Multiplication Operators between Lipschitz-Type Spaces on a Tree

Theorem 2.1. For a function ψ on T, the following statements are equivalent:

(a)  M_ψ : ℒ_w → ℒ is bounded.

(b)  M_ψ : ℒ_w,0 → ℒ₀ is bounded.

(c)  τ_ψ and σ_ψ are finite.

Furthermore, under these conditions, we have

Proof. (a)⇒(c) Assume M_ψ : ℒ_w → ℒ is bounded. Applying M_ψ to the constant function 1, we have ψ ∈ ℒ, so that, by Lemma 1.1, we have σ_ψ < ∞. Next, consider the function f on T defined by f(v) = log (1+|v|). Then f(o) = 0; for v ≠ o, a straightforward calculation shows that

and lim _|v|→∞ |v | Df(v) = 1. Thus, ∥f∥_w = 1 and so ∥M_ψf∥_ℒ ≤ ∥M_ψ∥. Therefore, for v ∈ T^*, noting that one has Hence τ_ψ < ∞.

(c)⇒(a) Assume τ_ψ and σ_ψ are finite. Then, by Lemma 1.3, for f ∈ ℒ_w and v ∈ T^*, we have

Since |ψ(o)| ≤ σ_ψ, we obtain proving the boundedness of M_ψ : ℒ_w → ℒ and the upper estimate.

(b)⇒(c) Suppose M_ψ : ℒ_w,0 → ℒ₀ is bounded. The finiteness of σ_ψ follows again from the fact that ψ = M_ψ1 ∈ ℒ₀ and from Lemma 1.1. To prove that τ_ψ < ∞, let 0 < α < 1 and, for v ∈ T, define f_α(v) = (log (1+|v|))^α. Then f_α(o) = 0 and |v | Df_a(v) → 0 as |v | → ∞; so f_α ∈ ℒ_w,0. Since for 0 < α < 1, the function x ↦ x − x^α is increasing for x ≥ 1, the function Df_α(v) is increasing in α, and Df_α(v) ≤ Df(v) for v ∈ T^*, where f(v) = log (1+|v|), for v ∈ T. Thus, ∥f_α∥_w ≤ ∥f∥_w = 1. Therefore, for v ∈ T^*, we have

Letting α → 1, we obtain Hence τ_ψ < ∞.

(c)⇒(b) Assume σ_ψ and τ_ψ are finite, and let f ∈ ℒ_w,0. Then, by Lemma 1.3, for v ∈ T^*, we have

as |v | → ∞. Thus, ψf ∈ ℒ₀. The boundedness of M_ψ and the estimate ∥M_ψf∥_ℒ ≤ τ_ψ + σ_ψ can be shown as in the proof of (c)⇒(a).

Finally we show that, under boundedness assumptions on M_ψ, ∥M_ψ∥≥max {τ_ψ, σ_ψ}. For v ∈ T^*, let f_v = 1/(|v | + 1)χ_v, where χ_v denotes the characteristic function of {v}. Then ∥f_v∥_w = 1 and

Furthermore, letting f_o = (1/2)χ_o, we see that $$ and ∥ψf_o∥_ℒ = |ψ(o)|. Therefore, we deduce that ∥M_ψ∥≥σ_ψ.

Next, fix v ∈ T^* and for w ∈ T, define

Then, g_v ∈ ℒ_w and Observe that, for w ∈ T^*, we have Hence Define f_v = g_v/∥g_v∥_w. Then ∥f_v∥_w = 1 and Taking the limit as |v | → ∞, we obtain ∥M_ψ∥≥τ_ψ. Therefore, ∥M_ψ∥≥max {τ_ψ, σ_ψ}.

Theorem 2.2. There are no isometries M_ψ from ℒ_w to ℒ or ℒ_w,0 to ℒ₀, respectively.

Lemma 2.3. A bounded multiplication operator M_ψ from ℒ_w to ℒ is compact if and only if for every bounded sequence {f_n} in ℒ_w converging to 0 pointwise, the sequence {∥ψf_n∥_ℒ} → 0 as n → ∞.

Proof. Assume M_ψ is compact, and let {f_n} be a bounded sequence in ℒ_w converging to 0 pointwise. Without loss of generality, we may assume ∥f_n∥_w ≤ 1 for all n ∈ ℕ. Then the sequence {M_ψf_n} = {ψf_n} has a subsequence $$ which converges in the ℒ-norm to some function f ∈ ℒ. Clearly $$, and by part (a) of Lemma 1.1, for v ∈ T^*, we have

Thus, $$ pointwise on T. Since f_n → 0 pointwise, it follows that f must be identically 0, which implies that $$. With 0 being the only limit point of {ψf_n} in ℒ, it follows that $$ as n → ∞.

