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in particular, f ( y) = for all y T . Then Proposition 5.4.3.iv, we have in .NET Printer USS Code 39 in .NET in particular, f ( y) = for all y T . Then Proposition 5.4.3.iv, we have

in particular, f ( y) = for all y T . Then Proposition 5.4.3.iv, we have use visual .net code 39 full ascii maker toincoporate barcode 39 in .net Jasper Reports ( y) y T Gy . . Using f ,f 1 2 1 2. (g) . x T ,y T ,g Sx ( y) ( y, gx) Gy ( y) ( y, gx) dg Gy G x T ,y T 1 ( y) ( y, gx) = dg (g) 2 2 S Gx Gy x T ,y T 1 ( y) ( y, gx) max (g) 2 dg. 2 g S Gx Gy S x T ,y T 1 Gx (g) . A spectral crite .net vs 2010 bar code 39 rion for Property (T). On the other hand, we have ( y) ( y, gx) dg = Gx Gy S x T ,y T = ( y) Gy 1 . Gx ( y, gx)dg ( y) ( y, x) = f Gy x X It follows that 1 max (g) 2 g S ,. namely that (S , 2 ) is a Kazhdan pair for G. Since S is compact, the group G has Property (T). Let us now show that (i) implies (iii).

Let x0 T . Recall from (Pvi) that Gx0 S is a compact generating set for G. Let > 0 be such that (Gx0 S, ) is a Kazhdan pair for G.

Let = min ( y, z) = min ( y, z) ( y) : y T , z X , ( y, z) > 0 . Gy ( y) : y, z X , ( y, z) > 0 . Gy Observe that & gt; 0, since has nite range. Let N = max d ( y, x0 ) + d (x0 , sy) : y T , s Gx0 S , where d is the distance on X associated to the graph G . Let ( , H) be a unitary representation of G without non-zero invariant 0 vectors.

We claim that, for every f E (X ), we have 1 2 2 f ,f 2N 2 2 ( y) f . Gy 0 Indeed, assume by contradiction that there exists f E (X ) with f such that 2 ( y) 1 2 2 . f ,f < . Gy 2N 2 y T 5.4 G-equivariant random walks Since f ( y). y T 2 ( y). Gy = 1,. there exists y0 T such that f ( y0 ). 1 ( y) . . Gy Fix s Gx0 S. Code 39 for .NET Choose paths of minimal length between y0 and x0 and between x0 and sy0 , that is, choose z0 , .

. . , zn X and w0 , .

. . , wm X such that z0 = y0 , zn = x0 , n = d ( y0 , x0 ), for 0 i n 1 and w0 = x0 , wm = sy0 , m = d (x0 , sy0 ), for 0 i m 1.

By the choice of , we have f ( y0 ) f (sy0 ) f ( y0 ) f (x0 ) + f (x0 ) f (sy0 ) . (zi , zi+1 ) > 0 (wi , wi+1 ) > 0 f (zi ) f (zi+ 1 ) (zi , zi+1 ). (zi ) . Gzi (wi ) . . Gwi f (wi ) f (wi+ .net vs 2010 Code 3 of 9 1 ) (wi , wi+1 ). Each of the term s on the right-hand side is equal to a term of the form f ( y) f (z) ( y, z) ( y) . Gy for some y T , z X . Since n + m N , it follows that f ( y0 ) f (sy0 ) N y T z X f ( y) f (z) ( y, z). ( y) . . Gy A spectral crite .net framework ANSI/AIM Code 39 rion for Property (T). By the Cauchy Schwarz inequality, we have y T z X 2 ( y) f ( y ) f (z) ( y, z) . Gy f ( y) f ( z) 2 ( y, z) ( y) . Gy y T ( y) f ( y) f (z) 2 ( y, z) . Gy ( y) . Gy ( y, z) . = . y T z X y T z X ( y) . . Gy Together with Proposition 5.4.3.iv, this implies f ( y0 ) f (sy0 ) 2 2 2N 2 y T ( y) 2 , y T Gy 2 . ( y) . Gy f ,f Since f ,f 1 2 2 2N 2. it follows that f ( y0 ) (s)f ( y0 ). = f ( y0 ) f ( visual .net barcode code39 sy0 ) 2 1 ( y) < 2 2 f ( y0 ) . Gy for all s Gx0 Code 39 Extended for .NET S. This is a contradiction, since is a Kazhdan constant for Gx0 S.

Remark 5.4.7 Assume that there exists > 0 such that.

f ,f f. for every unitar y representation ( , H) of G without non-zero invariant vectors 0 and for every f E (X ). The proof above shows that (S, 2 ) is a Kazhdan pair, where S is the set of all s G for which (x, sy) > 0 for some x, y T . In the next proposition, (ii) and (iii) display variants of the Poincar inequality.

Proposition 5.4.8 Let X be a set and G a unimodular locally compact group acting on X .

Let be a reversible random walk on X and let be a stationary.
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