variables $Y_n \in E$ and $R_{Y_n}^{(n)} \in \mathcal{M}_R(E)$ for all $n \ge 1$ satisfying the description in Section 2; then $m_n$ is given by

$$m_n := m_0 + \sum_{i=1}^{n} R_{Y_i}^{(i)}. \qquad (A.2)$$

Equivalently, we want to construct the joint distribution of all $(Y_n, R_{Y_n}^{(n)})$, $n \ge 1$, as a probability measure on $(E \times \mathcal{M}_R(E))^{\infty}$. We will achieve this using the construction (A.1) twice. However, we have (so far) only been able to do so assuming one of the following assumptions (or both).

(i) $R_x^{(1)}$ is always a positive measure, so there are no subtractions in the urn, or

(ii) $E$ is a Borel space. (See e.g. [26, Chapter 1 and Appendix A]and [7, Theorem 8.3.6]; note in particular that any Polish space is a Borel space, and that any Borel space is either countable (possibly finite), or Borel isomorphic to $[0, 1]$.)

The reasons for the technical assumption (ii) will be discussed below.

(i): Consider first the simple case when $R_x^{(1)}$ always is a positive measure, i.e., $R_x^{(1)} \in \mathcal{M}_+(E)$. In this case, there is no need to consider signed measures. Write $\mathcal{X} := E \times \mathcal{M}_+(E)$. Let $n \ge 0$ and assume that we have constructed the distribution $\mu_n$ of $(Y_i, R_{Y_i}^{(i)})_1^n$, as a probability measure on $\mathcal{X}^n$. (This assumption is void for $n = 0$.) We write an element of $\mathcal{X}^n$ as $(y_i, r_i)_1^n$; then we can realize $Y_i$ and $R_{Y_i}^{(i)}$ for $i \le n$ as the coordinate functions $y_i$ and $r_i$ on the probability space $(\mathcal{X}^n, \mu_n)$. By (A.2), $m_n$ then is given by the function $m_n: \mathcal{X}^n \to \mathcal{M}_{>0}(E)$ defined by

$$m_n((y_i, r_i)_1^n) := m_0 + \sum_{i=1}^{n} r_i. \qquad (A.3)$$

Thus, the normalized measure $\tilde{m}_n$ is given by the function $\gamma_n: \mathcal{X}^n \to \mathbb{P}(E)$ defined by

$$\gamma_n(\xi_n) := m_n(\xi_n)/m_n(\xi_n)(E). \qquad (A.4)$$

Nota that $\gamma_n$ is a probability kernel from $\mathcal{X}^n$ to $E$.

We want $Y_{n+1}$ to be a random element of $E$ such that, conditioned on the history up to time $n$, $Y_{n+1}$ has the distribution $\tilde{m}_n$. In other words, conditioned on $(Y_i, R_{Y_i}^{(i)})_1^n = \xi_n \in \mathcal{X}^n$, $Y_{n+1}$ has the conditional distribution $\gamma_n(\xi_n)$. This means that

$$((Y_i, R_{Y_i}^{(i)})_1^n, Y_{n+1}) \sim \mu_n \otimes \gamma_n, \qquad (A.5)$$

and we may take this as a formal definition of (the distribution of) $Y_{n+1}$.

Next, the replacement kernel $\mathcal{R}$ is now assumed to be a probability kernel from $E$ to $\mathcal{M}_+(E)$. We may (trivially) regard it as a kernel from $\mathcal{X}^n \times E$ by letting $\mathcal{R}_{(\xi_n, x)} := \mathcal{R}_x$. Hence, (A.1) defines the probability measure $(\mu_n \otimes \gamma_n) \otimes \mathcal{R}$ on $\mathcal{X}^n \times E \times \mathcal{M}_+(E) = \mathcal{X}^{n+1}$. We want $R_{Y_{n+1}}^{(n+1)}$ to have the conditional distribution, given the previous history, $\mathcal{R}_{Y_{n+1}}$, and thus

$$((Y_i, R_{Y_i}^{(i)})_1^n, Y_{n+1}, R_{Y_{n+1}}^{(n+1)}) \sim \mu_n \otimes \gamma_n \otimes \mathcal{R}. \qquad (A.6)$$

(Note that $\otimes$ is associative: $(\mu_n \otimes \gamma_n) \otimes \mathcal{R} = \mu_n \otimes (\gamma_n \otimes \mathcal{R})$, so we may omit the brackets.) We may take (A.6) as a formal definition of $R_{Y_{n+1}}^{(n+1)}$.

In other words, our formal construction is

$$\mu_{n+1} := \mu_n \otimes \gamma_n \otimes \mathcal{R} \in \mathcal{P}(\mathcal{X}^{n+1}). \qquad (A.7)$$
This completes the inductive step, and starting from the trivial probability measure $\mu_0$ on a one-point space, we obtain recursively a probability measure $\mu_n$ on $\mathcal{X}^n$ for every $n \ge 1$. Finally, since $\mu_n$ are obtained recursively by composing with the probability kernels $\gamma_n \otimes \mathcal{R}$, the Ionescu Tulcea theorem [26, Theorem 6.17] now shows the existence of a probability space and infinite sequences $Y_n$ and $R_{Y_n}^{(n)}$ with the desired distribution; this defines also $m_n$ by (A.2). Equivalently, the Ionescu Tulcea theorem shows the existence of a probability measure on $\mathcal{X}^\infty$ with the desired projection $\mu_n$ to $\mathcal{X}^n$ for each $n$. This completes the construction in the special case when $R_x^{(1)} \in \mathcal{M}_+(E)$. It follows from the construction that $(m_n, Y_n, R_{Y_n}^{(n)})_{n \ge 1}$ is a Markov chain.