Conversely, assume every bounded sequence {f_n} in ℒ_w converging to 0 pointwise has the property that $$ as n → ∞. Let {g_n} be a sequence in ℒ_w with $$ for all n ∈ ℕ. Then |g_n(o)| ≤ 1 for all n ∈ ℕ, and by part (a) of Lemma 1.2, for v ∈ T^*, we obtain

Thus, {g_n} is uniformly bounded on finite subsets of T. So some subsequence $$ converges pointwise to some function g. Fix v ∈ T^* and ɛ > 0. Then for k sufficiently large, we have We deduce for all k sufficiently large. So g ∈ ℒ_w. The sequence defined by $$ is bounded in ℒ_w and converges to 0 pointwise. Thus by hypothesis, we obtain $$ as k → ∞. It follows that $$ in the ℒ-norm, thus proving the compactness of M_ψ.

Lemma 2.4. A bounded multiplication operator M_ψ from ℒ_w,0 to ℒ₀ is compact if and only if for every bounded sequence {f_n} in ℒ_w,0 converging to 0 pointwise, the sequence {∥ψf_n∥_ℒ} → 0 as n → ∞.

Lemma 2.5. For f ∈ ℒ_w and  v ∈ T

where $$.

Proof. Fix v ∈ T and argue by induction on n = |v|. For n = 0, inequality (2.22) is obvious. So assume |v | = n > 0 and |f(u)|≤|f(o)| + 2log (1+|u|)s_w(f) for all vertices u such that |u | < n. Then

Next, observe that 1/(|v | + 1)≤log ((|v | + 1)/|v|), so Hence Inequality (2.22) now follows immediately from (2.23) and (2.25).

Theorem 2.6. Let M_ψ be a bounded multiplication operator from ℒ_w to ℒ (or equivalently from ℒ_w,0 to ℒ₀). Then the following statements are equivalent:

(a)  M_ψ : ℒ_w → ℒ is compact.

(b)  M_ψ : ℒ_w,0 → ℒ₀ is compact.

(c)  lim _|v|→∞ | ψ(v)|/(|v | + 1) = 0 and lim _|v|→∞Dψ(v)log  | v | = 0.

Proof. We first prove (a)⇒(c). Assume M_ψ : ℒ_w → ℒ is compact. It suffices to show that, for any sequence {v_n} in T such that 2≤|v_n | → ∞, we have lim _n→∞ | ψ(v_n)|/(|v_n | + 1) = 0 and lim _n→∞Dψ(v_n)log  | v_n | = 0. Let {v_n} be such a sequence, and for each n ∈ ℕ, define $$. Then f_n(o) = 0, f_n → 0 pointwise as n → ∞, and ∥f_n∥_w = 1. By Lemma 2.3, it follows that ∥ψf_n∥_ℒ → 0 as n → ∞. Furthermore

Thus lim _n→∞ |ψ(v_n)|/(|v_n | + 1) = 0.

Next, for each n ∈ ℕ and v ∈ T, define

Then Dg_n(v) = 0   if $$ or |v | >|v_n | − 1. In addition, if $$, then |v | Dg_n(v) < 4. Indeed, there are two possibilities. Either $$, in which case or $$, in which case Thus {∥g_n∥_w} is bounded, and {g_n} converges to 0 pointwise. By Lemma 2.3, it follows that ∥ψg_n∥_ℒ → 0 as n → ∞. Moreover Therefore lim _n→∞  Dψ(v_n)log |v_n| = 0.

To prove the implication (c)⇒(a), suppose lim _|v|→∞Dψ(v)log  | v | = 0 and lim _|v|→∞ | ψ(v)|/(|v | + 1) = 0. Clearly, if ψ is identically 0, then M_ψ is compact. So assume M_ψ : ℒ_w → ℒ is bounded with ψ not identically 0. By Lemma 2.3, it suffices to show that if {f_n} is bounded in ℒ_w converging to 0 pointwise, then ∥ψf_n∥_ℒ → 0 as n → ∞. Let {f_n} be such a sequence, $$, and fix ɛ > 0. Note that

Thus there exists an M ∈ ℕ such that for |v | ≥ M. Using Lemma 2.5, for |v| > M, we have

On the other hand, on the set B_M = {v ∈ T:|v | ≤ M}, {f_n} converges to 0 uniformly, and thus Df_n does as well. Moreover

uniformly on B_M. Therefore D(ψf_n) → 0 uniformly on T. Furthermore, the sequence {(ψf_n)(o)} converges to 0 as n → ∞. Hence ∥ψf_n∥_ℒ → 0 as n → ∞, proving that M_ψ is compact.

Finally, note that the functions f_n and g_n defined in the proof of (a)⇒(c) are in ℒ_w,0. So the equivalence of (b) and (c) is proved analogously.

