Multilinear singular integrals on non-commutative $L^p$ spaces

Abstract

We prove $L^p$ bounds for the extensions of standard multilinear Calderón–Zygmund operators to tuples of ${\text {UMD}}$ spaces tied by a natural product structure. The product can, for instance, mean the pointwise product in ${\text {UMD}}$ function lattices, or the composition of operators in the Schatten-von Neumann subclass of the algebra of bounded operators on a Hilbert space. We do not require additional assumptions beyond ${\text {UMD}}$ on each space—in contrast to previous results, we e.g. show that the Rademacher maximal function property is not necessary. The obtained generality allows for novel applications. For instance, we prove new versions of fractional Leibniz rules via our results concerning the boundedness of multilinear singular integrals in non-commutative $L^p$ spaces. Our proof techniques combine a novel scheme of induction on the multilinearity index with dyadic-probabilistic techniques in the ${\text {UMD}}$ space setting.

Appendix A: Iterated mixed-norm non-commutative \(L^p\) spaces

Let ${\mathcal {M}}$ be a von Neumann algebra equipped with a n.s.f. trace as described in Example 3.11. Recall in particular that ${\mathcal {A}}=L^0({\mathcal {M}})$ is an associative $*$-algebra endowed with a compatible complete metrizable topology, induced by the metric $d_{{\mathcal {A}}}$ of convergence in measure. For an integer $S\ge 1$, let $(M_s,\mu _s)$, $s=1,\ldots , S$, be $\sigma$-finite measure spaces and $(\Omega _S,\omega _S)$ the product measure space

$$\begin{aligned} \Omega _S=\prod _{s=1}^S M_s,\quad \omega _S= \prod _{s=1}^S \mu _s. \end{aligned}$$

Let ${\mathscr {A}}_{0,S}$ be the vector space of simple functions $f:\Omega _S\rightarrow {\mathcal {A}}$, namely

$$\begin{aligned} f (t)=\sum _{j=1}^{J } A_{j } {\mathbf {1}}_{E^{j}}(t), \quad t=(t_1,\ldots , t_s)\in \Omega _S, \end{aligned}$$

with $A_{j}\in {\mathcal {A}},$ $E^{j} \subset \Omega _S$ with $\omega _S(E^j)<\infty$. Then ${\mathscr {A}}_{0,S}$ is an associative algebra with respect to the pointwise product: for $f,g \in {\mathscr {A}}_{0,S}$, the function fg defined by $(fg)(t)=f(t) g(t)$, where the latter is the strong product in ${\mathcal {A}}$, belongs to ${\mathscr {A}}_{0,S}$. We denote by

$$\begin{aligned} {\mathscr {A}}_S {{:}{=}} \text { closure of } {\mathscr {A}}_0 \text { w.r.t. sequential } \ d_{\mathcal A}\text {-pointwise convergence} \end{aligned}$$

namely, $f \in {\mathscr {A}}_S$ if there exists a sequence $f_n \in {\mathscr {A}}_{0,S}$ such that

$$\begin{aligned} \lim _{n} d_{{\mathcal {A}}} (f(t),f_n(t)) =0 \quad a.e. \ \ t \in \Omega _S. \end{aligned}$$

Then ${\mathscr {A}}_S$, the class of strongly measurable $\mathcal A$-valued functions on $\Omega _S$, is an associative algebra with respect to the same product. Furthermore, ${\mathscr {A}}_S$ is complete with respect to the topology of convergence in measure, namely $f_n\rightarrow f$ if for all $\varepsilon >0$

$$\begin{aligned} \lim _{n} \mu \left( \left\{ t\in \Omega _S: d_{{\mathcal {A}}} (f(t),f_n(t))>\varepsilon \right\} \right) =0, \end{aligned}$$

and the product operation is continuous. Note that the latter topology is also metrizable, proceeding in an analogous way to [24, Proposition A.2.4].

Recall that ${\mathcal {M}}$ is equipped with the n.s.f. trace $\tau$, which is a linear bounded functional on $L^1(\mathcal M)$. Then the functional

$$\begin{aligned} \tau _S(f) {:}{=}\int _{\Omega _S} \tau (f(t)) \, \mathrm {d} \omega _S(t) \end{aligned}$$

is linear and bounded on the Bochner space $L^1(\Omega _S,\omega _S; L^1({\mathcal {M}}))$, which is a subspace of ${\mathscr {A}}_S$. With this definition, ${\mathscr {A}}_S$ is endowed with the trace $\tau _S$. Under these assumptions, we have the following proposition.

