A Hilbert space lemma: Difference between revisions

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New page: This is a finitised version of the lemma which is key to the ergodic proof of DHJ(2). :'''Lemma''' Let H be a Hilbert space with nested subspaces :<math>V_0 \subset V_1 \subset \ldots \su...
 
 
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This is a finitised version of the lemma which is key to the ergodic proof of DHJ(2).
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== Preliminary lemma ==
 
A preliminary lemma in this direction is as follows.


:'''Lemma''' Let H be a Hilbert space with nested subspaces
:'''Lemma''' Let H be a Hilbert space with nested subspaces
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* (Boundedness) <math>1, f</math> have norm at most 1.
* (Boundedness) <math>1, f</math> have norm at most 1.
* (Positive density) <math> \langle 1, f \rangle \geq \delta > 0</math>.
* (Positive density) <math> \langle 1, f \rangle \geq \delta > 0</math>.
* (Constancy of 1) <math>U_i^* 1 = 1</math> for all i.
* (Measure preservation) <math>U_i^* 1 = 1</math> for all i.
* (Substationarity) Each <math>U_i</math> has operator norm at most <math>1+O(\varepsilon)</math>.
* (Substationarity) Each <math>U_i</math> has operator norm at most <math>1+O(\varepsilon)</math>.
* (Measurability) For each i, <math>U_i</math> maps <math>V_{i-1}</math> to <math>V_i</math>.   
* (Measurability) For each i, <math>U_i</math> maps <math>V_{i-1}</math> to <math>V_i</math>.   
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: Then there exists <math>1 \leq i \leq m</math> such that <math> \langle f, U_i f \rangle > \delta^2 - o(1)</math>, where o(1) goes to zero as <math>\varepsilon \to 0</math>.
: Then there exists <math>1 \leq i \leq m</math> such that <math> \langle f, U_i f \rangle > \delta^2 - o(1)</math>, where o(1) goes to zero as <math>\varepsilon \to 0</math>.


== Proof ==
== Sketch of infinitary proof ==
 
By a compactness argument it suffices to treat the case <math>\varepsilon=0, m = \infty</math>.  We may then restrict H to the closure of <math>\bigcup_i V_i</math>.  The stability hypothesis is then asserting that the <math>U_i</math> converge in the weak operator topology to some limit P, and that <math>U_k U_j</math> also converge in this topology.  By computing the limit <math>\lim_{k \to \infty} \lim_{j \to \infty} \langle U_k U_j v, w \rangle</math> in two different ways, we conclude that <math>P^2 = P</math>, i.e. P is idempotent.  On the other hand, from substationarity we know that P has operator norm at most 1.  The two facts are only compatible when P is an orthonormal projection.  It now suffices to show that <math> \langle f, Pf \rangle \geq \delta^2,</math>.  But the left-hand side is <math>\|Pf\|^2</math>; meanwhile, positive density and measure preservation give us <math>P^* 1 = 1</math> and thus <math>\langle Pf, 1 \rangle \geq \delta</math>, and the claim follows from Cauchy-Schwarz.
 
== Finitary proof ==


By the stability hypothesis, we have an operator <math>P_i = \Pi_{V_i} P_i \Pi_{V_i}</math> such that
By the stability hypothesis, we have an operator <math>P_i = \Pi_{V_i} P_i \Pi_{V_i}</math> such that
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which by (3) is equal to
which by (3) is equal to
:<math> \| P_{i_0} f \|^2 + o(1).</math>
:<math> \| P_{i_0} f \|^2 + o(1).</math>
On the other hand, from positive density and constancy we have
On the other hand, from positive density and measure preservation we have
:<math> \langle U_{i_0+1} f, 1 \rangle \geq \delta</math>
:<math> \langle U_{i_0+1} f, 1 \rangle \geq \delta</math>
and thus by (1)
and thus by (1)

Latest revision as of 12:55, 18 May 2009

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Preliminary lemma

A preliminary lemma in this direction is as follows.