(ii): Consider now the general case, when $R_x^{(1)} \in \mathcal{M}_R(E)$ is a signed measure, but we assume that the urn is tenable. Assume now also that $E$ is a Borel space. We may now define $\mathcal{X} := E \times \mathcal{M}_R(E)$, and try to argue as above. The only problem is that $\gamma_n$ defined by (A.4) is not a probability kernel, since $m_n(\xi_n)$ is not a positive measure for all $\xi_n \in \mathcal{X}^n$. (We will even have $m_n(\xi_n)(E) = 0$ for some $\xi_n$, and then $\gamma_n(\xi_n)$ is not even defined.) We thus have to modify the definition of $\gamma_n$. Consider again some $n \ge 0$ and assume that we have constructed $\mu_n \in \mathcal{P}(\mathcal{X}^n)$. Note that for a Borel space $E$, $\mathcal{M}_{>0}(E)$ is a measurable subset of $\mathcal{M}_R(E)$, as may easily be verified. Let $\Upsilon_n := m_n^{-1}(\mathcal{M}_{>0}(E))$, where $m_n: \mathcal{X}^n \to \mathcal{M}_R(E)$ is the function defined in (A.3); thus $\Upsilon_n$ is a measurable subset of $\mathcal{X}^n$. We assume that the urn is tenable, which means that $m_n$ a.s. satisfies $m_n \in \mathcal{M}_{>0}(E)$. In other words, $m_n(\xi) \in \mathcal{M}_{>0}(E)$ for $\mu_n$-a.e. $\xi$; equivalently, $\mu_n(\Upsilon_n) = 1$. We may now modify (A.4) and define a probability kernel $\gamma_n$ from $\mathcal{X}^n$ to $E$ by

$$ \gamma_n(\xi_n) := \begin{cases} \tilde{m}_n(\xi_n) := m_n(\xi_n)/m_n(\xi_n)(E), & \text{if } \xi_n \in \Upsilon_n, \\ \nu, & \text{if } \xi_n \in \mathcal{X}^n \setminus \Upsilon_n, \end{cases} \quad (A.8) $$

where $\nu$ is an arbitrary, fixed probability measure on $E$. Then the construction proceeds as above. (The choice of $\nu$ does not affect $\mu_{n+1}$, since $\mu_n(\Upsilon) = 1$.) This completes the construction in case (ii), when $E$ is a Borel space.

What happens when $E$ is not a Borel space? In some cases it might be possible to modify the construction above; for example if (for each $n \ge 1$) there exists a measurable subset $\Upsilon_n$ of $m_n^{-1}(\mathcal{M}_{>0}(E))$ such that $\mu_n(\Upsilon_n) = 1$. However, we will see in Example A.2 that in general no such $\Upsilon_n$ exists. In general, unless (i) ar (ii) above holds, we have to assume that the process $m_n$ is defined by some external construction. (See Example A.2 for an example where a construction is trivial.)

**Example A.2.** Let $E := \{0, 1\}^A$ for some uncountable set $A$. Let $Z$ be a random element of $E$, with some distribution $\nu_Z \in \mathcal{P}(E)$, and let

$$ R_x := -\delta_x + 2\delta_Z, \quad x \in E; \quad (A.9) $$

also, let $m_0 := \delta_{x_0}$ for some $x_0 \in E$. This describes an urn with balls (corresponding to point masses) labelled by elements of $E$; we start with a single ball $x_0$, and in each step we remove one randomly chosen ball, and add two new balls with label $Z_n$, where $(Z_n)_1^\infty$ are i.i.d. This process is obviously well defined and tenable. Nevertheless, we will see that there is no measurable set $\Upsilon_1$ such that the construction (A.8) works for $n = 1$. (In particular, $\mathcal{M}_+(E)$ is not a measurable subset of $\mathcal{M}_R(E)$.) Note that necessarily $Y_1 = x_0$, and thus $R_{Y_1}^{(1)} = -\delta_{x_0} + 2\delta_{Z_1}$. Hence, the distribution $\mu_1$ of $(Y_1, R_{Y_1}^{(1)})$ is the product measure $\delta_{x_0} \times \mathcal{L}(R_{Y_1}^{(1)})$. Suppose that $\Upsilon_1 \subseteq \mathcal{X} = E \times \mathcal{M}_R(E)$ is measurable and such that $\mu_1(\Upsilon_1) = 1$ and $m_1(y, r) = m_0 + r \in \mathcal{M}_{>0}(E)$ for every $(y, r) \in \Upsilon_1$. We will show that this leads to a contradiction.

Let $\Lambda \subseteq \mathcal{M}_R(E)$ be a non-empty measurable set. Recall that the $\sigma$-field on $\mathcal{M}_R(E)$ is generated by the mappings $\mu \mapsto \mu(B)$ for $B \in \mathcal{E}$, where $\mathcal{E}$ is the $\sigma$-field on $E$. It is well
known that this implies that there exists a countable family $(B_i) \subset \mathcal{E}$ such that $\Lambda$ belongs to the $\sigma$-field generated by the mappings $\mu \mapsto \mu(B_i)$, $i \in \mathbb{N}$. (Because the union of these $\sigma$-field over all countable families $(B_i)$ is a $\sigma$-field.)

Similarly, since the product $\sigma$-field $\mathcal{E}$ is generated by the coordinate maps $(x_a)_{a \in \mathcal{A}} \mapsto x_a$ for $a \in \mathcal{A}$, for each $B \in \mathcal{E}$ there is a countable subset $\mathcal{A}_B \subset \mathcal{A}$ and a (measurable) set $\tilde{B} \subseteq \{0, 1\}^{\mathcal{A}_B}$ such that

$$B = \tilde{B}_i \times \{0, 1\}^{\mathcal{A} \setminus \mathcal{A}_B}. \qquad (A.10)$$

Fix a coordinate $a' \in \mathcal{A} \setminus \bigcup_i \mathcal{A}_{B_i}$. Define, for $j \in \{0, 1\}$, the elements $z^j = (z_a^j) \in E$ by

$$z_a^j := \begin{cases} j, & a = a', \\ 0, & a \neq a'. \end{cases} \qquad (A.11)$$

Take a signed measure $\lambda \in \Lambda$, and for $N \ge 0$, let $\lambda_N := \lambda + N(\delta_{z^0} - \delta_{z^1})$. For each $B_i$, we have $a' \notin \mathcal{A}_{B_i}$, and thus, by (A.10), $z^0 \in B_i \iff z^1 \in B_i$. Consequently, for every $N \ge 0$,

$$\lambda_N(B_i) = \lambda(B_i) + N(\mathbf{1}\{z^0 \in \mathcal{A}_{B_i}\} - \mathbf{1}\{z^1 \in \mathcal{A}_{B_i}\}) = \lambda(B_i) \qquad (A.12)$$

for every $B_i$. Since $\Lambda$ is in the $\sigma$-field generated by the maps $\mu \mapsto \mu(B_i)$, and $\lambda \in \Lambda$, it follows that $\lambda_N \in \Lambda$. On the other hand, if $B := \{(x_a) \in E : x_a' = 1\}$, then $B \in \mathcal{E}$ and $\lambda_N(B) = \lambda(B) - N$; hence, if $N$ is large enough, $\lambda_N(B) < 1$ and thus $\lambda_N + \mathbf{m}_0 \notin \mathcal{M}_{>0}(E)$.