Theorem 2.7. Let M_ψ be a bounded multiplication operator from ℒ_w to ℒ. Then

Proof. For each n ∈ ℕ, define f_n = (1/(n + 1))χ_n, where χ_n denotes the characteristic function of the set {v ∈ T:|v | = n}. Then f_n ∈ ℒ_w,0, ∥f_n∥_w = 1, and f_n → 0 pointwise. Thus, by Lemma 1.4, {f_n} converges to 0 weakly in ℒ_w,0. Let 𝒦 be the set of compact operators from ℒ_w,0 to ℒ₀, and let K ∈ 𝒦. Then K is completely continuous [8], and so ∥Kf_n∥_ℒ → 0 as n → ∞. Thus

Now note that Hence We will now show that $$. This estimate is clearly true if B(ψ) = 0. So assume {v_n} is a sequence in T such that 2≤|v_n | → ∞ as n → ∞ and For n ∈ ℕ and v ∈ T, define Then h_n(o) = 0, $$, and

The supremum of |v | Dh_n(v) is attained at the vertices of length |v_n | − 1 and is given by

Since (|v_n| − 1)log (|v_n|/(|v_n| − 1)) ≤ 1, we have

By letting $$, we have g_n ∈ ℒ_w,0, ∥g_n∥_w = 1, and g_n → 0 pointwise. By Lemma 1.4, the sequence {g_n} converges to 0 weakly in ℒ_w,0. Thus ∥Kg_n∥_ℒ → 0 as n → ∞. Therefore,

For each n ∈ ℕ, we have $$. So

Since lim _n→∞s_n = 1, we have Therefore, ∥M_ψ∥_e ≥ max {A(ψ), B(ψ)}.

Theorem 2.8. Let M_ψ be a bounded multiplication operator from ℒ_w to ℒ. Then

Proof. For n ∈ ℕ, define the operator K_n on ℒ_w by

where f ∈ ℒ_w and v_n is the ancestor of v of length n. For f ∈ ℒ_w, (K_nf)(o) = f(o), and K_nf ∈ ℒ_w,0. Let B_n = {v ∈ T : |v| ≤ n}, and note that K_nf attains finitely many values, whose number does not exceed the cardinality of B_n. Let {g_k} be a sequence in ℒ_w such that ∥g_k∥_w ≤ 1 for each k ∈ ℕ. Then a = sup _k∈ℕ | g_k(o)| ≤ 1, and |K_ng_k(o)| ≤ a. Furthermore, by part (a) of Lemma 1.3, for each v ∈ T^* and for each k ∈ ℕ, we have |K_ng_k(v)| ≤ 1 + log n. Thus, some subsequence of {K_ng_k} _k∈ℕ must converge to a function g on T attaining constant values on the sectors determined by the vertices of length n. It follows that this subsequence converges to g in ℒ_w as well, proving that K_n is a compact operator on ℒ_w. Since M_ψ is bounded as an operator from ℒ_w to ℒ, it follows that M_ψK_n : ℒ_w → ℒ is compact for all n ∈ ℕ.

Define the operator J_n = I − K_n, where I denotes the identity operator on ℒ_w. Then J_nf(o) = 0, and for v ∈ T^*, we have

By part (a) of Lemma 1.3, we see that

Using (2.51) and (2.52), we obtain

Since taking the limit as n → ∞, we obtain as desired.

Theorem 3.1. For a function ψ on T, the following statements are equivalent:

(a)  M_ψ : ℒ → ℒ_w is bounded.

(b)  M_ψ : ℒ₀ → ℒ_w,0 is bounded.

(c)  θ_ψ and ω_ψ are finite.

Furthermore, under the above conditions, one has

Proof. (a)⇒(c) Assume M_ψ is bounded from ℒ to ℒ_w. The function f_o = (1/2)χ_o ∈ ℒ and $$. Thus

Next, fix v ∈ T^*. Then χ_v ∈ ℒ and ∥χ_v∥_ℒ = 1; so

Taking the supremum over all v ∈ T, from (3.3) and (3.4) we see that ω_ψ is finite and

With v ∈ T^*, we now define

Then f_v ∈ ℒ, f_v(o) = 0 and $$. By the boundedness of M_ψ we obtain Therefore, Taking the supremum over all v ∈ T^*, we obtain θ_ψ < ∞. From this and (3.5), we deduce the lower estimate

(c)⇒(a) Assume θ_ψ and ω_ψ are finite. Then, ψ ∈ ℒ_w, and by Lemma 1.1, for f ∈ ℒ with ∥f∥_ℒ = 1 and v ∈ T^*, we have

Thus, ψf ∈ ℒ_w. Note that |f(o)| + ∥Df∥_∞ = 1 and From this, we have proving the boundedness of M_ψ : ℒ → ℒ_w and the upper estimate

(b)⇒(c) The proof is the same as for (a)⇒(c); since for v ∈ T^*, the functions χ_v and f_v used there belong to ℒ₀.

(c)⇒(b) Assume θ_ψ and ω_ψ are finite, and let f ∈ ℒ₀. Then, by Lemma 1.1, for v ∈ T^*, we have

as |v | → ∞. Thus, ψf ∈ ℒ_w,0. The proof of the boundedness of M_ψ is similar to that in (c)⇒(a).

