Absolute continuity and singularity of measures without measure - PowerPoint PPT Presentation

Absolute continuity and singularity of measures without measure theory R. B. Burckel Department of Mathematics Kansas State University Manhattan, KS 66506 February Fourier Talks Norbert Wiener Center University of Maryland February 17th, 2011

F. & M. Riesz Theorem Every non-zero analytic measure ν on T is mutually absolutely continuous with respect to Lebesgue measure λ . Corollary (Szeg¨ o’s Theorem) Let σ be a Borel probability measure on T that annihilates some set of positive Lebesgue measure. Then the powers z n , n ∈ N , span L 2 ( σ ) . 2 / 19

Proof of the F. & M. Riesz Theorem Let µ := | ν | and f the Radon-Nikodym derivative d ν d µ . That is, d ν = fd µ and | f | = 1 µ − a.e . Then the analyticity hypothesis on ν can be written � z n f ( z ) d µ ( z ) = 0 ∀ n ∈ N . (1) T Let � , � and � · � denote inner product and norm in L 2 ( µ ) and U the unitary operator of multiplication by z . By “span” will be meant “closed linear span in L 2 ( µ ).” 3 / 19

According to (1) the constant function 1 is orthogonal to every U n f ( n ∈ N ), so the set M := (closed) span { U n f : n ∈ N } in L 2 ( µ ) (2) is a proper subspace of L 2 ( µ ), evidently U -invariant. In fact, (3) UM � M . 4 / 19

Now the linear span of { z n : n ∈ Z } is dense in C ( T ) (why?), hence also in L 2 ( µ ). So, given g ∈ L 2 ( µ ), since | f | = 1 µ -a.e. ¯ f g ∈ L 2 ( µ ) and accordingly some P n in this linear span satisfy � P n − ¯ f g � → 0. That is, � P n f − g � → 0, showing that span { z n f : n ∈ Z } = L 2 ( µ ) . (4) 5 / 19

If we suppose, contrary to (3), that UM = M , then U ∗ M = U ∗ UM = M , so M contains, along with z f , ( U ∗ ) m z f = (¯ z ) m z f = z − m +1 f ∀ m ∈ N , and consequently z n f ∈ M ∀ n ∈ Z , which with (4) contradicts the proper inclusion M � L 2 ( µ ). This contradiction confirms (3). 6 / 19

Form the orthocomplement M ⊖ UM � = { 0 } and note that the closed subspaces U n ( M ⊖ UM ) are orthogonal, which is pretty clear when they are written as U n M ⊖ U n +1 M . As a special case { U n h } n ∈ Z is an orthonormal sequence in L 2 ( µ ) (5) for every unit vector h ∈ M ⊖ UM . 7 / 19

Note that � U k M is orthogonal to U n ( M ⊖ UM ) ∀ n ∈ Z . k ≥ 0 For if m 1 ∈ M ⊖ UM and m 0 lies in this intersection, then m 0 = U | n | +1 m 2 for some m 2 ∈ M , and so � m 0 , U n m 1 � = � U | n | +1 m 2 , U n m 1 � = � U | n |− n +1 m 2 , m 1 � = 0 , since U | n |− n +1 m 2 ∈ UM . The same argument shows that � U k M is orthogonal to U n 1 ∀ n ∈ Z . (6) k ≥ 0 8 / 19

The Wold decomposition says that M is the orthogonal sum � U k M ⊕ � U n ( M ⊖ UM ) . M = (7) k ≥ 0 n ≥ 0 Again, this is pretty transparent when the right side is written out as � U k M ⊕ ( M ⊖ UM ) ⊕ ( UM ⊖ U 2 M ) ⊕ ( U 2 M ⊖ U 3 M ) ⊕ · · · k ≥ 0 As previously noted, vectors U n 1 = z n ( n ∈ Z ) span a dense subspace of L 2 ( µ ). From (6), then ∩ k ≥ 0 U k M must be { 0 } and (7) reads � U n ( M ⊖ UM ) . M = (8) n ≥ 0 9 / 19

Next we aim to show the non-zero space M ⊖ UM is 1-dimensional . (9) If not, ∃ orthogonal unit vectors g , h ∈ M ⊖ UM . By familiar maneuvers, U m h ⊥ U k g ∀ m , k ∈ N 0 , so 0 = � U m h , U k g � = � U m − k h , g � ∀ m , k ∈ N 0 , whence � 0 = � U n h , g � = z n h ¯ g d µ ∀ n ∈ Z . T Again, due to denseness of the powers z n , this entails h ¯ g = 0 µ -a.e. That is, | h || g | = 0 µ -a.e. (10) 10 / 19

