# Download Applied functional analysis by Balakrishnan A.V. PDF

By Balakrishnan A.V.

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**Example text**

Tn . The problem is to show that for these positive ti (1 − λ + λti ) ≥ λ ti . This inequality is the special case of the Brunn–Minkowski inequality in which the set A is a unit cube and the set B is a cuboid with sides t1 , . . , tn , aligned in the same way as the cube. It is immediate because the arithmetic/geometric mean inequality shows that for each i (1 − λ) + λti ≥ tλi . As long ago as 1957, Knothe [K] gave a proof of the Brunn–Minkowski inequality which involved a kind of mass transportation.

A simple transport problem The problem is to ﬁnd a partition of R2 into sets A and B with µ(A) = µ(B) = 1/2 so as to minimise the cost of transporting A to (−1, 0) and B to (1, 0): x − (−1, 0) A 2 x − (1, 0) dµ + 2 dµ. B I claim that the best thing to do is to divide the measure µ using a line in the direction (0, 1), as shown in Figure 1. To establish the claim, we need to check that given two points (a, u) and (b, v) with a < b, it is better to move the leftmost point to (−1, 0) and the rightmost point to (1, 0), than it is to swap the order.

Let us recall the construction of the Cartan imbedding. Let θ : G → G be an involutive automorphism of a group G. Let K be the subgroup of ﬁxed points of θ. The Cartan imbedding of G/K → G is given as follows: gK → g(θg)−1 . It is easy to see that this map is well deﬁned and injective. Let us write down the Cartan imbedding explicitly in our case: f (qT ) = q(iqi−1 )−1 = [q, i]. Note that Q/T = CP = S . Thus f : Q/T = S → Q. 1 2 2 Proposition 3. The image of f coincides with S 2 = {q ∈ Q| q = t + jy + kz} ⊂ R ⊕ Cj.