Theorem 3.2. There are no isometries M_ψ from ℒ to ℒ_w or from ℒ₀ to ℒ_w,0.

Lemma 3.3. A bounded multiplication operator M_ψ from ℒ to ℒ_w (ℒ₀ to ℒ_w,0) is compact if and only if for every bounded sequence {f_n} in ℒ (ℒ₀) converging to 0 pointwise, the sequence $$ converges to 0 as n → ∞.

Proof. Suppose M_ψ is compact from ℒ to ℒ_w and {f_n} is a bounded sequence in ℒ converging to 0 pointwise. Without loss of generality, we may assume $$ for all n ∈ ℕ. Since M_ψ is compact, the sequence {ψf_n} has a subsequence $$ that converges in the ℒ_w-norm to some function f ∈ ℒ_w.

By Lemma 1.3, for v ∈ T^* we have

Thus, $$ pointwise on T^*. Furthermore, since $$, $$ as k → ∞. Thus $$ pointwise on T. Since by assumption, f_n → 0 pointwise, it follows that f is identically 0, and thus $$. Since 0 is the only limit point in ℒ_w of the sequence {ψf_n}, we deduce that $$ as n → ∞.

Conversely, suppose that every bounded sequence {f_n} in ℒ that converges to 0 pointwise has the property that $$ as n → ∞. Let {g_n} be a sequence in ℒ such that $$ for all n ∈ ℕ. Then |g_n(o)| ≤ 1, and by part (a) of Lemma 1.1, for v ∈ T^* we have |g_n(v)|≤|v|. So {g_n} is uniformly bounded on finite subsets of T. Thus there is a subsequence $$, which converges pointwise to some function g.

Fix ɛ > 0 and v ∈ T^*. Then $$ as well as $$ for k sufficiently large. Therefore, for all k sufficiently large, we have

Thus g ∈ ℒ. The sequence $$ is bounded in ℒ and converges to 0 pointwise. So $$ as k → ∞. Thus $$ in the ℒ_w-norm. Therefore, M_ψ is compact.

The proof for the case of M_ψ : ℒ₀ → ℒ_w,0 is similar.

Theorem 3.4. Let M_ψ be a bounded multiplication operator from ℒ to ℒ_w (or equivalently from ℒ₀ to ℒ_w,0). Then the following are equivalent:

(a)  M_ψ : ℒ → ℒ_w is compact.

(b)  M_ψ : ℒ₀ → ℒ_w,0 is compact.

(c)  lim _|v|→∞ | v|²Dψ(v) = 0 and lim _|v|→∞(|v | + 1) | ψ(v)| = 0.

Proof. (a)⇒(c) Suppose M_ψ : ℒ → ℒ_w is compact. We need to show that if {v_n} is a sequence in T such that 2≤|v_n| increasing unboundedly, then $$ and lim _n→∞(|v_n | + 1) | ψ(v_n)| = 0. Let {v_n} be such a sequence, and for n ∈ ℕ define $$. Clearly f_n → 0 pointwise, and $$. Using Lemma 3.3, we see that

On the other hand, since f_n(o) = 0 for all n ∈ ℕ, we have

Hence lim _n→∞(|v_n | + 1) | ψ(v_n)| = 0.

Next, for n ∈ ℕ, define

where ⌊x⌋ denotes the largest integer less than or equal to x. Then g_n → 0 pointwise, and $$. Since $$, we have By Lemma 3.3 we obtain lim _n→∞ | v_n|²Dψ(v_n) ≤ lim _n→∞∥ψg_n∥_w = 0.

(c)⇒(a) Suppose lim _|v|→∞|v|²Dψ(v) = 0 and lim _|v|→∞(|v| + 1)|ψ(v)| = 0. Assume ψ is not identically zero, otherwise M_ψ is trivially compact. By Lemma 3.3, to prove that M_ψ is compact, it suffices to show that if {f_n} is a bounded sequence in ℒ converging to 0 pointwise, then $$ as n → ∞. Let {f_n} be such a bounded sequence, $$, and fix ɛ > 0. There exists M ∈ ℕ such that (|v| + 1)|ψ(v)| < ɛ/2s and |v|²Dψ(v) < ɛ/2s for |v | ≥ M. For v ∈ T^* and by Lemma 1.1, we have

Since f_n → 0 uniformly on {v ∈ T:|v | ≤ M} as n → ∞, so does Df_n. So, on the set {v ∈ T:|v | ≤ M}, |v | D(ψf_n)(v) → 0 as n → ∞. On the other hand, on {v ∈ T:|v | ≥ M}, we have So |v | D(ψf_n)(v) → 0 as n → ∞. Since f_n → 0 pointwise, ψ(o)f_n(o) → 0 as n → ∞. Thus $$ as n → ∞. The compactness of M_ψ follows at once from Lemma 3.3.

The proof of the equivalence of (b) and (c) is analogous.

