Note that the integrand x2 1 1 n n2 n 2 ex1 2 not only

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Unformatted text preview: 2 n |S | a n n − x1 ￿ n−2 ￿ ￿1 ￿￿ 2 1 Γ(n/2) b x2 n x2 1 1 =√ 1− 1+ dx1 n−1 n n − x2 π n Γ( 2 ) a 1 ￿ n−2 ￿ b￿ 1 Γ(n/2) x2 n 1 1 /2 (by M V T ) = √ (1 + ηn ) 1− dx1 n π n Γ( n−1 ) a 2 for all large enough n for some constant 0 ≤ ηn ≤ max(a, b)2 n − max(a, b)2 1 so consequently limn→∞ ηn = 0. Note that the integrand ￿ x2 1− 1 n ￿ n−2 n 2 → e−x1 /2 not only pointwise but also uniformly over the closed interval [a, b]. (Which can be deduced by taking logarithm of both sides and using the inequality ￿￿ ￿ ￿ 2 2￿ ￿ ￿ln 1 − x1 − x1 ￿ ≤ C ￿ n 2 ￿ n2 where C depends on a and b.) Cosequently we can use the dominated convergence theorem and the result follows. Similirlarly this results holds for projections onto higher dimensional subspaces. √ ˜ 6.2.2 Corollary. Let S n−1 = {x ∈ Rn : ...
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This note was uploaded on 01/16/2014 for the course MATH 6397 taught by Professor Staff during the Spring '08 term at University of Houston.

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