We have shown that there is no nonempty measurable set $\Lambda \subseteq \mathcal{M}_R(E)$ such that

$$\lambda \in \Lambda \implies \lambda + \mathbf{m}_0 \in \mathcal{M}_{>0}(E). \qquad (A.13)$$

However, if $\Upsilon_1$ is as above, then the section $\Lambda := \{r \in \Upsilon_1 : (x_0, r) \in \Upsilon_1\}$ is measurable, satisfies (A.13), and also $\mathbb{P}(R_{\Upsilon_1}^{(1)} \in \Lambda) = 1$, a contradiction.

Note that the proof shows that $\mathcal{M}_{>0}(E)$ is not a measurable subset of $\mathcal{M}_R(E)$, and, moreover, that it does not contain any non-empty measurable subset. (The same holds for $\mathcal{M}_+(E)$.)

## B Some functional analysis

In this appendix we state some general results on spectra of operators in Banach spaces; these are used in our examples in Section 6. The results are simple and have presumably been known for a long time, but since we have not found references to the results in the form that we need, we give full proofs for completeness.

Recall that if $T$ is a bounded operator on $\mathcal{X}$, and $T^*$ is its adjoint acting on the dual space $\mathcal{X}^*$, then [8, Proposition VII.6.1]

$$\sigma(T^*) = \sigma(T). \qquad (B.1)$$

Our first lemma deals with the situation when we instead consider $T^*$ as acting on a subspace of $\mathcal{X}^*$.

**Definition B.1.** If $K$ is a compact subset of $C$, define $K^\wedge$ as the union of $K$ and all bounded connected components of $C \setminus K$; in other words, its complement $C \setminus K^\wedge$ is the unbounded component of $C \setminus K$. ($K^\wedge$ is known as the polynomially convex hull of $K$, see [8, Definition VII.5.2 and Proposition VII.5.3].) In particular, if $T$ is a bounded operator on a Banach space and $\rho_\infty(T)$ denotes the unbounded component of the resolvent set $\rho(T) = C \setminus \sigma(T)$, then

$$\sigma(T)^\wedge := C \setminus \rho_\infty(T). \qquad (B.2)$$
We let $\langle x^*, x \rangle$ denote the pairing of elements $x^* \in \mathcal{X}^*$ and $x \in \mathcal{X}$, for any Banach space $\mathcal{X}$.

**Lemma B.2.** Let $T$ be a bounded operator on a complex Banach space $\mathcal{X}$, and suppose that $\mathcal{Y} \subseteq \mathcal{X}^*$ is a closed subspace of the dual space $\mathcal{X}^*$ such that the adjoint operator $T^*$ maps $\mathcal{Y}$ into itself.

(i) Then

$$ \sigma(T^*|_y) \subseteq \sigma(T)^\wedge. \qquad (B.3) $$

(ii) Suppose further that $\mathcal{Y}$ is norm-determining, i.e., that if $x \in \mathcal{X}$, then

$$ \|x\| = \sup\{\langle x^*, x \rangle : x^* \in \mathcal{Y}, \|x^*\| = 1\}. \qquad (B.4) $$

Then also

$$ \sigma(T) \subseteq \sigma(T^*|_y)^\wedge \qquad (B.5) $$

and thus

$$ \sigma(T^*|_y)^\wedge = \sigma(T)^\wedge. \qquad (B.6) $$

**Proof.** (i): As said above, the spectrum $\sigma(T^*)$ of $T^*$ as an operator on $\mathcal{X}^*$ equals $\sigma(T)$, and the resolvent is simply given by $(z - T^*)^{-1} = ((z - T)^{-1})^*$, $z \in \rho(T) = \mathbb{C} \setminus \sigma(T)$.

We first show that this resolvent maps $\mathcal{Y}$ into itself, at least when $z \notin \sigma(T)^\wedge$. To do so, we take $y \in \mathcal{Y}$ and let $x^{**} \in \mathcal{X}^{**}$ be such that $x^{**} \perp \mathcal{Y}$, i.e., $\langle x^{**}, y' \rangle = 0$ for every $y' \in \mathcal{Y}$. Consider the function

$$ g(z) := \langle x^{**}, (z - T^*)^{-1}y \rangle, \quad z \in \rho(T) = \rho(T^*). \qquad (B.7) $$

The function $g$ is analytic on $\rho(T)$, see [8, Theorem VII.3.6]. Furthermore, if $|z| > \|T\|$, then $(z - T^*)^{-1} = \sum_{k=0}^{\infty} z^{-k-1}(T^*)^k$ with an absolutely convergent sum, and thus, because $T^*(\mathcal{Y}) \subseteq \mathcal{Y}$,

$$ (z - T^*)^{-1}y = \sum_{k=0}^{\infty} z^{-k-1}(T^*)^k y \in \mathcal{Y}. \qquad (B.8) $$

Consequently, (B.7) and (B.8) imply that if $|z| > \|T\|$, then $g(z) = 0$. By analytic continuation, $g(z) = 0$ in the unbounded connected component $\rho_\infty(T)$ of $\rho(T)$.

This holds for any $x^{**} \perp \mathcal{Y}$, and thus, by definition of $g$ in (B.7), it follows that $(z - T^*)^{-1}y \in \mathcal{Y}$ for all $z \in \rho_\infty(T)$. In other words, for all $z \in \rho_\infty(T)$, we have $(z - T^*)^{-1}: \mathcal{Y} \to \mathcal{Y}$, which means that it is the inverse of the induced operator $(z - T^*)|_y$. Hence, for all $z \in \rho_\infty(T)$, $z$ belongs to the resolvent set $\rho(T^*|_y)$; in other words, $\rho_\infty(T) \subseteq \rho(T^*|_y)$, and thus (B.3) holds by (B.2).

(ii): The canonical embedding $\mathcal{X} \to \mathcal{X}^{**}$ induces a linear map $\mathcal{X} \to \mathcal{Y}^*$, which is an isometric embedding by the assumption (B.4). Hence, we may regard $\mathcal{X}$ as a subspace of $\mathcal{Y}^*$. We may thus apply part (i) with $\mathcal{X}$ and $\mathcal{Y}$, and also $T$ and $T^*$, interchanged. This yields (B.5), and (B.6) then easily follows from (B.3) and (B.5). $\square$

**Corollary B.3.** Let $T$ be a bounded operator on a complex Banach space $\mathcal{X}$, and suppose that $\mathcal{Y} \subseteq \mathcal{X}^*$ is a closed subspace of the dual space $\mathcal{X}^*$ such that the adjoint operator $T^*$ maps $\mathcal{Y}$ into itself. Suppose further that $\mathcal{Y}$ is norm-determining. Then

(i) $T$ is an slqc operator on $\mathcal{X}$ if and only if $T^*$ is an slqc operator on $\mathcal{Y}$.
(ii) $T$ is a strictly small operator on $\mathcal{X}$ if and only if $T^*$ is a strictly small operator on $\mathcal{Y}$.