Theorem 3.5. Let M_ψ be a bounded multiplication operator from ℒ to ℒ_w. Then

Proof. Fix k ∈ ℕ, and for each n ∈ ℕ, consider the sets

Define the functions $$ and $$. Then f_n,k, g_n,k ∈ ℒ₀, $$, and f_n,k and g_n,k approach 0 pointwise as n → ∞. By Lemma 1.2, the sequences {f_n,k} and {g_n,k} approach 0 weakly in ℒ₀ as n → ∞. Let 𝒦₀ be the set of compact operators from ℒ₀ to ℒ_w,0, and note that every operator in 𝒦₀ is completely continuous. Thus, if K ∈ 𝒦₀, then $$ and $$, as n → ∞.

Therefore, if K ∈ 𝒦₀, then

Similarly,

Therefore, combining (3.28) and (3.29), we obtain Letting k → ∞, we obtain $$.

Next, we wish to show that $$. The result is clearly true if ℬ(ψ) = 0. So assume there exists a sequence {v_n} in T such that 2<|v_n | → ∞ as n → ∞ and

For n ∈ ℕ, define Clearly, $$, and The supremum of Dh_n(v) is attained on the set {v ∈ T : |v| = |v_n| − 1}. Thus $$. Define $$, and observe that g_n ∈ ℒ₀, ∥g_n∥_ℒ = 1, and g_n → 0 pointwise on T. By Lemma 1.2, g_n → 0 weakly in ℒ₀. Thus $$ as n → ∞ for any K ∈ 𝒦₀.

For each n ∈ ℕ, $$. Thus

We deduce that

Therefore,

Theorem 3.6. Let M_ψ be a bounded multiplication operator from ℒ to ℒ_w. Then

Proof. For each n ∈ ℕ, consider the operator K_n defined by

for f ∈ ℒ, where v_n is the ancestor of v of length n. Then (K_nf)(o) = f(o), and K_nf ∈ ℒ_w,0. Arguing as in the proof of Theorem 2.8, by the boundedness of M_ψ, it follows that M_ψK_n is a compact operator from ℒ to ℒ_w.

Define the operator J_n = I − K_n, where I is the identity operator I on ℒ. Then,

Since (J_nf)(v) = 0 for |v | ≤ n, by Lemma 1.1, we obtain From these two estimates, we arrive at Since from (3.41), taking the limit as n → ∞, we obtain as desired.

Theorem 4.1. For a function ψ on T, the following statements are equivalent:

(a)  M_ψ : ℒ_w → L^∞ is bounded.

(b)  M_ψ : ℒ_w,0 → L^∞ is bounded.

(c)$$ is finite.

Furthermore, under the above conditions, one has ∥M_ψ∥ = γ_ψ.

Proof. The implication (a)⇒(b) is obvious.

(b)⇒(a) We begin by showing that, for each f ∈ ℒ_w, the function ψf is bounded. Since M_ψ is bounded on ℒ_w,0, ψ = M_ψ1 ∈ L^∞. Thus, if f is constant, then ψf ∈ L^∞. Fix f ∈ ℒ_w, f nonconstant, v ∈ T, and set n = |v|. For w ∈ T, define

where w_n is the ancestor of w of length n. Then f_n ∈ ℒ_w,0, and $$. Thus, ψf_n ∈ L^∞ and So |ψ(v)f(v)| = |ψ(v)f_n(v)| ≤ ∥M_ψ∥∥f∥_w. Therefore, ψf ∈ L^∞ and proving the boundedness of M_ψ as an operator from ℒ_w to L^∞.

(a)⇒(c) Assume M_ψ : ℒ_w → L^∞ is bounded. Then ψ = M_ψ1 ∈ L^∞, and

For v ∈ T, define f(v) = log (1+|v|). Then f(o) = 0, and since for x ≥ 1 the function x ↦ xlog ((x + 1)/x) is increasing and has limit 1 as x → ∞, f ∈ ℒ_w and ∥f∥_w = 1. Thus proving (c). Furthermore, from (4.5) and (4.6), we obtain

(c)⇒(a) Assume $$. Let f ∈ ℒ_w such that ∥f∥_w = 1. Then |ψ(o)f(o)| ≤ |ψ(o)|, and by Lemma 1.3, for v ∈ T^*, we have

Thus, ψf ∈ L^∞ and proving the boundedness of M_ψ as an operator from ℒ_w to L^∞. Taking the supremum over all functions f ∈ ℒ_w such that ∥f∥_w = 1, from (4.9) we obtain ∥M_ψ∥ ≤ γ_ψ. Therefore, from (4.7) we conclude that ∥M_ψ∥ = γ_ψ.

Theorem 4.2. A bounded multiplication operator M_ψ from ℒ_w or ℒ_w,0 to L^∞ is bounded below if and only if

Proof. Assume M_ψ is bounded below, and, arguing by contradiction, assume there exists v ∈ T such that ψ(v) = 0. Then M_ψχ_v is identically 0. Since operators that are bounded below are necessarily injective [8], it follows that M_ψ is not bounded below. Therefore, if M_ψ is bounded below, then ψ is nonvanishing.

