By Blumberg H.
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Extra resources for New Properties of All Real Functions
Example text
Fig. 1. 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.
Am )) whose coordinates are the measures of the sets on which φ is linear. As ti approaches 0, the ith linear function drops to negative infinity and so µ(Ai ) decreases to zero. In view of the hypothesis on µ, H can be defined continuously on 46 K. Ball the simplex and maps each face of the simplex into itself. It is a well-known consequence (or reformulation) of Brouwer’s fixed point theorem, that such a map is surjective. (If the map omits a point, say in the interior of the simplex, then we can follow it by a projection of the punctured simplex onto its boundary, to obtain a continuous map from the simplex to its boundary which fixes each face.
Let us denote by T the circle {z ∈ C| |z| = 1} ⊂ Q. + Proposition 2. (1) The stabilizer of E0 ∈ Gr2,4 in Q × Q is equal to T × T . (2) The stabilizer of span{1, i} in non-oriented Grassmannian Gr2,4 is equal to T × T ∪ j(T × T ). Proof. Clearly T × T ⊂ StabE0 . Hence we have a covering + (Q × Q)/(T × T ) → Gr2,4 . + Since Gr2,4 is simply connected, this covering is an isomorphism. This proves part (1). Let us prove part (2). Clearly (T × T ) ∪ j(T × T ) is a subgroup contained in the stabilizer of span{1, i} ∈ Gr2,4 .