Lemma Let H be a Hilbert space with nested subspaces
[math]\displaystyle{ V_0 \subset V_1 \subset \ldots \subset V_m \subset H }[/math].
Suppose [math]\displaystyle{ V_0 }[/math] contains two elements 1, f.
For each [math]\displaystyle{ 1 \leq i \leq m }[/math], let [math]\displaystyle{ U_i: H \to H }[/math] be a bounded linear operator. We assume the following axioms:
  • (Boundedness) [math]\displaystyle{ 1, f }[/math] have norm at most 1.
  • (Positive density) [math]\displaystyle{ \langle 1, f \rangle \geq \delta \gt 0 }[/math].
  • (Measure preservation) [math]\displaystyle{ U_i^* 1 = 1 }[/math] for all i.
  • (Substationarity) Each [math]\displaystyle{ U_i }[/math] has operator norm at most [math]\displaystyle{ 1+O(\varepsilon) }[/math].
  • (Measurability) For each i, [math]\displaystyle{ U_i }[/math] maps [math]\displaystyle{ V_{i-1} }[/math] to [math]\displaystyle{ V_i }[/math].
  • (Stability) For each [math]\displaystyle{ 0 \leq i \lt m-1 }[/math], the operators [math]\displaystyle{ \Pi_{V_i} U_j \Pi_{V_i} }[/math] for [math]\displaystyle{ i \lt j \leq m }[/math] and [math]\displaystyle{ \Pi_{V_i} U_k U_j \Pi_{V_i} }[/math] for [math]\displaystyle{ i \lt j \lt k \leq m }[/math] differ from each other by at most [math]\displaystyle{ O(\varepsilon) }[/math] in the operator norm, where [math]\displaystyle{ \Pi_{V_i} }[/math] is the orthogonal projection onto [math]\displaystyle{ V_i }[/math].
  • (Parameter sizes) [math]\displaystyle{ \varepsilon }[/math] is sufficiently small depending on [math]\displaystyle{ \delta }[/math], and m is sufficiently large depending on [math]\displaystyle{ \delta, \varepsilon }[/math].
Then there exists [math]\displaystyle{ 1 \leq i \leq m }[/math] such that [math]\displaystyle{ \langle f, U_i f \rangle \gt \delta^2 - o(1) }[/math], where o(1) goes to zero as [math]\displaystyle{ \varepsilon \to 0 }[/math].

Sketch of infinitary proof

By a compactness argument it suffices to treat the case [math]\displaystyle{ \varepsilon=0, m = \infty }[/math]. We may then restrict H to the closure of [math]\displaystyle{ \bigcup_i V_i }[/math]. The stability hypothesis is then asserting that the [math]\displaystyle{ U_i }[/math] converge in the weak operator topology to some limit P, and that [math]\displaystyle{ U_k U_j }[/math] also converge in this topology. By computing the limit [math]\displaystyle{ \lim_{k \to \infty} \lim_{j \to \infty} \langle U_k U_j v, w \rangle }[/math] in two different ways, we conclude that [math]\displaystyle{ P^2 = P }[/math], i.e. P is idempotent. On the other hand, from substationarity we know that P has operator norm at most 1. The two facts are only compatible when P is an orthonormal projection. It now suffices to show that [math]\displaystyle{ \langle f, Pf \rangle \geq \delta^2, }[/math]. But the left-hand side is [math]\displaystyle{ \|Pf\|^2 }[/math]; meanwhile, positive density and measure preservation give us [math]\displaystyle{ P^* 1 = 1 }[/math] and thus [math]\displaystyle{ \langle Pf, 1 \rangle \geq \delta }[/math], and the claim follows from Cauchy-Schwarz.

Finitary proof

By the stability hypothesis, we have an operator [math]\displaystyle{ P_i = \Pi_{V_i} P_i \Pi_{V_i} }[/math] such that

[math]\displaystyle{ \Pi_{V_i} U_j \Pi_{V_i} = P_i + o(1) }[/math] (1)

for all [math]\displaystyle{ i \lt j \leq m }[/math] and

[math]\displaystyle{ \Pi_{V_i} U_k U_j \Pi_{V_i} = P_i + o(1) }[/math] (2)

for all [math]\displaystyle{ i \lt j \lt k \leq m }[/math], where o(1) denotes a quantity that goes to zero as [math]\displaystyle{ \varepsilon \to 0 }[/math] (in the operator norm for operators, or the Hilbert space norm for vectors). By the substationarity hypothesis, [math]\displaystyle{ P_i }[/math] has operator norm [math]\displaystyle{ 1+o(1) }[/math].