Proposition A.1

For a Hölder tuple $\{p_j^{0}:1\le j\le m\}$ as in (3.1), let

$$\begin{aligned} X_{j}^0= L^{p_j^0}({\mathcal {M}}). \end{aligned}$$

Let $\{p_j^{s}:1\le j\le m\}$ be further Hölder tuples of exponents, for $1\le s\le S$. Then the Banach subspaces of ${\mathscr {A}}_s$

$$\begin{aligned} X_j^s = L^{p_j^{s}} (M_s,\mu _s; X_j^{s-1} ), \quad s=1,\ldots ,S, \end{aligned}$$

(A.2)

are a ${\text {UMD}}$ Hölder m-tuple.

Before the proof proper, we need to set some notation, and develop suitable auxiliary lemmata. For $1\le k \le m-1$, $\mathcal J=\{j_1<j_2<\cdots <j_k\}\subset {\mathcal {J}}_m$, and $0\le s\le S$ we write

$$\begin{aligned} \frac{1}{q_{{\mathcal {J}}}^s}= \sum _{u=1}^k \frac{1}{p_{j_u}^s}, \quad \frac{1}{p_{{\mathcal {J}}}^s}= 1 -\frac{1}{q_{{\mathcal {J}}}^s}. \end{aligned}$$

It will be convenient to introduce the auxiliary mixed norm spaces

$$\begin{aligned} \begin{aligned}&E_{j}^1 = L^{p_{j}^1}(M_1,\mu _1), \, \\&E_{j}^s = L^{p_{j}^s} (M_s,\mu _s;E_{j}^{s-1}), \quad s=2,\ldots , S, \end{aligned} \end{aligned}$$

for $j=1,\ldots , m$ and similarly

$$\begin{aligned} \begin{aligned}&E_{{\mathcal {J}}}^0 = {\mathbb {C}}, \, \\&E_{{\mathcal {J}}}^s = L^{q_{{\mathcal {J}}}^s} (M_s,\mu _s;E_{{\mathcal {J}}}^{s-1} ), \quad s=1,\ldots , S. \end{aligned} \end{aligned}$$

In general we write S(X) for the unit sphere in the Banach space X.

Lemma A.3

Let ${\mathcal {J}}=\{j_u:1\le u\le k\}$. There exists maps $B_u^{s}:S(E_{{\mathcal {J}}}^s)\rightarrow S(E_{j_u}^s)$ such that

$$\begin{aligned} f = \prod _{u=1}^k B_u^s(f) \quad \forall f\in S(E_{{\mathcal {J}}}^s) \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\Vert f_n -f \Vert _{E_{{\mathcal {J}}}^{s }}\rightarrow 0, \; \Vert f_n(t_s)-f(t_s)\Vert _{E_{{\mathcal {J}}}^{s-1}} \rightarrow 0 \; \mathrm {a.e.} \; t_s\in M_s \implies \\&\Vert B_u^s(f) -B_u^s(f_n) \Vert _{E_{j_u}^{s }}\rightarrow 0, \; \Vert B_u^s(f_n) (t_s)-B_u^s(f_n) (t_s)\Vert _{E_{j_u}^{s-1}} \rightarrow 0 \; \mathrm {a.e.} \; t_s\in M_s, \\&\quad 1\le u\le k. \end{aligned}\nonumber \\ \end{aligned}$$

(A.4)

Proof

We deal with the case $j_u=u,u=1,\ldots , k$ which is generic. We prove the statement by induction on s. If $s\ge 2$, assume inductively that maps $B_u^{s-1}$ as in the statement have been constructed; for the base case $s=1$, we run the argument below with $B^{0}_u$ the identity map. In both cases, we need to define $B^s_u:S(E_{{\mathcal {J}}}^s)\rightarrow S(E_{u}^s)$. We use that each $f\in S(E_{{\mathcal {J}}}^s)$ is $E_{{\mathcal {J}}}^{s-1}$-valued. So for each $t_s\in M_s$, write