*Proof.* (i): Suppose that $T$ is an slqc operator. Let $\theta := \sup \Re(\sigma(T) \setminus \{1\})$ and note that $\theta < 1$ as in Lemma 2.10(i). We then have

$$ U := \{\lambda : \Re\lambda > \theta\} \setminus \{1\} \subset \rho(T), \qquad (B.9) $$

which implies, since the set $U$ is connected and unbounded,

$$ \{\lambda : \Re\lambda > \theta\} \setminus \{1\} \subset \rho_{\infty}(T), \qquad (B.10) $$

and thus

$$ \sigma(T)^{\wedge} \subset \{\lambda : \Re\lambda \le \theta\} \cup \{1\}. \qquad (B.11) $$

Hence Lemma B.2 yields

$$ \sigma(T^*|_Y)^{\wedge} = \sigma(T)^{\wedge} \subset \{\lambda : \Re\lambda \le \theta\} \cup \{1\}, \qquad (B.12) $$

which implies (QC2) for $T^*|_Y$, and also that 1 is isolated in $\sigma(T^*|_Y)$ if 1 belongs to this spectrum at all. It remains to show only that 1 is an eigenvalue of $T^*|_Y$ and that the corresponding spectral projection $\Pi_1(T^*|_Y)$ has rank 1.

We can regard $T^*$ as an operator on $\mathcal{X}^*$ or on its subspace $\mathcal{Y}$. In both cases we have, see [8, Equation VII.6.9],

$$ \Pi_1(T^*) = \frac{1}{2\pi i} \oint_{\Gamma} (z - T^*)^{-1} dz \qquad (B.13) $$

where we choose $\Gamma$ to be a small circle around 1 inside $\rho_{\infty}(T)$, cf. (B.10). By the proof of Lemma B.2, if $z \in \Gamma$, then $(z - T^*)^{-1}$ maps $\mathcal{Y}$ into itself, and its induced operator on $\mathcal{Y}$ is $(z - T^*|_Y)^{-1}$. Hence, (B.13) shows that $\Pi_1(T^*)$ maps $\mathcal{Y}$ into itself, and that its induced operator on $\mathcal{Y}$ is $\Pi_1(T^*|_Y)$.

Moreover, $(z - T^*)^{-1} = ((z - T)^{-1})^*$ for $z \in \Gamma$, and thus by (B.13) and the same formula for $T$, we have $\Pi_1(T^*) = \Pi_1(T)^*$. By Assumption (QC1), $\Pi_1(T)$ has rank 1, and is thus given by

$$ \Pi_1(T)x = \langle x_0^*, x \rangle x_0 \qquad (B.14) $$

for some non-zero $x_0 \in \mathcal{X}$ and $x_0^* \in \mathcal{X}^*$ with $\langle x_0^*, x_0 \rangle = 1$. It follows that, for any $x^* \in \mathcal{X}^*$,

$$ \Pi_1(T^*)(x^*) = \Pi_1(T)^*(x^*) = \langle x^*, x_0 \rangle x_0^*. \qquad (B.15) $$

Since $\mathcal{Y}$ is norm-determining, there exists $y \in \mathcal{Y}$ such that $\langle y, x_0 \rangle \ne 0$. Since $\Pi_1(T^*) : \mathcal{Y} \to \mathcal{Y}$, we have $\Pi_1(T^*)(y) \in \mathcal{Y}$, and (B.15) then shows that $x_0^* \in \mathcal{Y}$.

We have shown that $\Pi_1(T^*|_Y)$ is the rank 1 operator defined by (B.15) restricted to $x^* \in \mathcal{Y}$. In particular, $x_0^* \in \mathcal{Y}$ is an eigenvector with $T^*x_0^* = x_0^*$. Hence, (QC1) in Definition 2.7 holds for $T^*|_Y$, which concludes the proof that $T^*|_Y$ is slqc if $T$ is.

The converse follows, as in the proof of Lemma B.2(ii), by interchanging the roles of $\mathcal{X}$ and $\mathcal{Y}$, noting that $\mathcal{X}$ always is norm-determining as a subspace of $\mathcal{Y}^*$.

(ii): Now suppose that $T$ is strictly small. This means that in the proof of (i), we have $\theta < 1/2$. Hence, (B.12) shows that $T^*$ is strictly small. The converse follows as above. $\square$

In the following lemma, we compare the spectra of the “same” operator in two different spaces. When necessary, we use subscripts such as $T_{\mathcal{X}}$ to denote the space where we consider the operator.
**Lemma B.4.** Let $\mathcal{X}$ and $\mathcal{Y}$ be two complex Banach spaces and suppose that $\mathcal{Y} \subseteq \mathcal{X}$ with a continuous, but not necessarily isometric, inclusion. Suppose that $T$ is a bounded operator on $\mathcal{X}$ such that $T(\mathcal{X}) \subseteq \mathcal{Y}$.

(i) Then

$$ \sigma(T_{\mathcal{X}}) = \begin{cases} \sigma(T_{\mathcal{Y}}), & \mathcal{Y} = \mathcal{X}, \\ \sigma(T_{\mathcal{Y}}) \cup \{0\}, & \mathcal{Y} \subsetneq \mathcal{X}. \end{cases} \qquad (B.16) $$

(We do not make any claims on whether $0 \in \sigma(T_{\mathcal{Y}})$ or not.)

(ii) If $\lambda \neq 0$ is an isolated point in $\sigma(T_{\mathcal{X}})$, then $\Pi_{\lambda}(T_{\mathcal{Y}})$ equals the operator induced by $\Pi_{\lambda}(T_{\mathcal{X}})$ on $\mathcal{Y}$. (Thus we can use the notation $\Pi_{\lambda}$ for both without confusion.) Moreover, $\Pi_{\lambda}\mathcal{X} = \Pi_{\lambda}\mathcal{Y} \subseteq \mathcal{Y}$.

(iii) $T_{\mathcal{X}}$ is slqc if and only if $T_{\mathcal{Y}}$ is slqc. $T_{\mathcal{X}}$ is strictly small if and only if $T_{\mathcal{Y}}$ is strictly small.

*Proof.* (i): Note first that by the closed graph theorem, $T: \mathcal{X} \to \mathcal{Y}$ is a bounded operator. Hence, the operator $T_{\mathcal{Y}}$ induced on $\mathcal{Y}$ is a bounded operator on $\mathcal{Y}$, and the spectra $\sigma(T_{\mathcal{X}})$ and $\sigma(T_{\mathcal{Y}})$ are both defined.