From (1) we see that [math]\displaystyle{ \| P_i f \|^2 }[/math] is monotone increasing in i up to errors of [math]\displaystyle{ o(1) }[/math]. By the pigeonhole principle, one can thus find [math]\displaystyle{ i_0 }[/math] such that

[math]\displaystyle{ \| P_{i_0 + \lfloor 1/\varepsilon \rfloor} f\|^2 = \| P_{i_0} f \|^2 + o(1) }[/math]

which then implies that

[math]\displaystyle{ P_i f = P_j f + o(1) }[/math] (3)

for [math]\displaystyle{ i_0 \leq i, j \leq i_0 + 1/\varepsilon }[/math]. A second application of the pigeonhole principle then locates [math]\displaystyle{ i_0 \leq i_1 \leq i_0 + 1/\varepsilon - 3 }[/math] such that

[math]\displaystyle{ \| P_{i_1+3}^* P_{i_0} f\|^2 = \| P_{i_1}^* P_{i_0} f \|^2 + o(1) }[/math]

and thus

[math]\displaystyle{ P_i^* P_{i_0} f = P_j^* P_{i_0} f + o(1) }[/math] (4)

for [math]\displaystyle{ i_1 \leq i,j \leq i_1+3 }[/math].

Now we compute the expression

[math]\displaystyle{ \| P_{i_0} f \|^2 = \langle P_{i_0} f, P_{i_0} f \rangle. }[/math]

From (2), this is equal to

[math]\displaystyle{ \langle U_{i_1+2} U_{i_1+1} f, P_{i_0} f \rangle + o(1). }[/math]

Applying (1), this is equal to

[math]\displaystyle{ \langle P_{i_1+1} U_{i_1+1} f, P_{i_0} f \rangle + o(1). }[/math]

Moving the [math]\displaystyle{ P_{i_1+1} }[/math] temporarily to the other side of the inner product, applying (4), and then moving it back, this is equal to

[math]\displaystyle{ \langle P_{i_1} U_{i_1+1} f, P_{i_0} f \rangle + o(1). }[/math]

which by (1) is equal to

[math]\displaystyle{ \langle P_{i_1} P_{i_1} f, P_{i_0} f \rangle + o(1). }[/math]

Aplying (3), this is equal to

[math]\displaystyle{ \langle P_{i_1} P_{i_0} f, P_{i_0} f \rangle + o(1). }[/math]

which is in turn equal to

[math]\displaystyle{ \langle P_{i_0} f, P_{i_1}^* P_{i_0} f \rangle + o(1). }[/math]

On the other hand, by Cauchy-Schwarz and the boundedness of [math]\displaystyle{ P_{i_1} }[/math], this is bounded by [math]\displaystyle{ \|P_{i_0} f \|^2 + o(1) }[/math]. In order for Cauchy-Schwarz to be sharp, we are thus forced to conclude that

[math]\displaystyle{ P_{i_1}^* P_{i_0} f = P_{i_0} f + o(1) }[/math]. (5)

Now, we look at the expression

[math]\displaystyle{ \langle f, P_{i_0} f \rangle. }[/math]

By (5), this is

[math]\displaystyle{ \langle P_{i_1} f, P_{i_0} f \rangle + o(1) }[/math]

which by (3) is equal to

[math]\displaystyle{ \| P_{i_0} f \|^2 + o(1). }[/math]

On the other hand, from positive density and measure preservation we have

[math]\displaystyle{ \langle U_{i_0+1} f, 1 \rangle \geq \delta }[/math]

and thus by (1)

[math]\displaystyle{ \langle P_{i_0} f, 1 \rangle \geq \delta. }[/math]

By Cauchy-Schwarz we thus have

[math]\displaystyle{ \| P_{i_0} f \|^2 \geq \delta^2 }[/math]

and the claim follows. [math]\displaystyle{ \Box }[/math]