Compatible pairs and complex interpolation

Self-checked by the writing AI.

Interpolation arguments in modular theory act on two endpoint spaces at once. The endpoints need not be nested, reflexive, separable, or dense in one another. What is needed is a common Hausdorff ambient space, a precise sum norm, and an analytic strip class whose boundary values decay in the endpoint norms.

This unit constructs that framework from the ground up. It proves completeness of the intersection, sum, strip, and interpolation spaces and proves the exact geometric-mean bound for a linear map bounded at both endpoints. The mathematical antecedent is Takesaki, Theory of Operator Algebras II, Appendix A.12. The organization, proofs, checks, and exercises here are independently written.

Compatible pairs, intersections, and sums

A compatible pair is a pair \(\mathbf X=(X_0,X_1)\) of complex Banach spaces with continuous injective linear maps into one Hausdorff topological vector space \(\mathcal V\). We identify each endpoint with its image. This identification matters: it gives a definite meaning to equality between an element of \(X_0\) and an element of \(X_1\).

Set

inside \(\mathcal V\), with

and

The infimum runs over \(x_j\in X_j\). In particular, the canonical maps \(X_j\to\Sigma(\mathbf X)\) are contractions.

Proposition. Both \(\Delta(\mathbf X)\) and \(\Sigma(\mathbf X)\) are Banach spaces.

Proof. Let \((x_n)\) be Cauchy in \(\Delta(\mathbf X)\). It converges to some \(x_j\) in \(X_j\) for \(j=0,1\). Continuity of the two ambient embeddings makes the same sequence converge to \(x_0\) and \(x_1\) in \(\mathcal V\). Since \(\mathcal V\) is Hausdorff, \(x_0=x_1\). The common vector lies in the intersection, and convergence holds in the maximum norm.

For the sum, give \(X_0\oplus X_1\) the norm

This is a Banach space. The addition map

is continuous. Its kernel is closed because \(\mathcal V\) is Hausdorff. Hence \((X_0\oplus X_1)/\ker A\) is Banach. The induced bijection from this quotient to \(\Sigma(\mathbf X)\) has quotient norm exactly (CI.2), so it is an isometric isomorphism. \(\square\)

No norm on the ambient space is used. Different compatible realizations can therefore lead to different intersections and sums even when the abstract endpoint Banach spaces are isomorphic.

The bounded-strip estimate

Write

We use the following form of the three-lines argument.

Lemma. Let \(h:\mathbb S\to\mathbb C\) be bounded and continuous, and holomorphic on \(\mathbb S^\circ\). If

then, for \(0\leq s\leq1\),

At an interior point, the right side is interpreted as zero when one endpoint bound is zero.

Proof. First suppose \(a_0a_1>0\). Fix \(\varepsilon>0\) and use the real logarithms of \(a_0,a_1\) to define

On either vertical boundary, its modulus is at most \(\exp(-\varepsilon t^2)\). If \(|h|\leq C\) on the strip, then on either horizontal edge \(z=s\pm iR\),

because \(s^2-s\leq0\). For sufficiently large \(R\), the boundary maximum principle on the rectangle with vertices \(\pm iR\) and \(1\pm iR\) gives \(|H_\varepsilon|\leq1\) throughout that rectangle. Evaluating at \(s+it\), then letting \(\varepsilon\downarrow0\), proves (CI.3). Replacing \(a_j\) by \(a_j+\delta\) and sending \(\delta\downarrow0\) handles a zero endpoint bound. The endpoint cases follow directly from the hypotheses. \(\square\)

The proof uses boundedness on the full strip. Without a growth condition, the two boundary lines alone do not control a holomorphic function on an unbounded strip.

The strip space is complete

For a compatible pair \(\mathbf X=(X_0,X_1)\), let \(\mathcal F(\mathbf X)\) consist of functions

with all of the following properties:

1. \(f\) is bounded and continuous in the sum norm on \(\mathbb S\);
2. \(f\) is holomorphic in the sum norm on \(\mathbb S^\circ\);
3. \(f(j+it)\in X_j\) for \(j=0,1\) and every real \(t\);
4. each boundary map \(t\mapsto f(j+it)\) is continuous as an \(X_j\)-valued map and tends to zero in \(X_j\) as \(|t|\to\infty\).

Put

Let

For every \(z=s+it\in\mathbb S\),

Indeed, if \(\ell\in\Sigma(\mathbf X)^*\) has norm at most one, then \(\ell\circ f\) satisfies OA-MOD-CI-02, since

The complex Hahn–Banach theorem identifies the norm of \(f(z)\) with the supremum over these functionals and gives (CI.5).

Theorem. The normed space \(\mathcal F(\mathbf X)\) is complete.

Proof. Let \((f_n)\) be Cauchy for (CI.4). On each boundary it is uniformly Cauchy in \(X_j\). Completeness of \(X_j\) gives a uniform limit

Applying (CI.5) to \(f_n-f_m\) shows that \((f_n)\) is uniformly Cauchy on the whole strip in \(\Sigma(\mathbf X)\). Let \(f\) be its uniform sum-norm limit. It is bounded and continuous. Since the endpoint inclusions into the sum space are contractive, its boundary values agree with \(g_j\), so conditions 3 and 4 hold.