As noted in (5) � 1 � if n = 0 z n | h | 2 d µ ( z ) = � U n h , h � = 0 if n ∈ Z \ { 0 } , T that is, the measure | h | 2 d µ has exactly the same Fourier coefficients as λ , so | h | 2 d µ = d λ. And the same is true for g . Thus, | h | 2 d µ = d λ = | g | 2 d µ, (11) whence, by (10), | h | 3 d µ = | h || g | 2 d µ = 0 , contrary to � h � 2 = � | h | 2 d µ = 1. Thus g = 0, and (9) is confirmed. 11 / 19

That is, for any h ∈ M ⊖ UM of norm 1 M ⊖ UM = C h , so (8) says span { U n h : n ∈ N 0 } = M . In particular, since Uf = z f ∈ M , we see that z f lies in the span of z n h . It follows that span { z n f : n ∈ Z } ⊂ span { z n h : n ∈ Z } . Combined with (4) this says 12 / 19

span { z n f : n ∈ Z } = span { z n h : n ∈ Z } = L 2 ( µ ) . (12) A little thought shows the equality of these two spans entails f d µ ≪ h d µ ≪ f d µ, and thanks to (11) d λ ≪ | h | d µ ≪ d λ. Thus fd µ = d ν is mutually absolutely continuous with respect to d λ . 13 / 19

Proof of Szeg¨ o Denote by M the (closed) span in L 2 ( σ ) of the z n , n ∈ N , and assume M � = L 2 ( σ ). There is then a non-zero function g ∈ L 2 ( σ ) orthogonal to M : � 0 = � z n , g � L 2 ( σ ) = z n ¯ ∀ n ∈ N . g d σ T This says that ¯ gd σ is a (non-zero) analytic measure. Hence d λ ≪ ¯ gd σ . So for every Borel B , � σ ( B ) = 0 ⇒ g d σ = 0 ⇒ λ ( B ) = 0 . ¯ B Contrary to the hypothesis on σ , which has σ annihilating a B with λ ( B ) > 0. 14 / 19

Holland [1974] σ is a Borel probability measure on T which is singular with respect to λ . � u + z F ( z ) := u − z d σ ( u ) , T a holomorphic self-map of D . A k := k th Taylor coefficient of F ( z ) − 1 F ( z ) + 1 . Then ∞ | A k | 2 = 1 � (i) k =1 n A k z k , n ∈ N , satisfy and the polynomials P n ( z ) := � k =1 n � | 1 − P n | 2 d σ = 1 − � | A k | 2 ∀ n ∈ N . (ii) T k =1 15 / 19

(i) and (ii) show (very constructively!) that 1 ∈ span { z n : n ∈ N } , i . e ., span { z n : n ∈ N 0 } = span { z n : n ∈ N } . By induction it follows span { z n : n ∈ Z } = span { z n : n ∈ N } , i . e ., L 2 ( σ ) = span { z n : n ∈ N } (Note the stronger hypothesis than in Szeg¨ o.) 16 / 19

Øksendal [1971] A C -valued Borel measure ν on T satisfying (A) is given and what has to be shown is that ν ( K ) = 0 for every λ -null Borel K . Because Borel measures are inner regular, it suffices to consider only compact K . Clearly it further suffices to do this for the modified measure ν 0 := ν − ν ( T ) λ. 17 / 19

The measure ν 0 is also analytic but in addition annihilates 1. That is, � z n d ν 0 ( z ) = 0 ν 0 ( − n ) = ˆ ∀ n ∈ N 0 . (A*) T For each n ∈ N , an N ∈ N , z j ∈ K and ρ j > 0 are chosen appropriately and the rational functions N z − z j � g n ( z ) := 1 − z − (1 + ρ j ) z j j =1 are introduced. 18 / 19

They are bounded by 2 on T and converge there to the indicator function of K . Since g n is holomorphic in a neighborhood of D , the partial sums of its Taylor series at 0 approximate it uniformly on T , and each sum has ν 0 -integral 0, thanks to (A*). Consequently, � g n d ν 0 = 0 ∀ n ∈ N . T It follows from the Lebesgue Dominated Convergence Theorem that � ν 0 ( K ) = lim g n d ν 0 = 0 , n →∞ T as wanted. 19 / 19