If $\mathcal{Y} = \mathcal{X}$, then the norms on $\mathcal{X}$ and $\mathcal{Y}$ are equivalent, again by the closed graph theorem, and thus $\sigma(T_{\mathcal{X}}) = \sigma(T_{\mathcal{Y}})$.

Assume in the sequel that $\mathcal{Y} \neq \mathcal{X}$. In particular, since $T(\mathcal{X}) \subseteq \mathcal{Y}$, $T$ is not onto $\mathcal{X}$, and thus $T_{\mathcal{X}}$ is not invertible; hence $0 \in \sigma(T_{\mathcal{X}})$.

Suppose that $\lambda \in \rho(T_{\mathcal{X}})$. This means that the resolvent $R_{\lambda} := (\lambda - T)^{-1}$ exists as a bounded operator on $\mathcal{X}$. We have

$$ I = (\lambda - T)R_{\lambda} = \lambda R_{\lambda} - T R_{\lambda}. \qquad (B.17) $$

Hence, if $y \in \mathcal{Y}$, then, using again $T(\mathcal{X}) \subseteq \mathcal{Y}$,

$$ \lambda R_{\lambda} y = y + T R_{\lambda} y \in \mathcal{Y}. \qquad (B.18) $$

Since $0 \notin \rho(T_{\mathcal{X}})$, as remarked above, we have $\lambda \neq 0$. Hence (B.18) implies $R_{\lambda}y \in \mathcal{Y}$, and thus $R_{\lambda}: \mathcal{Y} \to \mathcal{Y}$. It follows immediately that the operator induced by $R_{\lambda}$ on $\mathcal{Y}$ is an inverse to $\lambda - T_{\mathcal{Y}}$, and thus $\lambda \in \rho(T_{\mathcal{Y}})$. We have shown that

$$ \rho(T_{\mathcal{X}}) \subseteq \rho(T_{\mathcal{Y}}) \setminus \{0\}. \qquad (B.19) $$

Conversely, suppose that $\lambda \in \rho(T_{\mathcal{Y}})$, and let $R'_{\lambda} := (\lambda - T_{\mathcal{Y}})^{-1}: \mathcal{Y} \to \mathcal{Y}$ denote the corresponding resolvent. Since $T: \mathcal{X} \to \mathcal{Y}$, we may define the operator $Q := I + R'_{\lambda}T$ on $\mathcal{X}$. For any $x \in \mathcal{X}$, we then have, since $Tx \in \mathcal{Y}$,

$$ (\lambda - T)Qx = (\lambda - T)x + (\lambda - T)R'_{\lambda}Tx = \lambda x - Tx + Tx = \lambda x \qquad (B.20) $$

and

$$ \begin{aligned} Q(\lambda - T)x &= (\lambda - T)x + R'_{\lambda}T(\lambda - T)x = (\lambda - T)x + R'_{\lambda}(\lambda - T)Tx = \lambda x - Tx + Tx \\ &= \lambda x. \end{aligned} \qquad (B.21) $$

Hence, if also $\lambda \neq 0$, then $\lambda^{-1}Q$ is an inverse of $\lambda - T$ on $\mathcal{X}$, and thus $\lambda \in \rho(T_{\mathcal{X}})$. Consequently, $\rho(T_{\mathcal{Y}}) \setminus \{0\} \subseteq \rho(\mathcal{X})$. Thus equality holds in (B.19), and thus (B.16) holds.

(ii): Let $\Gamma$ be a sufficiently small circle around $\lambda$, such that $\Gamma$ and its interior are disjoint from $\sigma(T_{\mathcal{X}}) \setminus \{\lambda\}$. Then the spectral projections $\Pi_{\lambda}(T_{\mathcal{X}})$ and $\Pi_{\lambda}(T_{\mathcal{Y}})$ are both
obtained by integrating the respective resolvents along $\Gamma$, as in (B.13). If $\lambda' \in \Gamma$, then, as shown in the proof of (i), $(\lambda' - T_Y)^{-1}$ is the operator induced by $(\lambda' - T_\chi)^{-1}$ on $\mathcal{Y}$; hence it follows that $\Pi_\lambda(T_Y)$ is the operator induced on $\mathcal{Y}$ by the projection $\Pi_\lambda(T_\chi)$. Consequently,

$$ \Pi_{\lambda}\mathcal{Y} = (\Pi_{\lambda}\mathcal{X}) \cap \mathcal{Y}. \qquad (B.22) $$

Moreover, $T$ maps $\Pi_\lambda\mathcal{X}$ into itself, and the operator induced by $T$ on $\Pi_\lambda\mathcal{X}$ is invertible (since its spectrum is $\{\lambda\}$, and $\lambda \neq 0$), and thus onto. Since $T: \mathcal{X} \to \mathcal{Y}$, it follows that $\Pi_\lambda\mathcal{X} \subseteq \mathcal{Y}$. Combined with (B.22), this yields $\Pi_\lambda\mathcal{X} = \Pi_\lambda\mathcal{Y}$ as asserted.

(iii): An immediate consequence of (i) and (ii). $\square$

**Lemma B.5.** Let $\mathcal{N}$ be a closed subspace of a complex Banach space $\mathcal{X}$, and let $\mathcal{Z} := \mathcal{X}/\mathcal{N}$. Suppose that $T$ is a bounded operator on $\mathcal{X}$ such that $Tn = 0$ for every $n \in \mathcal{N}$. Then $T$ can also be regarded as an operator on $\mathcal{Z}$, and the following holds.

(i) If $\mathcal{N} \neq \{0\}$, then $\sigma(T_\chi) = \sigma(T_\mathcal{Z}) \cup \{0\}$. (If $\mathcal{N} = \{0\}$, then trivially $\sigma(T_\chi) = \sigma(T_\mathcal{Z})$.)

(ii) If $\lambda \neq 0$ is an isolated point in $\sigma(T_\chi)$, then $\Pi_\lambda\mathcal{N} = \{0\}$, and thus $\Pi_\lambda(T_\chi)$ induces an operator on $\mathcal{Z} = \mathcal{X}/\mathcal{N}$; this induced operator equals $\Pi_\lambda(T_\mathcal{Z})$. Moreover, the quotient map $\mathcal{X} \to \mathcal{Z}$ is a bijection $\Pi_\lambda(T_\chi)\mathcal{X} \to \Pi_\lambda(T_\mathcal{Z})\mathcal{Z}$.