Next assume ψ is nonvanishing and inf _v∈T |ψ(v)|/(|v | + 1) = 0. Then, there exists a sequence {v_n} in T with 1≤|v_n | → ∞, such that |ψ(v_n)|/(|v_n | + 1)→0 as n → ∞. For n ∈ ℕ, define $$. Then ∥f_n∥_w = 1, but

Thus, M_ψ is not bounded below.

Conversely, assume inf _v∈T |ψ(v)|/(|v| + 1) = c > 0 and that M_ψ is not bounded below. Then, for each n ∈ ℕ, there exists f_n ∈ ℒ_w such that ∥f_n∥_w = 1 and ∥ψf_n∥_∞ < 1/n. Then, for each v ∈ T, we have

so that the sequence {g_n} defined by g_n(v) = (|v | + 1)f_n(v) converges to 0 uniformly.

On the other hand, for v ∈ T^*, we have

uniformly as n → ∞. Since |ψ(o)f_n(o)| < 1/n, yet ∥f_n∥_w = 1, this yields a contradiction.

Theorem 4.3. There are no isometric multiplication operators M_ψ from ℒ_w or ℒ_w,0 to L^∞.

Lemma 4.4. A bounded multiplication operator M_ψ from ℒ_w to L^∞ is compact if and only if for every bounded sequence {f_n} in ℒ_w converging to 0 pointwise, the sequence ∥ψf_n∥_∞ approaches 0 as n → ∞.

Proof. Assume M_ψ is compact on ℒ_w, and let {f_n} be a bounded sequence in ℒ_w converging to 0 pointwise. By rescaling the sequence, if necessary, we may assume ∥f_n∥_w ≤ 1 for all n ∈ ℕ. By the compactness of M_ψ, {f_n} has a subsequence $$ such that $$ converges in the norm to some function f ∈ L^∞. In particular, $$ pointwise. Since by assumption, f_n → 0 pointwise, it follows that f must be identically 0. Thus, the only limit point of sequence {ψf_n} in L^∞ is 0. Hence ∥ψf_n∥_∞ → 0.

Conversely, assume that for every bounded sequence {f_n} in ℒ_w converging to 0 pointwise, the sequence $$ approaches 0 as n → ∞. Let {g_n} be a sequence in ℒ_w with ∥g_n∥_w ≤ 1. Fix w ∈ T, and, by replacing g_n with g_n − g_n(w), assume g_n(w) = 0 for all n ∈ ℕ. Then, for each v ∈ T, |g_n(v)| = |g_n(v) − g_n(w)| ≤ d(v, w). Therefore, g_n is uniformly bounded on finite subsets of T, and so some subsequence $$ converges pointwise to some function g on T. Fix ɛ > 0 and v ∈ T^*. Then, $$, $$, and $$ for all k sufficiently large. Thus,

for k sufficiently large. Consequently, g ∈ ℒ_w, we have Since ɛ was arbitrary, it follows that ∥g∥_w ≤ 1. Therefore, the sequence {f_k} defined by $$ is bounded in ℒ_w and converges to 0 pointwise, hence, by the hypothesis, $$ as n → ∞. We conclude that $$ in L^∞, proving the compactness of M_ψ.

Lemma 4.5. A bounded multiplication operator M_ψ from ℒ_w,0 to L^∞ is compact if and only if for every bounded sequence {f_n} in ℒ_w,0 converging to 0 pointwise, the sequence ∥ψf_n∥_∞ approaches 0 as n → ∞.

Theorem 4.6. For a bounded operator M_ψ from ℒ_w to L^∞ (or equivalently from ℒ_w,0 to L^∞) the following statements are equivalent:

(a)  M_ψ : ℒ_w → L^∞ is compact.

(b)  M_ψ : ℒ_w,0 → L^∞ is compact.

(c)  lim _|v|→∞ log |v||ψ(v)| = 0.

Proof. (a)⇒(b) is trivial.

(b)⇒(c): Let {v_n} be a sequence of vertices such that 1≤|v_n | → ∞ as n → ∞. We need to show that

For n ∈ ℕ define Then {f_n} converges to 0 pointwise. Using the fact that (|v | − 1)(log  | v | − log (|v | − 1)) ≤ 1 for any choice of v in T^* with |v | > 1, we have for 2≤|v | ≤|v_n|. Moreover, |v | Df_n(v) = 0 for |v | = 1 and for |v | >|v_n|. Thus, f_n ∈ ℒ_w,0, and $$ is bounded. By the compactness of M_ψ as an operator from ℒ_w,0 to L^∞ and by Lemma 4.5, we deduce as n → ∞.