$$\begin{aligned} f (t_s ) = |f(t_s )|_{ E_{{\mathcal {J}}}^{s-1}} g(t_s) = \prod _{u=1}^k \left( |f (t_s ) |_{E_{{\mathcal {J}}}^{s-1}}^{\frac{q_{{\mathcal {J}}}^s }{p_u^s}} g_{u}(t_s) \right) {=}{:}\prod _{u=1}^k B^s_u(f)(t_s ) \end{aligned}$$

where g is $S( E_{{\mathcal {J}}}^{s-1})$-valued, so that each $g_u = B_u^{s-1}(g)$ is $S(E_{u}^{s-1})$-valued. Notice that each $f_{u}=B^s_u(f)$ is (strongly) $\mu _s$-measurable with values in $E_{u}^{s-1 }$: in fact $|f (\cdot ) |_{E_{{\mathcal {J}}}^{s-1}}$ is $\mu _s$-measurable and each $g_u$ is $\mu _s$-measurable, as $B_u^{s-1}$ is (norm) continuous from $E_{\mathcal {J}}^{s-1} \rightarrow E_{u}^{s-1 }$ and g is $\mu _s$-measurable with values in $E_{\mathcal {J}}^{s-1}$ . A direct calculation reveals that

$$\begin{aligned} \Vert f_{u}\Vert _{E_{u}^s}=1, \quad 1\le u\le k. \end{aligned}$$

It remains to show that the thus defined maps $B_u^s$ are continuous in the sense of (A.4) by assuming the same properties hold for the maps $B_u^{s-1}$. Let $f_n, f$ be as in the first line of (A.4) and write $f_n(t_s)=|f_n(t_s)|_{E_{{\mathcal {J}}}^{s-1}} g_n(t_s)$. We first show the pointwise convergence: for each we have

$$\begin{aligned} \begin{aligned}&\Vert {B}^s_u(f)(t_s)-B^s_u(f_n)(t_s)\Vert _{E_{u}^{s-1}} \le \Vert f(t_s)\Vert _{E_{{\mathcal {J}}}^{s-1}}\Vert {B}^{s-1}_u(g(t_s))-B^{s-1}_u(g_n(t_s)) \Vert _{E_{u}^{s-1}}\\&+ \quad \Vert B^{s-1}_u(g_n(t_s))\Vert _{E_{u}^{s-1}} \big |\Vert f(t_s)\Vert _{E_{{\mathcal {J}}}^{s-1}}^{\frac{q_{{\mathcal {J}}}^s}{p_u^s}} - \Vert f_n(t_s)\Vert _{E_{{\mathcal {J}}}^{s-1}}^{\frac{q_{{\mathcal {J}}}^s}{p_u^s}} \big | \end{aligned} \end{aligned}$$

Relying on the norm continuity of ${B}^{s-1}_u$ we obtain that both summands in the previous display converge to zero for each $t_s$ such that $\Vert f_n(t_s)\Vert _{E_{{\mathcal {J}}}^{s-1}} \rightarrow \Vert f(t_s)\Vert _{E_{{\mathcal {J}}}^{s-1}}, \Vert g_n(t_s)- g(t_s)\Vert _{E_{{\mathcal {J}}}^{s-1}}\rightarrow 0$; this is a set of full $\mu _s$ measure, so that this part of the proof is complete. We come to the norm continuity in (A.4). We have

$$\begin{aligned} \begin{aligned}&\Vert {B}^s_u(f)-B^s_u(f_n)\Vert _{E_{u}^s}^{p_u^s}\lesssim \int _{M_s} |f(t_s) |_{E_{{\mathcal {J}}}^{s-1}}^{q_{{\mathcal {J}}}^s} |B_u^{s-1}(g(t_s))- B_u^{s-1}(g_n(t_s))|^{p_u^s}_{E_{u}^{s-1}}\, \mathrm {d} \mu _s(t_s) \\&\quad + \int _{M_s} \left| |f(t_s)|_{E_{{\mathcal {J}}}^{s-1}}^{\frac{q_{{\mathcal {J}}}^s}{p_u^s}} -|f_n(t_s)|_{E_{{\mathcal {J}}}^{s-1}}^{\frac{q_{{\mathcal {J}}}^s}{p_u^s}} \right| ^{p_u^s}|B_u^{s-1}(g_n(t_s))|^{p_u^s}_{E_{u}^{s-1}}\, \mathrm {d} \mu _s(t_s). \end{aligned} \end{aligned}$$

