Banach tensor cross norms: complete finite-rank and extremal-norm proofs

Original text by Claude Opus 5.5 (Anthropic), September 2026; public domain (CC0). Scoped selection by GPT-6.1 Sol (OpenAI), Ultra, October 2026; original contributions CC0.

This supplement proves finite-rank separation, the injective and projective norm formulas, the cross-norm identities and their extremal bounds. Its proof inputs are the algebraic tensor universal property, finite-dimensional linear algebra, Banach completeness and Hahn–Banach. Hilbert tensor notation in the conventions is supplied by Sections 1–2 of the companion spatial supplement and is not used in the Banach cross-norm proof.

Conventions

• Banach spaces are complex. $E^*$ is the dual of $E$, with $\|f\|=\sup\{|f(x)|:\|x\|\le1\}$, and $j_E:E\to E^{**}$ is the canonical isometry, $j_E(x)(f)=f(x)$. An operator is a bounded linear map. $B(E,F)$ is the Banach space of operators $E\to F$ with the operator norm, $B(E)=B(E,E)$, and a contraction is an operator of norm at most one.
• $E\odot F$ is the algebraic tensor product, spanned by the elementary tensors $x\otimes y$. Every bilinear map $b$ on $E\times F$ defines a unique linear map on $E\odot F$ with $x\otimes y\mapsto b(x,y)$. We use this universal property without comment to define linear maps on $E\odot F$.
• Hilbert spaces are complex, and inner products are linear in the first variable. $H\otimes K$ is the Hilbert tensor product: the completion of $H\odot K$ for the inner product $\langle\xi\otimes\eta,\xi'\otimes\eta'\rangle=\langle\xi,\xi'\rangle\langle\eta,\eta'\rangle$.
• $C^*$-algebras need not have a unit. $A_h$ is the set of self-adjoint elements of $A$, and $C^*(S)$ is the $C^*$-subalgebra generated by a subset $S$. For operators on a Hilbert space, $S'$ is the commutant and $S''$ the bicommutant. We write $[x,y]=xy-yx$ and $x\circ y=xy+yx$.

1. Tensors as operators of finite rank

Lemma 1.1. Let $E$ and $F$ be normed spaces, and $u\in E\odot F$.

1. $u=\sum_{i=1}^nx_i\otimes y_i$ for some $x_i\in E$ and linearly independent $y_1,\ldots,y_n\in F$, with $n=0$ when $u=0$.
2. If $y_1,\ldots,y_n$ are linearly independent and $\sum_ix_i\otimes y_i=0$, then every $x_i$ is $0$.
3. For $f\in E^*$ and $g\in F^*$, the number $\langle u,f\otimes g\rangle=\sum_if(x_i)g(y_i)$ does not depend on the representation $u=\sum_ix_i\otimes y_i$. Moreover $u=0$ if and only if $\langle u,f\otimes g\rangle=0$ for all $f\in E^*$ and $g\in F^*$.

Proof. (1) Start with any representation of $u$. Choose a basis of the span of its second factors, expand each second factor in this basis, and collect terms by bilinearity.

(2) Since the $y_i$ are linearly independent, there are linear functionals $g_1,\ldots,g_n$ on their span with $g_j(y_i)=\delta_{ij}$. They are bounded, because the span is finite-dimensional, and the Hahn–Banach theorem extends them to elements of $F^*$. The linear map $E\odot F\to E$ given by $x\otimes y\mapsto g_j(y)x$ sends $\sum_ix_i\otimes y_i$ to $x_j$. So $x_j=0$.

(3) The number is the value at $u$ of the linear functional defined by the bilinear form $(x,y)\mapsto f(x)g(y)$, so it depends only on $u$. If it vanishes for all $f$ and $g$, write $u$ as in (1) and take $g=g_j$ from (2). Then $f(x_j)=0$ for every $f\in E^*$, so $x_j=0$ by the Hahn–Banach theorem, and $u=0$. $\square$

Extending (3) linearly in the second variable, $\langle u,v\rangle=\sum_{i,k}f_k(x_i)g_k(y_i)$ for $v=\sum_kf_k\otimes g_k\in E^*\odot F^*$ is a well-defined bilinear pairing of $E\odot F$ with $E^*\odot F^*$. By (3), $E^*\odot F^*$ separates the points of $E\odot F$. The pairing also separates the points of $E^*\odot F^*$: if $\langle x\otimes y,v\rangle=0$ for all $x,y$, write $v=\sum_kf_k\otimes g_k$ with linearly independent $g_k$. For each $x$ the functional $\sum_kf_k(x)g_k$ is then zero, so every $f_k(x)=0$, and $v=0$.

Proposition 1.2 (tensors as operators). Let $E$ and $F$ be normed spaces. For $u=\sum_ix_i\otimes y_i\in E\odot F$ define $$T_u:E^*\to F,\qquad T_u(f)=\sum_if(x_i)\,y_i . \tag{1.1}$$

1. $T_u$ is a well-defined operator of finite rank, $\|T_u\|\le\sum_i\|x_i\|\|y_i\|$, and $g(T_uf)=\langle u,f\otimes g\rangle$ for $f\in E^*$, $g\in F^*$. The map $u\mapsto T_u$ is linear and injective.
2. Put $u^{\mathrm t}=\sum_iy_i\otimes x_i\in F\odot E$. The adjoint $T_u^*:F^*\to E^{**}$ takes its values in $j_E(E)$, and $T_u^*=j_E\circ T_{u^{\mathrm t}}$.