(c)⇒(a) Assume {f_n} is a sequence in ℒ_w converging to 0 pointwise and such that a = sup _n∈ℕ ∥f_n∥_w < ∞. By Lemma 1.3, for all v ∈ T^* and all n ∈ ℕ, we have

Fix ɛ > 0. There exists N ∈ ℕ such that N ≥ 3, and for |v | ≥ N, log  |v||ψ(v)| < ɛ/2a. Thus, for |v | ≥ N and for all n ∈ ℕ, |ψ(v)f_n(v)| ≤ 2alog |v||ψ(v)| < ɛ. On the other hand, since f_n → 0 pointwise, for each vertex v such that |v | < N and ψ(v) ≠ 0, we obtain |f_n(v)| < ɛ/|ψ(v)| for all n sufficiently large. Hence |ψ(v)f_n(v)| < ɛ for all v ∈ T and all n sufficiently large. Therefore, $$ as n → ∞, which, by Lemma 4.4, proves the compactness of M_ψ.

Theorem 4.7. Let M_ψ be a bounded multiplication operator from ℒ_w or ℒ_w,0 to L^∞. Then

Proof. Define A(ψ) = lim _n→∞  sup _|v|≥n log |v||ψ(v)|. If A(ψ) = 0, then by Theorem 4.6, M_ψ is compact, hence its essential norm is 0. So assume A(ψ) > 0. We first show that $$. Let {v_n} be a sequence in T such that 1≤|v_n | → ∞ as n → ∞ and

Fix p ∈ (0,1), and for each n ∈ ℕ, define Then {f_n,p} converges to 0 pointwise, f_n,p ∈ ℒ_w,0, f_n,p(v_n) = log  | v_n|, and By Lemma 1.4, {f_n,p} converges to 0 weakly in ℒ_w,0. Let K be a compact operator from ℒ_w,0 (or equivalently, from ℒ_w) to L^∞. Since compact operators are completely continuous, it follows that $$ as n → ∞. Thus, Taking the infimum over all such compact operators K and passing to the limit as p approaches 0, we obtain

To prove the estimate $$, for each n ∈ ℕ and for f ∈ ℒ_w, define

where v_n is the ancestor of v of length n. In the proof of Theorem 2.8, it is was shown that K_n is a compact operator on ℒ_w. Since M_ψ : ℒ_w → L^∞ is bounded, it follows that M_ψK_n is also compact as an operator from ℒ_w to L^∞.

Let v ∈ T, and let w be a vertex in the path from o to v of length k ≥ 1. Label the vertices from w to v by v_j, j = k, …, |v|. Then for f ∈ ℒ_w with ∥f∥_w = 1, we have

Thus We deduce

Taking the limit as n → ∞, we obtain $$.

Theorem 5.1. For a function ψ on T, the following statements are equivalent:

(a)  M_ψ : L^∞ → ℒ_w is bounded.

(b)  sup _v∈T |v||ψ(v)| < ∞.

Furthermore, under these conditions, one has

Proof. (a)⇒(b) Assume M_ψ : L^∞ → ℒ_w is bounded. Fix v ∈ T^*. Since χ_v ∈ L^∞ and $$, the function ψχ_v ∈ ℒ_w, so

Thus, sup _v∈T | v∥ψ(v)| is finite.

(b)⇒(a) Suppose sup _v∈T|v||ψ(v)| < ∞. Let f ∈ L^∞ such that ∥f∥_∞ = 1. Then

Thus, M_ψ is bounded and ∥M_ψ∥ ≤ η_ψ.

We next show that ∥M_ψ∥≥η_ψ. The inequality is obvious if ψ is identically 0. For ψ not identically 0 and for v ∈ T, define

Then ∥f∥_∞ = 1, and for v ∈ T^*, D(ψf)(v)   = |ψ(v)|+|ψ(v⁻)|, so that Thus, ∥M_ψ∥≥η_ψ, completing the proof.

Theorem 5.2. For a function ψ on T, the following statements are equivalent:

(a)  M_ψ : L^∞ → ℒ_w,0 is bounded.

(b)  lim _|v|→∞ |v||ψ(v)| = 0.

Furthermore, under these conditions, one has,

Proof. (a)⇒(b) Assume M_ψ : L^∞ → ℒ_w,0 is bounded. Applying M_ψ to the constant function 1, we obtain ψ = M_ψ1 ∈ ℒ_w,0. On the other hand, if 𝒪 = {v ∈ T : |v|  is  odd}, then ψχ_𝒪 ∈ ℒ_w,0, so for v ∈ T^*, we have

as |v | → ∞, proving (b).

(b)⇒(a) Suppose |v||ψ(v)| → 0 as |v | → ∞. First observe that

as |v | → ∞. Then for f ∈ L^∞ and v ∈ T^*, we have as |v | → ∞. Thus, ψf ∈ ℒ_w,0. The proof of the boundedness of M_ψ and of the formula ∥M_ψ∥ = η_ψ is similar to the case when M_ψ : L^∞ → ℒ_w.

Theorem 5.3. There are no isometric multiplication operators M_ψ from L^∞ to ℒ_w or ℒ_w,0.