The first integrand converges to zero pointwise a.e. and is dominated by $|f(t_s) |_{E_{{\mathcal {J}}}^{s-1}}^{q_{{\mathcal {J}}}^s}$, so the integral converges to zero by dominated convergence. The second integral is equal to

$$\begin{aligned} \Vert F -F_n \Vert ^{p_u^s}_{L^{p_u^s}(M_s,\mu _s)}, \quad F(t_s)=|f (t_s)|_{E_{{\mathcal {J}}}^{s-1}}^{\frac{q_{{\mathcal {J}}}^s}{p_u^s}}, \quad F_n(t_s)=|f_n(t_s)|_{E_{{\mathcal {J}}}^{s-1}}^{\frac{q_{{\mathcal {J}}}^s}{p_u^s}}. \end{aligned}$$

Notice that $\Vert F \Vert _{p_u^s}= \Vert f\Vert _{E_{{\mathcal {J}}}^{s}}^{{q_{\mathcal {J}}^s}/{p_u^s}}$, $\Vert F_n \Vert _{p_u}= \Vert f_n\Vert _{E_{{\mathcal {J}}}^{s}}^{{q_{\mathcal {J}}^s}/{p_u^s}}$. As $F_n\rightarrow F$ pointwise, $F_n,F\in L^{p_u^s}(M_s,\mu _s)$ and $\Vert F_n \Vert _{p_u^s}\rightarrow \Vert F \Vert _{p_u^s}$, then $\Vert F -F_n \Vert _{p_u^s}$ converges to zero by a well-known variation of the proof of the $L^p$ dominated convergence theorem. $\square$

Lemma A.5

Let²

$$\begin{aligned} \begin{aligned}&X_{{\mathcal {J}}}^0 = L^{q_{{\mathcal {J}}}^0}({\mathcal {M}}), \quad X_{{\mathcal {J}},+}^0= L^{q_{{\mathcal {J}}}^0}({\mathcal {M}})_+, \\&X_{{\mathcal {J}}}^s = L^{q_{{\mathcal {J}}}^s} (M_s,\mu _s;X_{{\mathcal {J}}}^{s-1}),\quad X_{{\mathcal {J}},+}^s = L^{q_{\mathcal J}^s} (M_s,\mu _s;X_{{\mathcal {J}},+}^{s-1}), \quad s=1,\ldots , S. \end{aligned} \end{aligned}$$

Let $f\in X_{{\mathcal {J}},+}^s$ be a simple function with $\Vert f\Vert _{X_{{\mathcal {J}}}^s}=1$. Then there exist $f_u \in X_{j_u,+}^s$, $u=1,\ldots , k$ with

$$\begin{aligned} f=\prod _{u=1}^k f_u, \quad \Vert f_u\Vert _{X_{j_u}^S}=1. \end{aligned}$$

Proof

Again we deal with the generic case $j_u=u,u=1,\ldots , k$. First of all, we make a remark about the case $s=0$. Fix $A\in X_{{\mathcal {J}},+}^0$ with $\Vert A\Vert _{X_{\mathcal {J}}^0}=1$. Using the Borel functional calculus for positive closed densely defined operators to define $A^\theta$ for $\theta >0$

$$\begin{aligned} A= \prod _{u=1}^k B_u(A), \quad B_u(A)= A^{\frac{q_{\mathcal {J}}^0}{p_{u}^0}}, \quad u=1,\ldots ,k. \end{aligned}$$

(A.6)

Trivially

$$\begin{aligned} \Vert B_u(A)\Vert _{X_u^0}=\Vert A\Vert ^{\frac{q_{\mathcal {J}}^0}{p_{u}^0}}_{X_{\mathcal {J}}^0}=1, \quad u=1,\ldots ,k. \end{aligned}$$

We now prove the main statement. Let $f\in X_{{\mathcal {J}},+}^s$ be a simple function with $\Vert f\Vert _{X_{{\mathcal {J}}}^s}=1$. We factor

$$\begin{aligned} f(t) = F(t) A(t), \quad F(t)= |f(t)|_{X_{\mathcal {J}}^0}, \quad t \in \Omega _s. \end{aligned}$$