Proof. (1) The operator $f\mapsto f(x)y$ depends bilinearly on $(x,y)$, and this gives the linear map $u\mapsto T_u$. The norm bound is the triangle inequality, and the range lies in the span of the $y_i$. If $T_u=0$, then $\langle u,f\otimes g\rangle=g(T_uf)=0$ for all $f,g$, so $u=0$ by Lemma 1.1(3).

(2) For $g\in F^*$ and $f\in E^*$, $(T_u^*g)(f)=g(T_uf)=\sum_if(x_i)g(y_i)=f\big(\sum_ig(y_i)x_i\big)=j_E(T_{u^{\mathrm t}}g)(f)$. $\square$

So $E\odot F$ is a space of operators of finite rank from $E^*$ to $F$. Through $u\mapsto u^{\mathrm t}$ it is also a space of operators of finite rank from $F^*$ to $E$, and the adjoint of each of the two operators of $u$ is the other one followed by $j_E$ or $j_F$.

2. Cross norms: the injective and the projective norm

From now on $E$, $F$, $G$ are Banach spaces.

Definition 2.1. A norm $\beta$ on $E\odot F$ is a cross norm if $\beta(x\otimes y)=\|x\|\|y\|$ for all $x\in E$ and $y\in F$. We write $E\otimes_\beta F$ for the normed space $(E\odot F,\beta)$ and $E\hat\otimes_\beta F$ for its completion.

Definition 2.2. For $u\in E\odot F$ put $$\lambda(u)=\sup\big\{|\langle u,f\otimes g\rangle|:\ f\in E^*,\ g\in F^*,\ \|f\|\le1,\ \|g\|\le1\big\}, \tag{2.1}$$ $$\gamma(u)=\inf\Big\{\sum_{i=1}^n\|x_i\|\|y_i\|:\ u=\sum_{i=1}^nx_i\otimes y_i\Big\}. \tag{2.2}$$ They are the injective and the projective norm. [Ryan] writes $\varepsilon$ and $\pi$ for them.

Theorem 2.3.

1. $\lambda(u)=\|T_u\|$ for every $u\in E\odot F$. So $\lambda$ is a norm, and $u\mapsto T_u$ extends to an isometry of $E\hat\otimes_\lambda F$ onto the norm closure of $\{T_u:u\in E\odot F\}$ in $B(E^*,F)$.
2. $\lambda\le\gamma$, and both are cross norms.
3. If $\beta$ is a seminorm on $E\odot F$ with $\beta(x\otimes y)\le\|x\|\|y\|$ for all $x,y$, then $\beta\le\gamma$. In particular $\gamma$ is the largest cross norm.
4. $|\langle u,v\rangle|\le\lambda(u)\,\gamma(v)$ for $u\in E\odot F$ and $v\in E^*\odot F^*$. Here $\gamma(v)$ is the projective norm of $v$, formed with the norms of $E^*$ and $F^*$.

Proof. (1) By Proposition 1.2(1) and the Hahn–Banach theorem in the form $\|y\|=\sup_{\|g\|\le1}|g(y)|$, $$\lambda(u)=\sup_{\|f\|\le1}\ \sup_{\|g\|\le1}|g(T_uf)|=\sup_{\|f\|\le1}\|T_uf\|=\|T_u\| .$$ The operator norm is a norm on $B(E^*,F)$, and $u\mapsto T_u$ is linear and injective, so $\lambda$ is a norm. Since $B(E^*,F)$ is complete, the isometry extends to the completion, with the closure of its range as image.

(2) For every representation of $u$ and all $\|f\|,\|g\|\le1$, $|\sum_if(x_i)g(y_i)|\le\sum_i\|x_i\|\|y_i\|$. Hence $\lambda\le\gamma$. The function $\gamma$ is a seminorm: scaling the $x_i$ of a representation of $u$ gives $\gamma(cu)\le|c|\gamma(u)$, with equality for $c\ne0$ by applying this to $c^{-1}$; and joining representations of $u$ and $u'$ gives $\gamma(u+u')\le\gamma(u)+\gamma(u')$. As $\lambda\le\gamma$ and $\lambda$ is a norm, so is $\gamma$. Finally, $\lambda(x\otimes y)=\sup_{f,g}|f(x)||g(y)|=\|x\|\|y\|$ by the Hahn–Banach theorem, and $\lambda(x\otimes y)\le\gamma(x\otimes y)\le\|x\|\|y\|$.

(3) For every representation, $\beta(u)\le\sum_i\beta(x_i\otimes y_i)\le\sum_i\|x_i\|\|y_i\|$.

(4) By (2.1) and homogeneity, $|\langle u,f\otimes g\rangle|\le\lambda(u)\|f\|\|g\|$. Summing over a representation $v=\sum_kf_k\otimes g_k$ and taking the infimum gives the claim. $\square$

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