(iii) $T_\chi$ is slqc if and only if $T_\mathcal{Z}$ is slqc. $T_\chi$ is strictly small if and only if $T_\mathcal{Z}$ is strictly small.

*Proof.* That $T$ can be regarded as an operator on the quotient space $\mathcal{Z}$ is well known. Moreover, $\mathcal{Z}^*$ is identified with the closed subspace $\{x^* \in X^*: x^*(\mathcal{N}) = 0\}$ of $X^*$.

If $x^* \in X^*$ and $n \in \mathcal{N}$, then $\langle T^*x^*n \rangle = \langle x^*, Tn \rangle = 0$; thus $T^*x^* \in \mathcal{Z}^*$ for every $x^* \in X^*$. Hence, we can apply Lemma B.4 to $T^*$ on the spaces $X^*$ and $\mathcal{Z}^* \subseteq X^*$.

(i): If $\mathcal{N} \neq \{0\}$, then (B.16) yields $\sigma(T_{\chi^*}^*) = \sigma(T_{\mathcal{Z}^*}^*) \cup \{0\}$, and thus $\sigma(T_\chi) = \sigma(T_\mathcal{Z}) \cup \{0\}$ by (B.1).

(ii): By (B.1), $\lambda$ is an isolated point of $\sigma(T_{\chi^*}^*)$. Recall also that (by the argument in the proof of Corollary B.3) $\Pi_\lambda(T)^* = \Pi_\lambda(T^*)$, for any of the spaces $\mathcal{X}$ and $\mathcal{Z}$. Lemma B.4(ii) thus shows that $\Pi_\lambda(T_\chi)^*: \mathcal{X}^* \to \mathcal{Z}^*$. Hence, if $n \in \mathcal{N}$, then for any $x^* \in \mathcal{X}^*$ we have $\langle x^*, \Pi_\lambda n \rangle = \langle \Pi_\lambda^* x^*, n \rangle = 0$, and thus $\Pi_\lambda n = 0$. Hence $\Pi_\lambda\mathcal{N} = \{0\}$ as claimed.

Moreover, if $\pi: \mathcal{X} \to \mathcal{Z}$ is the quotient mapping, then $\pi^*: \mathcal{Z}^* \to \mathcal{X}^*$ is the inclusion mapping, and Lemma B.4(ii) shows also that $\Pi_\lambda(T_\chi)^*\pi^* = \pi^*\Pi_\lambda(T_\mathcal{Z})^*$. Hence, by taking adjoints,

$$ \pi\Pi_\lambda(T_\chi) = \Pi_\lambda(T_\mathcal{Z})\pi, \qquad (B.23) $$

which shows that $\Pi_\lambda(T_\chi)$ induces $\Pi_\lambda(T_\mathcal{Z})$ on $\mathcal{Z}$. Furthermore, (B.23) also implies

$$ \pi\Pi_\lambda(T_\chi)\mathcal{X} = \Pi_\lambda(T_\mathcal{Z})\pi\mathcal{X} = \Pi_\lambda(T_\mathcal{Z})\mathcal{Z}, \qquad (B.24) $$

and thus $\pi$ maps $\Pi_\lambda(T_\chi)\mathcal{X}$ onto $\Pi_\lambda(T_\mathcal{Z})\mathcal{Z}$. Moreover, $\pi$ is injective on $\Pi_\lambda(T_\chi)\mathcal{X}$, since $\pi x = 0$ for some $x \in \Pi_\lambda(T_\chi)\mathcal{X}$ means that $x \in \mathcal{N}$, and thus $x = \Pi_\lambda(T_\chi)x = 0$ as shown above.

(iii): An immediate consequence of (i) and (ii). $\square$

We end this appendix with a standard definition.

**Definition B.6.** Let $T$ be a bounded operator in a complex Banach space $B$. Let $r(T)$ denote the spectral radius of $T$. Furthermore, consider all decompositions $B = F \oplus H$ as a direct sum of two closed $T$-invariant subspaces such that $\dim(F) < \infty$, and define the essential spectral radius of $T$ by

$$ r_e(T) := \inf\{r(T|_H) : B = F \oplus H \text{ as above}\}. \qquad (B.25) $$
**Remark B.7.** It is easily seen that the definition (B.25) is equivalent to [19, Definition XIV.1]. There are several other, equivalent, definitions; for example, $r_e(T)$ equals the spectral radius of $T$ in the Banach algebra $B(B)/K(B)$, where $B(B)$ is the Banach algebra of bounded linear operators and $K(B)$ is the ideal of compact operators. For this, and the relation to the essential spectrum (which has several, non-equivalent, versions), see e.g. [11, §1.4] and [28, p. 243].

**Remark B.8.** Taking $F = \{0\}$ and $H = B$ in (B.25) shows that $r_e(T) \le r(T)$ for every $T$. An operator $T$ is quasi-compact if $r_e(T) < r(T)$. (See [19, Definition II.1] for another, equivalent, definition.)

## C A technical lemma

We state an elementary lemma that is used in the proof of Theorem 2.14.

**Lemma C.1.** Let $\alpha \in \mathbb{R}$ and $k \ge 0$. Then, as $n \to \infty$,

$$ \sum_{j=1}^{n} j^{-1-i\alpha} \log^k(n/j) = \begin{cases} (1+o(1)) \frac{\log^{k+1} n}{k+1} & \text{if } \alpha = 0, \\ O(\log^k n) & \text{if } \alpha \ne 0. \end{cases} \quad (C.1) $$

*Proof.* We first approximate the sum by an integral. Let $g_n(x) := x^{-1-i\alpha} \log^k(n/x)$, $x \ge 1$. Then, assuming in the sequel $n \ge 2$,

$$ g'_{n}(x) = (-1 - i\alpha)x^{-2-i\alpha} \log^{k}(n/x) - kx^{-2-i\alpha} \log^{k-1}(n/x) = O((\log^{k} n)x^{-2}), \quad x \ge 1. \quad (C.2) $$

Hence, for $j \ge 1$,

$$ g_n(j) - \int_j^{j+1} g_n(x) \, dx = \int_j^{j+1} (g_n(j) - g_n(x)) \, dx = O((\log^k n)j^{-2}). \quad (C.3) $$

Consequently, with the change of variables $x = n/y$,

$$ \begin{aligned} \sum_{j=1}^{n} j^{-1-i\alpha} \log^k(n/j) &= \sum_{j=1}^{n} g_n(j) = g_n(n) + \sum_{j=1}^{n-1} \left( \int_{j}^{j+1} g_n(x) \, dx + O((\log^k n)j^{-2}) \right) \\ &= \int_{1}^{n} g_n(x) \, dx + O(\log^k n) = \int_{1}^{n} \frac{\log^k(n/x)}{x^{1+i\alpha}} \, dx + O(\log^k n) \\ &= n^{-i\alpha} \int_{1}^{n} \frac{\log^k y}{y^{1-i\alpha}} \, dy + O(\log^k n). \end{aligned} \quad (C.4) $$

It thus suffices to consider the final integral in (C.4).

