A Hilbert space lemma

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

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

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

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

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

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

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

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

which then implies that

[math] P_i f = P_j f + o(1)[/math] (3)

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

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

and thus

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

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

Now we compute the expression

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

From (2), this is equal to

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

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

which by (1) is equal to

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

Aplying (3), this is equal to

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

which is in turn equal to

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

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

Now, we look at the expression

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

By (5), this is

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

which by (3) is equal to

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

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

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

and thus by (1)

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

By Cauchy-Schwarz we thus have

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

and the claim follows. [math]\Box[/math]