It remains to justify holomorphy without silently replacing weak convergence by norm convergence. Choose a closed disk contained in \(\mathbb S^\circ\). For each \(n\), the Banach-valued Cauchy formula on its boundary follows from the scalar formula: apply any \(\ell\in\Sigma(\mathbf X)^*\), and then use Hahn–Banach to recover equality of the vectors. The boundary integral exists as a Banach-valued Riemann integral because continuous curves are uniformly approximable by step functions. Uniform convergence permits passage to the limit in that formula. Expanding the Cauchy kernel on every smaller concentric disk gives a norm-convergent power series for \(f\). Thus \(f\) is holomorphic.

Finally, \(f_n\to f\) uniformly in \(X_j\) on each boundary, so \(\|f_n-f\|_{\mathcal F}\to0\). \(\square\)

This proof also explains why the topology in the definition cannot be left implicit: interior convergence takes place in \(\Sigma(\mathbf X)\), while the stronger endpoint norms control the boundary.

Evaluation spaces and quotient completeness

Fix \(0<\theta<1\). Define

and

The evaluation map

has norm at most one by (CI.5). Hence

is closed. The induced map

is an isometric bijection when the range is given (CI.6). Consequently \([X_0,X_1]_\theta\) is a Banach space. In particular, (CI.6) is a norm rather than only a seminorm, and

The intersection embeds in every interpolation space. More precisely, if \(x\in X_0\cap X_1\), then

For nonzero \(x\), put \(a=\|x\|_{X_0}\), \(b=\|x\|_{X_1}\). For \(\varepsilon>0\), the function

belongs to \(\mathcal F(\mathbf X)\), takes the value \(x\) at \(\theta\), and has boundary norm at most

Letting \(\varepsilon\downarrow0\) gives (CI.8). The Gaussian factor is needed: the constant analytic representative does not decay along the boundary and therefore does not belong to this version of \(\mathcal F\).

Interpolating a bounded linear map

Let \(\mathbf X=(X_0,X_1)\) and \(\mathbf Y=(Y_0,Y_1)\) be compatible pairs. Suppose

is linear and

for positive constants \(M_0,M_1\).

First, \(T\) is bounded between the sum spaces. If \(x=x_0+x_1\), then

Taking the infimum proves the assertion. Therefore \(Tf\in\mathcal F(\mathbf Y)\) whenever \(f\in\mathcal F(\mathbf X)\).

Interpolation theorem. For every \(0<\theta<1\),

and

Proof. Given \(f\in\mathcal F(\mathbf X)\), define

The scalar factor is bounded on the strip. On the left boundary its modulus is \(M_0^{-1}\), and on the right boundary it is \(M_1^{-1}\). Thus (CI.9) gives

If \(f(\theta)=x\), then

Definition (CI.6) therefore yields

Take the infimum over all representatives \(f\) of \(x\). \(\square\)

If an endpoint bound is zero, apply (CI.10) with \(M_j+\delta\) and let \(\delta\downarrow0\). No density of \(X_0\cap X_1\), reflexivity, or separability enters the proof.

Checks and solved exercises

Check 1: identical endpoints. Suppose \(X_0=X_1=X\), with the same norm and the same ambient embedding. Then

Inequality (CI.7) gives \(\|x\|_X\leq\|x\|_\theta\), while (CI.8) gives the reverse inequality. This also checks that the boundary-decay convention has not changed the expected constant-endpoint space.

Check 2: a one-dimensional weighted pair. Let both endpoints be \(\mathbb C\) in the usual ambient line, with

where \(a,b>0\). Then

The upper bound is (CI.8). For the lower bound, apply (CI.3) to any representative \(f\) after multiplying its endpoint bounds by \(a\) and \(b\):

Taking the infimum gives equality. This model verifies both exponents and the direction of the scaling in (CI.11).

Exercise 1. Let the bounded scalar function \(\varphi\) be continuous on the whole closed strip and holomorphic in its interior. Prove directly that multiplication by this function acts boundedly on \(\mathcal F(\mathbf X)\), with norm at most \(\max_{j=0,1}\sup_{t\in\mathbb R}|\varphi(j+it)|\).

Solution. Continuity of both factors gives product continuity on the whole closed strip, and the product rule gives holomorphy in its interior. The sum-space product is bounded by the bounds of its two factors. Its \(j\)-th boundary norm is at most

Boundary decay follows from boundedness of \(\varphi\) and decay of \(f\). Taking the maximum over the two boundary lines proves the bound; the supremum in the real parameter need not be attained.

Exercise 2. Let \(S:\Sigma(\mathbf X)\to\Sigma(\mathbf Y)\) and \(T:\Sigma(\mathbf Y)\to\Sigma(\mathbf Z)\) satisfy endpoint bounds \((A_0,A_1)\) and \((B_0,B_1)\). Show that the interpolated bound for \(TS\) obtained in one step agrees with the product of the two separate interpolated bounds.

Solution. The endpoint bounds for the composition are \((B_0A_0,B_1A_1)\). Formula (CI.10) gives

This is exactly the product of the separate operator-norm estimates.

The unit supplies the compatible-pair interpolation machinery only. Applications to noncommutative \(L^p\)-spaces must still prove that the proposed endpoints form compatible pairs and that the operator acts consistently on their sum.