If $\alpha = 0$, then

$$ \int_{1}^{n} \frac{\log^k y}{y^{1-i\alpha}} dy = \int_{1}^{n} \frac{\log^k y}{y} dy = \left[ \frac{\log^{k+1} y}{k+1} \right]_{1}^{n} = \frac{\log^{k+1} n}{k+1}, \quad (C.5) $$

and thus (C.1) follows in this case.

If $\alpha \ne 0$, we use integration by parts and get

$$ \begin{aligned} \int_{1}^{n} \log^{k}(y)y^{i\alpha-1} dy &= \left[ \log^{k}(y) \frac{y^{i\alpha}}{i\alpha} \right]_{1}^{n} - \frac{k}{i\alpha} \int_{1}^{n} \log^{k-1}(y)y^{i\alpha-1} dy \\ &= O(\log^{k}(n)) + \int_{1}^{n} O(\log^{k-1}(n)) \frac{dy}{y} = O(\log^{k}(n)). \end{aligned} \quad (C.6) $$

Hence, (C.1) follows from (C.4) in this case too. $\square$

<figure><img src="image_3.png" /></figure>
**Remark C.2.** It is possible to show that for $\alpha \neq 0$, the sum in (C.1) is asymptotic to $\zeta(1 + i\alpha) \log^k n$. Moreover, for any $\alpha$, an asymptotic expansion with an arbitrary number of terms may be obtained by singularity analysis as in similar examples in [13, Section 3.1].

## D $L^r$ convergence

We prove here the following two theorems, showing that the $L^2$ convergence in our main theorems extends to $L^r$ convergence, for any $r \le q$, where $q$ as above is as in Assumption (H)(ii). In particular, if Assumption (H) holds with any $q > 2$, then the results below hold for any $r > 2$.

**Theorem D.1.** Under the assumptions of Theorem 2.11, (2.24) holds also in $L^r$, for any $r \le q$.

**Theorem D.2.** Under the assumptions of Theorem 2.14, (2.38) holds also in $L^r$, with $\Lambda_j \in L^r$, for any $r \le q$.

To prove these theorems, we first prove $L^r$ versions of some of our lemmas; we will see that this only requires minor modifications of the proofs above.

We assume throughout this appendix that $r$ is fixed, with $2 \le r \le q$, Constants below may depend on $r$.

We assume as above that $m_0$ is fixed; for simplicity we state the results with constants $\mathcal{C}$ that may depend on $m_0$. We leave it to the reader to find explicit dependency on $m_0$, and thus as in Remark 2.16 extensions to random $m_0$.

We let $\|X\|_r := (E|X|^r)^{1/r}$ denote the $L^r$ norm of a random variable $X$.

**Lemma D.3.** Under the assumptions of Lemma 3.8, we have also

$$ \|v_n f\|_r \le \mathcal{C} n^{-(\frac{1}{2} \wedge (1-\theta'))} \|f\|_{B(W)}, \quad f \in D', n \ge 1. \quad (D.1) $$

*Proof.* We write (3.26) as $v_n f = \sum_{i=0}^n Y_i$, where $Y_0 := v_0 B_{0,n} f$ and $Y_i := \gamma_{i-1} \Delta M_i B_{i,n} f$, $i \ge 1$. Since (3.26) is a martingale, we can apply the Burkholder inequalities [16, Theorem 10.9.5], which together with Minkowski's inequality yield

$$ \begin{aligned} E[|v_n f|^r] &\le C E\left[\left(\sum_{i=0}^n |Y_i|^2\right)^{r/2}\right] = C \left\|\sum_{i=0}^n |Y_i|^2\right\|_{r/2}^{r/2} \le C \left(\sum_{i=0}^n \||Y_i|^2\|_{r/2}\right)^{r/2} \\ &= C \left(\sum_{i=0}^n (E[|Y_i|^r])^{2/r}\right)^{r/2} = C \left(\sum_{i=0}^n \|Y_i\|_r^2\right)^{r/2}. \end{aligned} \quad (D.2) $$

By (3.63),

$$ \|B_{m,n}\|_{D'} \le C \left( \frac{m_0(E) + n}{m_0(E) + (m \vee 1)} \right)^{\theta'-1} \le \mathcal{C} \left( \frac{n}{m \vee 1} \right)^{\theta'-1}. \quad (D.3) $$

Furthermore, $Y_0$ is non-random, with, using (D.3),

$$ |Y_0| \le \mathcal{C} \|B_{0,n} f\|_{B(W)} \le \mathcal{C} n^{\theta'-1} \|f\|_{B(W)}. \quad (D.4) $$

Moreover, for $i \ge 1$, using Lyapunov's inequality, (3.13) and (3.10), and (D.3) again,

$$ \|Y_i\|_r \le \|Y_i\|_q \le \frac{1}{i} \|\Delta M_i B_{i,n} f\|_q \le \mathcal{C} i^{-1} \|B_{i,n} f\|_{B(W)} \le \mathcal{C} i^{-\theta'} n^{\theta'-1} \|f\|_{B(W)} \quad (D.5) $$

By homogeneity, we may assume $\|f\|_{B(W)} = 1$, and then (D.2) and (D.4)-(D.5) yield

$$ (E[|v_n f|^r])^{2/r} \le \mathcal{C} n^{2(\theta'-1)} + \mathcal{C} \sum_{i=1}^n i^{-2\theta'} n^{2(\theta'-1)} \le \mathcal{C} \begin{cases} n^{-1} & \text{if } \theta' < 1/2 \\ n^{-1} \log n & \text{if } \theta' = 1/2 \\ n^{2\theta'-2} & \text{if } \theta' > 1/2. \end{cases} \quad (D.6) $$
This yields (D.1) when $\theta' \neq 1/2$. If $\theta' = 1/2$, then, as in the proof of Lemma 3.8, we replace $\theta'$ by some new $\theta' < 1/2$. $\square$

**Lemma D.4.** Under the assumptions of Lemma 3.11(i), (3.85) holds also in $L^r$.

*Proof.* Let $\theta := \sup \Re\sigma(\mathbf{R}_{D'})$ and choose $\theta'' \in (\theta \vee 1/2, \theta')$. Then Lemma D.3 applied with $\theta''$ yields, for any $f \in D'$,

$$ E [|n^{-\theta'} \mathfrak{v}_n f|^r] \le \mathfrak{C} n^{r(1-\theta')-r(1-\theta'')} \|f\|_{B(W)}^r = \mathfrak{C} n^{r(\theta''-\theta')} \|f\|_{B(W)}^r = o(1). \quad (D.7) $$

