solutions-3

# solutions-3 - Solution to Homework#3 36-754 25 February...

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Solution to Homework #3, 36-754 25 February 2006 Exercise 10.1 I need one last revision of the definition of a Markov operator: a linear operator on L 1 satisfying the following conditions. 1. If f 0 ( μ -a.e.), then Kf 0 ( μ -a.e.). 2. If f M ( μ -a.e.), then Kf M ( μ -a.e.). 3. K 1 Ξ = 1 Ξ . 4. If f n 0, then Kf n 0. First, we show that kernels induce operators, then that operators induce kernels. Given a kernel κ , we define the operator Kf ( x ) κ ( x, dy ) f ( y ). 1. Clearly, if f 0 a.e., then κ ( x, dy ) f ( y ) 0. 2. If f M a.e., then M - f ( x ) 0 a.e., and κ ( x, dy )( M - f ( y )) = κ ( x, dy ) M - κ ( x, dy ) f ( y ) = M - Kf ( x ) 0 a.e., so M Kf ( x ) a.e. 3. κ ( x, dy ) 1 Ξ ( y ) = 1 for all x , so K 1 Ξ ( x ) = 1 Ξ ( x ). 4. If f n ( x ) 0 pointwise, then, for each x , κ ( x, dy ) f n ( x ) 0 by monotone convergence, so Kf n ( x ) 0. Now for the converse: given an operator K , define κ ( x, B ) = K 1 B ( x ). For fixed x , we need this to be a probability measure. For every B ∈ X , 1 B L 1 , so K 1 B ( x ) is, for fixed x , a set-function defined over the whole of the σ -algebra in question. Since 1 B ( x ) 0, we know that K 1 B ( x ) 0, and this is a non- negative set-function. Next, we check that it’s finitely additive: If A and B are disjoint sets, 1 A B ( x ) = 1 A ( x ) + 1 B ( x ). Hence, by the linearity of K , K 1 A B = K 1 A + K 1 B , which by induction extends to any finite collection of disjoint sets. We notice that if A n ↓ ∅ , 1 A n ( x ) 0 pointwise, so K 1 A n ( x ) 0 (for fixed x ). But, by proposition 32 (ch. 3, p. 14), this implies that K 1 A is a measure. Finally, K 1 Ξ ( x ) = 1, for all x

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