algebraic Analysis of Solvable Lattice Models by M. Jimbo, T. Miwa PDF

By M. Jimbo, T. Miwa

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6. 1. 1, namely, isometry. We promulgate analogous criteria for normed and inner product spaces in this section. By this means, we can prove that for any n E N, there is essentially only one n-dimensional inner product space: Kn (Exercise 5). 9 that any infinite-dimensional separable Hilbert space is "inner product isomorphic" to £2' This leads to an interesting choice: One can view L 2 [0, 1] as a function space or as the sequence space £2. For any two closed intervals [a, b] and [e, d] , the spaces of continuous functions C [a, b] and C [e, d] are linearly isometric (Exercise 7), so there is no loss in focusing attention on C [0,1].

Is norm preserving: IIxll IIAxll. is inner product preserving: (x, y) = (Ax, Ay). = = Proof. If A is an isometry, then d (Ax, 0) = IIAxli = IIAx - 011 = d (x, 0) = IIx - 011 = IIxll, so (a) implies (b). We argue that (b) =} (c) for complex spaces only. 3-3), 4 (x, y) = = IIx + Yll2 -lix - Yll2 + i IIx + iyll2 - i IIx _ iyll2 IIAx + Ayll2 - IIAx - Ayll2 4 (Ax,Ay) , + i IIAx + iAyll2 - i IIAx - iAyll2 26 1. Metric and N ormed Spaces so A preserves inner products. The reverse implications are straightforward.

Show that lI(x, Y)lIoo = max (lIxll ,lIyll') defines a norm on the Cartesian product X x Y (cf. 9). If A is a bounded subset of X and B a bounded subset of Y, show that A x B is a bounded subset of X x Y. 2-7) is bounded. 2. Analysis 51 4. DIAMETER The DIAMETER d (A) of a nonempty subset A of a metric space (X, d) is d(A) =sup {d(x,y): X,y E A} if the set {d (x, y) : x, yEA} is bounded, d (A) = 00 otherwise. Show that a set is bounded if and only if it has finite diameter and that d(A) = 0 if and only if A is a singleton.

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