This shows the convergence (3.85) in $L^r$. $\square$

**Lemma D.5.** Under the assumptions of Lemma 4.2, the convergence (4.42) holds also in $L^r$ as $n \to \infty$. Moreover,

$$ \|\Lambda_f\|_r \le \mathfrak{C} \|f\|_{B(W)}, \quad (D.8) $$

and for any $0 < \varepsilon < \Re\lambda - 1/2$, there exists a constant $\mathfrak{C}_\varepsilon > 0$ such that

$$ \|\mathfrak{v}_n B_{0,n}^{-1} f - \Lambda_f\|_r \le \frac{\mathfrak{C}_\varepsilon}{n^{\Re\lambda - \varepsilon - 1/2}} \|f\|_{B(W)}. \quad (D.9) $$

*Proof.* Let $f \in D'$, and write (4.48) as $\mathfrak{v}_n B_{0,n}^{-1} f = \sum_{i=0}^n Y_i$, where $Y_0 := \mathfrak{v}_0 f$ and $Y_i := \gamma_{i-1} \Delta M_i B_{0,i}^{-1} f$ for $i \ge 1$. Note that $Y_0$ is non-random, and that (3.11) implies

$$ \|Y_0\|_r = |\mathfrak{v}_0 f| \le \mathfrak{C} \|f\|_{B(W)}. \quad (D.10) $$

For $i \ge 1$, we have, by (3.13) and (3.10),

$$ E |Y_i|^q = E [|\gamma_{i-1} \Delta M_i B_{0,i}^{-1} f|^q] \le \mathfrak{C} \gamma_{i-1}^q \|B_{0,i}^{-1} f\|_{B(W)}^q \quad (D.11) $$

and thus, using (4.51),

$$ \|Y_i\|_r \le \|Y_i\|_q \le \mathfrak{C} \gamma_{i-1} \|B_{0,i}^{-1} f\|_{B(W)} \le \mathfrak{C} i^{-1} i^{1-\Re\lambda+\varepsilon} \|f\|_{B(W)} = \mathfrak{C} i^{-(\Re\lambda-\varepsilon)} \|f\|_{B(W)}. \quad (D.12) $$

Since $2(\Re\lambda - \varepsilon) > 1$, it follows from (4.48) and Burkholder's inequality as in (D.2), now using (D.10) and (D.12), that

$$ E [|\mathfrak{v}_n B_{0,n}^{-1} f|^r] \le C \left( \sum_{i=0}^n \|Y_i\|_r^2 \right)^{r/2} \le \mathfrak{C} \|f\|_{B(W)}^r. \quad (D.13) $$

Consequently, the martingale (4.48) is bounded in $L^r$, which as is well known implies that it converges in $L^r$; hence (4.42) holds in $L^r$ as claimed.

Furthermore, this and (D.13) yield (D.8).

Similarly, by Burkholder's inequality and (D.12) again,

$$ E [|\mathfrak{v}_n B_{0,n}^{-1} f - \Lambda_f|^r] = E \left[ \left| \sum_{i=n+1}^{\infty} Y_i \right|^r \right] \le C \left( \sum_{i=n+1}^{\infty} \|Y_i\|_r^2 \right)^{r/2} \le \mathfrak{C} n^{\frac{r}{2}(1-2(\Re\lambda - \varepsilon))} \|f\|_{B(W)}^r, \quad (D.14) $$

which yields (D.9). $\square$

**Lemma D.6.** Under the assumptions of Lemma 4.4, (4.57) holds also in $L^r$.
Proof. Now, (D.13) implies that for every fixed $k \ge 0$,

$$ \|\mathfrak{v}_n B_{0,n}^{-1} f_k\|_r \le \mathfrak{C} \|f_k\|_{B(W)} \le \mathfrak{C} \|f\|_{B(W)}. \quad (D.15) $$

Moreover, taking $L^r$ norms in (4.67), we deduce using Minkowski's inequality, (D.15) and (D.9),

$$ \left\| \left( \frac{\mathfrak{m}_0(E) + n}{\mathfrak{m}_0(E) + 1} \right)^{1-\lambda} \frac{\mathfrak{v}_n f}{1 \vee \log^{\kappa-1} \left( \frac{\mathfrak{m}_0(E)+n}{\mathfrak{m}_0(E)+1} \right)} - (1+h_{0,n}(\lambda)) \Lambda_{f_{\kappa-1}} \right\|_r \le \frac{\mathfrak{C} \|f\|_{B(W)}}{1 \vee \log \left( \frac{\mathfrak{m}_0(E)+n}{\mathfrak{m}_0(E)+1} \right)}. \quad (D.16) $$

Furthermore, by (4.68), (4.70) and (D.8),

$$ \left\| (1 + h_{0,n}(\lambda)) \Lambda_{f_{\kappa-1}} - \frac{1}{(m_0(E) + 1)^{1-\lambda}} \Lambda \right\|_r \le \frac{C}{n} \|\Lambda_{f_{\kappa-1}}\|_r \le \frac{\mathfrak{C}}{n} \|f\|_{B(W)}. \quad (D.17) $$

Combining (4.72) and (4.73), we obtain

$$ \left\| (\mathfrak{m}_0(E) + n)^{1-\lambda} \frac{\mathfrak{v}_n f}{1 \vee \log^{\kappa-1} \left( \frac{\mathfrak{m}_0(E)+n}{\mathfrak{m}_0(E)+1} \right)} - \Lambda \right\|_r \le \mathfrak{C} \frac{1}{1 \vee \log \left( \frac{\mathfrak{m}_0(E)+n}{\mathfrak{m}_0(E)+1} \right)} \|f\|_{B(W)}. \quad (D.18) $$

Hence, (4.69) holds also in $L^r$, and thus so does (4.57). □

Proof of Theorem D.1. Let, as in the proof of Lemma 3.11(ii), $D' := (1 - \Pi_1)D$ and apply Lemma D.4 to $D'$ and $f - (\nu f)1 \in D'$ with $\theta' := 1 - \delta > \theta_D = \sup \Re \sigma(\mathbf{R}_{D'})$. □

Proof of Theorem D.2. This follows exactly as in the proof of Theorem 2.14(3) (after Lemma 4.4 in Section 4.2), now using Lemmas D.6 and D.4 instead of Lemmas 4.4 and 3.11. □

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