Lemma 5.4. A bounded multiplication operator M_ψ from L^∞ to ℒ_w is compact if and only if for every bounded sequence {f_n} in L^∞ converging to 0 pointwise, the sequence ∥ψf_n∥_w approaches 0 as n → ∞.

Proof. Assume M_ψ is compact, and let {f_n} be a bounded sequence in L^∞ converging to 0 pointwise. By rescaling the sequence, if necessary, we may assume ∥f_n∥_∞ ≤ 1 for all n ∈ ℕ. By the compactness of M_ψ, {f_n} has a subsequence $$ such that $$ converges in the ℒ_w-norm to some function f ∈ ℒ_w. Since by Lemma 1.3, for v ∈ T^*,

and $$, it follows that $$ pointwise. Since by assumption, f_n → 0 pointwise, the function f must be identically 0. Thus, the only limit point of the sequence {ψf_n} in ℒ_w is 0. Hence ∥ψf_n∥_w → 0 as n → ∞.

Conversely, suppose ∥ψf_n∥_w approaches 0 as n → ∞ for every bounded sequence {f_n} in L^∞ converging to 0 pointwise. Let {g_n} be a sequence in L^∞ with $$. Then some subsequence $$ converges to a bounded function g. Thus, the sequence $$ converges to 0 uniformly, and $$ is bounded. By the hypothesis, it follows that $$ as k → ∞. Thus, $$ in ℒ_w. Therefore, M_ψ is compact.

Lemma 5.5. A bounded multiplication operator M_ψ from L^∞ to ℒ_w,0 is compact if and only if for every bounded sequence {f_n} in L^∞ converging to 0 pointwise, the sequence $$ approaches 0 as n → ∞.

Theorem 5.6. For a bounded operator M_ψ from L^∞ to ℒ_w, the following statements are equivalent:

(a)  M_ψ is compact.

(b)  lim _|v|→∞ |v||ψ(v)| = 0.

Proof. (a)⇒(b) Assume M_ψ is compact. Let {v_n} be a sequence in T such that |v_n | → ∞ as n → ∞. For n ∈ ℕ, let f_n denote the characteristic function of the set {w ∈ T:|w | ≥|v_n|}. Then $$ and f_n → 0 pointwise. By Lemma 5.4 and the compactness of M_ψ, it follows that

as n → ∞.

(b)⇒(a) Assume lim _|v|→∞ |v∥ψ(v)| = 0 and that ψ is not identically 0. In particular, ψ is bounded. Let {f_n} be a sequence in L^∞ converging pointwise to 0 such that $$ is bounded above by some positive constant C. Then corresponding to ɛ > 0, there exists N ∈ ℕ such that |v||ψ(v)| < ɛ/4C for all vertices v such that |v | ≥ N. Therefore, for |v | > N and n ∈ ℕ, we have

Furthermore, the sequence {f_n} converges to 0 uniformly on the set {v ∈ T:|v | ≤ N} so that |f_n(v)| < ɛ/4N∥ψ∥_∞ for all n sufficiently large. Hence |v | D(ψf_n)(v) < ɛ for all v ∈ T^* and all n sufficiently large. Consequently, $$ as n → ∞. Using Lemma 5.4, we deduce that M_ψ is compact.

Corollary 5.7. For a function ψ on T, the following statements are equivalent:

(a)  M_ψ : L^∞ → ℒ_w is compact.

(b)  M_ψ : L^∞ → ℒ_w,0 is bounded.

(c)  M_ψ : L^∞ → ℒ_w,0 is compact.

(d) lim _|v|→∞ |v||ψ(v)| = 0.

Theorem 5.8. Let M_ψ : L^∞ → ℒ_w be bounded. Then

Proof. Set B(ψ) = lim _n→∞  sup _|v|≥n   | v | [|ψ(v)|+|ψ(v⁻)|]. In the case B(ψ) = 0, then lim _|v|→∞ |v | ψ(v) = 0, so by Theorem 5.6, M_ψ is compact, and thus ∥M_ψ∥_e = 0. So assume B(ψ) > 0. Then there exists a sequence {v_n} in T such that 1≤|v_n | → ∞ as n → ∞ and

For each n ∈ ℕ let f_n be the function on T defined by

Then $$, and {f_n} converges to 0 pointwise. Thus, for any compact operator K : L^∞ → ℒ_w, there exists a subsequence $$ such that $$ as k → ∞. Thus

Therefore, $$.

We now show that $$. For each n ∈ ℕ, define the operator K_n on L^∞ by

Then, for v ∈ T^*, we have Thus, K_nf ∈ ℒ_w with $$.

Assume {f_k} is a sequence in L^∞ with $$. Then there exists a subsequence $$ converging pointwise to some function f ∈ L^∞. Thus,

So $$ as j → ∞. Therefore, K_n is compact, and thus, since M_ψ is bounded, M_ψK_n is also compact.

For f ∈ L^∞, we have

Therefore, we obtain thus completing the proof.