Notice that $F\in E_{{\mathcal {J}}}^s$ of unit norm, so that using Lemma A.3

$$\begin{aligned} F= \prod _{u=1}^k B_u^s(F) , \quad \Vert B_u^s(F)\Vert _{E_{u}^s}=1, \end{aligned}$$

and we may write, also using (A.6)

$$\begin{aligned} f= \prod _{u=1}^k f_u, \quad f_u(t)=B_u^s(F)(t) B_u(A(t)), \end{aligned}$$

Notice that each $f_u$ is strongly measurable as $B_u(A(\cdot ))$ is a simple $X_{u,+}^0$-valued function and $B_u^s(F)$ is a measurable function in $E_u^s$. Also as $|B_u(A(t))|_{X_u^0}=1$ for all $t\in \Omega _s$

$$\begin{aligned} \Vert f_u\Vert _{X_u^s} = \Vert B_u^s(F)\Vert _{E_{u}^s}=1, \end{aligned}$$

which completes the proof of the claim. $\square$

We turn to the proof of the proposition. Namely we need to show that the tuple $X_j^s$ from (A.2) is a ${\text {UMD}}$ Hölder tuple for each $s=1,\ldots , S$. In proving this, by virtue of the case $s=0$ being already established in Example 3.11 we may argue inductively and assume the claim has been proved in the cases of $0,\ldots , s-1$.

Clearly each $X_j^s$ is a subspace of ${\mathscr {A}}_s$. Denoting by $q_j^s, s=0,\ldots , S$ the conjugate exponent of $p_j^s$, it is convenient to define the spaces

$$\begin{aligned}\begin{aligned}&Y_{j}^0 = L^{q_{j}^0}({\mathcal {M}}), \, \\&Y_{j}^s = L^{q_{j}^{s}} (M_s,\mu _s;Y_{j}^{s-1} ), \quad s=1,\ldots , S, \end{aligned} \end{aligned}$$

which are Banach subspaces of ${\mathscr {A}}_s$. Further, as each $X_j^s$ is a reflexive Banach space and enjoys the Radon-Nikodým property [24, Theorem 1.3.21], an inductive argument yields the Riesz representation theorem (cf. [24, Theorem 1.3.10]) then yields that

$$\begin{aligned} \left( X_j^s\right) ^*=Y_j^s, \quad 1\le j\le m \end{aligned}$$

through the identification

$$\begin{aligned} \lambda \in (X_j^s)^* \mapsto g_\lambda \in Y_j^s \quad \lambda (f) = \tau _s( g_\lambda f) , \quad f \in X_j^s. \end{aligned}$$

We have in particular shown that each $X_j^s$ is an admissible space for the algebra ${\mathscr {A}}_s$ with trace $\tau _s$ and $Y(X_j^s)=Y_j^s$ .

We verify that $\{ X_j^s:j \in {\mathcal {J}}_m\}$ is a ${\text {UMD}}$ Hölder tuple by induction on m. The case $m=2$ is actually immediate by virtue of the observation and the well known fact that each $X_j^s,Y_j^s$ is a ${\text {UMD}}$ space.

To obtain the inductive step, we fix $m\ge 3$ and verify the following equality. For each $1\le k \le m-1$, $\mathcal J=\{j_1<j_2<\cdots <j_k\}\subset {\mathcal {J}}_m$, there holds

$$\begin{aligned} Y( \{ X_{j}^s:j\in {\mathcal {J}}\}) \text { is isometrically isomorphic to } \left( X_{{\mathcal {J}}}^s\right) ^* , \end{aligned}$$

(A.7)

where we refer to the spaces defined in Lemma A.5. More explicitly, denoting

$$\begin{aligned} \begin{aligned}&Y_{{\mathcal {J}}}^0 = L^{p_{{\mathcal {J}}}^0}({\mathcal {M}}), \\&Y_{{\mathcal {J}}}^s = L^{p_{{\mathcal {J}}}^s} (M_s,\mu _s;Y_{{\mathcal {J}}}^{s-1}), \quad s=1,\ldots , S, \end{aligned}\end{aligned}$$

we have $Y(\{ X_{j}^s:j\in {\mathcal {J}}\})=Y_{{\mathcal {J}}}^s= \left( X_{{\mathcal {J}}}^s\right) ^*$.

