# A Hilbert space lemma

## Preliminary lemma

A preliminary lemma in this direction is as follows.

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

## Sketch of infinitary proof

By a compactness argument it suffices to treat the case $\varepsilon=0, m = \infty$. We may then restrict H to the closure of $\bigcup_i V_i$. The stability hypothesis is then asserting that the $U_i$ converge in the weak operator topology to some limit P, and that $U_k U_j$ also converge in this topology. By computing the limit $\lim_{k \to \infty} \lim_{j \to \infty} \langle U_k U_j v, w \rangle$ in two different ways, we conclude that $P^2 = P$, 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 $\langle f, Pf \rangle \geq \delta^2,$. But the left-hand side is $\|Pf\|^2$; meanwhile, positive density and measure preservation give us $P^* 1 = 1$ and thus $\langle Pf, 1 \rangle \geq \delta$, and the claim follows from Cauchy-Schwarz.

## Finitary proof

By the stability hypothesis, we have an operator $P_i = \Pi_{V_i} P_i \Pi_{V_i}$ such that

$\Pi_{V_i} U_j \Pi_{V_i} = P_i + o(1)$ (1)

for all $i \lt j \leq m$ and

$\Pi_{V_i} U_k U_j \Pi_{V_i} = P_i + o(1)$ (2)

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

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

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

which then implies that

$P_i f = P_j f + o(1)$ (3)

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

$\| P_{i_1+3}^* P_{i_0} f\|^2 = \| P_{i_1}^* P_{i_0} f \|^2 + o(1)$

and thus

$P_i^* P_{i_0} f = P_j^* P_{i_0} f + o(1)$ (4)

for $i_1 \leq i,j \leq i_1+3$.

Now we compute the expression

$\| P_{i_0} f \|^2 = \langle P_{i_0} f, P_{i_0} f \rangle.$

From (2), this is equal to

$\langle U_{i_1+2} U_{i_1+1} f, P_{i_0} f \rangle + o(1).$

Applying (1), this is equal to

$\langle P_{i_1+1} U_{i_1+1} f, P_{i_0} f \rangle + o(1).$

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

$\langle P_{i_1} U_{i_1+1} f, P_{i_0} f \rangle + o(1).$

which by (1) is equal to

$\langle P_{i_1} P_{i_1} f, P_{i_0} f \rangle + o(1).$

Aplying (3), this is equal to

$\langle P_{i_1} P_{i_0} f, P_{i_0} f \rangle + o(1).$

which is in turn equal to

$\langle P_{i_0} f, P_{i_1}^* P_{i_0} f \rangle + o(1).$

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

$P_{i_1}^* P_{i_0} f = P_{i_0} f + o(1)$. (5)

Now, we look at the expression

$\langle f, P_{i_0} f \rangle.$

By (5), this is

$\langle P_{i_1} f, P_{i_0} f \rangle + o(1)$

which by (3) is equal to

$\| P_{i_0} f \|^2 + o(1).$

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

$\langle U_{i_0+1} f, 1 \rangle \geq \delta$

and thus by (1)

$\langle P_{i_0} f, 1 \rangle \geq \delta.$

By Cauchy-Schwarz we thus have

$\| P_{i_0} f \|^2 \geq \delta^2$

and the claim follows. $\Box$