ell_p_not_convex - δ , there is r such that n 1-p...

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( May 10, 2008 ) ` p with 0 < p < 1 is not locally convex Paul Garrett garrett@math.umn.edu http: / /www.math.umn.edu/˜garrett/ That is, with 0 < p < 1, the topological vector space ` p = {{ x i C } : X i | x i | p < ∞} is not locally convex with the topology given by the metric d ( x, y ) = | x - y | p coming from | x | p = X i | x i | p (for 0 < p < 1 no p th root!) It is complete with respect to this metric. Note that | x | p fails to be a norm by failing to be homogeneous of degree 1. The failure of local convexity is as follows. Local convexity would require that the convex hull of the δ -ball at 0 be contained in some r -ball. That is, local convexity would require that, given δ , there is some r such that ± ± ± 1 n · ( δ, 0 , . . . ) + . . . + 1 n · (0 , . . . , 0 , δ | {z } n , 0 , . . . ) ± ± ± p = ² δ n ³ p + . . . + ² δ n ³ p < r (for n = 1 , 2 , 3 , . . . ) That is, local convexity would require that, given
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Unformatted text preview: δ , there is r such that n 1-p &lt; r δ p (for n = 1 , 2 , 3 , . . . ) This is impossible because 0 &lt; p &lt; 1. /// For contrast, to prove the triangle inequality for the alleged metric on ` p with 0 &lt; p &lt; 1, it suffices to prove that ( x + y ) p &lt; x p + y p (for 0 &lt; p &lt; 1 and x, y ≥ 0) To this end, take x ≥ y . By the mean value theorem, ( x + y ) p ≤ x p + pξ p-1 y (for some x ≤ ξ ≤ x + y ) and x p + pξ p-1 y ≤ x p + px p-1 y ≤ x p + py p-1 y = x p + py p ≤ x p + y p (since p-1 &lt; 0 and ξ ≥ x ≥ y ) This proves the triangle inequality for 0 &lt; p &lt; 1. /// 1...
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