Property P1 then corresponds to this equality in the cases $k= m-1$. Verifying property P2 amounts to checking that when $k<m-1$, the tuple $\{ X^s_j:j\in {\mathcal {J}}\} \cup \{ Y^s_{{\mathcal {J}}}\}$ is a ${\text {UMD}}$ Hölder $(k+1)$-tuple. As $k< m-1$, $\{ X_j^s:j\in {\mathcal {J}}\} \cup \{ Y^s_{{\mathcal {J}}}\}$ is a ${\text {UMD}}$ Hölder $(k+1)$-tuple and the exponents $\{p_j^s: j \in {\mathcal {J}}, p^s({{\mathcal {J}}})\}$ are a Hölder tuple, this check is made by a straightforward appeal to the induction assumption.

We are left with proving (A.7). To do this we will define a linear surjective isometry $\Phi :Y( \{ X_{j}^s:j\in {\mathcal {J}}\})\rightarrow Y_{{\mathcal {J}}}^{s }$. First of all note that

$$\begin{aligned} \Vert g\Vert _{ Y( \{ X_{j}^s:j\in {\mathcal {J}}\})} \le \Vert g\Vert _{L^{p_{\mathcal {J}}^s}(M_s,\mu _s; Y_{{\mathcal {J}}}^{s-1})} = \Vert g\Vert _{ Y_{{\mathcal {J}}}^{s }} \end{aligned}$$

(A.8)

descends immediately from Hölder’s inequality in $L^{p}(M_s,\mu _s)$-spaces and Lemma 3.1 applied to the ${\text {UMD}}$ Hölder tuple $X_{j_1}^{s-1}, X_{j_2}^{s-1},\ldots , X_{j_k}^{s-1}$. We will use this below.

Fix then $g\in Y( \{ X_{j}^s:j\in {\mathcal {J}}\})$. We claim that if f is a simple $X_{{\mathcal {J}},+}^0$-valued function on $\Omega _s$ with $\Vert f\Vert _{X_{{\mathcal {J}}}^s}=1$, then

$$\begin{aligned} |\tau _s(g f)|\le \Vert g\Vert _{ Y( \{ X_{j}^s:j\in \mathcal J\})} . \end{aligned}$$

(A.9)

Indeed, applying Lemma 3.1 we obtain

$$\begin{aligned}&\left| \tau _s(g f)\right| = \left| \tau _s\left( g \prod _{u=1}^k f_u\right) \right| \le \Vert g\Vert _{ Y( \{ X_{j}^s:j\in {\mathcal {J}}\})} \prod _{u=1}^k \Vert f_u\Vert _{X_{j_u}^s}, \\&\quad \Vert f_u\Vert _{X_{j_u}^s} = 1, \quad u=1,\ldots , k, \end{aligned}$$

which is (A.9). As $X_{\mathcal {J}}^s$ is the $X_{\mathcal {J}}^s$-norm closure of the linear span of simple $X_{{\mathcal {J}},+}^0$-valued function on $\Omega _s$, the linear bounded functional $f\mapsto \tau _s(g f)$ extends uniquely to an element $\Phi (g)$ of $(X_{\mathcal {J}}^s)^*\equiv Y_{\mathcal {J}}^s$ with

$$\begin{aligned} \Vert \Phi (g)\Vert _{ Y_{\mathcal {J}}^s} \le \Vert g\Vert _{ Y( \{ X_{j}^s:j\in {\mathcal {J}}\})}. \end{aligned}$$

It is easy to see that the map $\Phi :Y( \{ X_{j}^s:j\in \mathcal J\})\rightarrow Y_{{\mathcal {J}}}^{s }$ is linear. From (A.8) we gather that if $g \in Y_{{\mathcal {J}}}^{s }$ then $\Phi (g)$ is well-defined. In this case the linear bounded functionals $g\mapsto \tau _s(g f)$ and $\Phi (g)$ coincide on a dense set, it must be $\Phi (g)=g$. So $\Phi$ is obviously surjective. Furthermore using (A.8) again we obtain

$$\begin{aligned} \Vert \Phi (g)\Vert _{ Y_{\mathcal {J}}^s} \ge \Vert \Phi (g)\Vert _{ Y( \{ X_{j}^s:j\in {\mathcal {J}}\})} = \Vert g\Vert _{ Y( \{ X_{j}^s:j\in {\mathcal {J}}\})} \ge \Vert \Phi (g)\Vert _{ Y_{\mathcal {J}}^s} \end{aligned}$$

whence equality must hold throughout. So $\Phi$ is a linear isometric isomorphism and the proof of (A.7) is complete.