By C. Epstein
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This proves the opposite inequality Ln(f) 2: (£lk)Ln(w). 0 For an arbitrary function An(f) f let us denote = Ln(f) - Ln-1(f), n = 1,2, ... 2 says simply that the numbers An (fT) give precisely the shape of the tableau T. 1 For any measurable function f, the sequence (An(f))~=1 is decreasing. Proof. We need to prove that 2Ln(f) 2: L n- 1(f) + Ln+1 (f) for n 2: 1. Fix a positive number rJ and choose sets 0"1,0"2,··. ,O"n-1,0"~,0"2, ... ,O"n+1 such that (i) the restrictions flO"i and flO"j are increasing; (ii) O"i n O"j = 0 for i =F j, O"~ n O"j = 0 for i =F j; (iii) L:~~lIO"il2: Ln- 1(f) - rJ and L:j~IIO"jl 2: Ln+1(f) - rJ· Let us set 9 = flO", where 0" = (U~~} O"i) U (Uj~l O"j).
Let us just note that a Young tableau T is standard if the tableau function iT is an essentially invertible, measure preserving transformation of its domain (0, f). It is instructive to look at a simple example. Consider the function f : (0,2) - t R defined by f(t) = t 2 for t E (0,1] and f(t) = t - 1 for t E (1,2). This is not a Young tableau and one can calculate L 1 (J) = 5/4, L 2 (J) = 3/4. The longest set on which f is increasing is (0,1/2] U [5/4,2). Observe that the break points 1/2 and 5/4 have the property that f(1/2) = f(5/4) and f'(1/2) = f'(5/4); this is not an accident.