# 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

\[
\Delta(\mathbf X)=X_0\cap X_1,\qquad
\Sigma(\mathbf X)=X_0+X_1
\]

inside \(\mathcal V\), with

\[
\|x\|_{\Delta}
=\max\{\|x\|_{X_0},\|x\|_{X_1}\},
\tag{CI.1}
\]

and

\[
\|x\|_{\Sigma}
=\inf_{x=x_0+x_1}
 \bigl(\|x_0\|_{X_0}+\|x_1\|_{X_1}\bigr).
\tag{CI.2}
\]

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

\[
\|(x_0,x_1)\|_{\oplus}=\|x_0\|_{X_0}+\|x_1\|_{X_1}.
\]

This is a Banach space. The addition map

\[
A:X_0\oplus X_1\longrightarrow\mathcal V,\qquad
A(x_0,x_1)=x_0+x_1,
\]

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

\[
\begin{aligned}
\mathbb S
  &=\{z\in\mathbb C:0\leq\operatorname{Re}z\leq1\},\\
\mathbb S^\circ
  &=\{z\in\mathbb C:0<\operatorname{Re}z<1\}.
\end{aligned}
\]

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

\[
|h(it)|\leq a_0,\qquad |h(1+it)|\leq a_1
\quad(t\in\mathbb R),
\]

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

\[
|h(s+it)|\leq a_0^{\,1-s}a_1^{\,s}
\quad(t\in\mathbb R).
\tag{CI.3}
\]

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

\[
H_\varepsilon(z)
=h(z)a_0^{z-1}a_1^{-z}
 \exp\bigl(\varepsilon(z^2-z)\bigr).
\]

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 \(z=s\pm iR\),

\[
|H_\varepsilon(z)|
\leq
C\max(a_0^{-1},a_1^{-1})e^{-\varepsilon R^2},
\]

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

\[
f:\mathbb S\longrightarrow\Sigma(\mathbf X)
\]

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

\[
\|f\|_{\mathcal F}
=\max_{j=0,1}\sup_{t\in\mathbb R}\|f(j+it)\|_{X_j}.
\tag{CI.4}
\]

Let

\[
A_j(f)=\sup_t\|f(j+it)\|_{X_j}.
\]

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

\[
\|f(z)\|_{\Sigma}
\leq A_0(f)^{\,1-s}A_1(f)^{\,s}
\leq\|f\|_{\mathcal F}.
\tag{CI.5}
\]

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

\[
|\ell(f(j+it))|
\leq\|f(j+it)\|_{\Sigma}
\leq\|f(j+it)\|_{X_j}.
\]

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

\[
g_j\in C_0(\mathbb R;X_j).
\]

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

\[
[X_0,X_1]_\theta
=\{f(\theta):f\in\mathcal F(\mathbf X)\}
\subseteq\Sigma(\mathbf X)
\]

and

\[
\|x\|_\theta
=\inf\{\|f\|_{\mathcal F}:f\in\mathcal F(\mathbf X),\
f(\theta)=x\}.
\tag{CI.6}
\]

The evaluation map

\[
E_\theta:\mathcal F(\mathbf X)\longrightarrow\Sigma(\mathbf X),
\qquad E_\theta f=f(\theta),
\]

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

\[
K_\theta=\ker E_\theta
\]

is closed. The induced map

\[
\mathcal F(\mathbf X)/K_\theta
\longrightarrow [X_0,X_1]_\theta
\]

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

\[
\|x\|_{\Sigma}\leq\|x\|_\theta.
\tag{CI.7}
\]

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

\[
\|x\|_\theta
\leq
\|x\|_{X_0}^{\,1-\theta}\|x\|_{X_1}^{\,\theta}.
\tag{CI.8}
\]

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

\[
f_\varepsilon(z)
=e^{\varepsilon(z-\theta)^2}
 a^{z-\theta}b^{\theta-z}x
\]

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

\[
a^{1-\theta}b^\theta
\max\{e^{\varepsilon\theta^2},
      e^{\varepsilon(1-\theta)^2}\}.
\]

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

\[
T:\Sigma(\mathbf X)\longrightarrow\Sigma(\mathbf Y)
\]

is linear and

\[
\|Tx\|_{Y_j}\leq M_j\|x\|_{X_j}
\quad(x\in X_j,\ j=0,1)
\tag{CI.9}
\]

for positive constants \(M_0,M_1\).

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

\[
\|Tx\|_{\Sigma(\mathbf Y)}
\leq M_0\|x_0\|_{X_0}+M_1\|x_1\|_{X_1}
\leq\max(M_0,M_1)
       \bigl(\|x_0\|_{X_0}+\|x_1\|_{X_1}\bigr).
\]

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

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

\[
T[X_0,X_1]_\theta\subseteq[Y_0,Y_1]_\theta
\]

and

\[
\|Tx\|_{[Y_0,Y_1]_\theta}
\leq M_0^{\,1-\theta}M_1^{\,\theta}
      \|x\|_{[X_0,X_1]_\theta}.
\tag{CI.10}
\]

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

\[
g(z)=M_0^{z-1}M_1^{-z}Tf(z).
\tag{CI.11}
\]

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

\[
\|g\|_{\mathcal F(\mathbf Y)}
\leq\|f\|_{\mathcal F(\mathbf X)}.
\]

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

\[
g(\theta)=M_0^{\theta-1}M_1^{-\theta}Tx.
\]

Definition (CI.6) therefore yields

\[
\|Tx\|_{[Y_0,Y_1]_\theta}
\leq M_0^{1-\theta}M_1^\theta\|f\|_{\mathcal F(\mathbf X)}.
\]

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

\[
[X,X]_\theta=X
\quad\text{isometrically}.
\]

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

\[
\|z\|_{X_0}=a|z|,\qquad
\|z\|_{X_1}=b|z|,
\]

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

\[
\|z\|_\theta=a^{1-\theta}b^\theta|z|.
\]

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 \(b\):

\[
a^{1-\theta}b^\theta|f(\theta)|
\leq\|f\|_{\mathcal F}.
\]

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

\[
\sup_t|\varphi(j+it)|
\sup_t\|f(j+it)\|_{X_j}.
\]

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

\[
(B_0A_0)^{1-\theta}(B_1A_1)^\theta
=
\bigl(B_0^{1-\theta}B_1^\theta\bigr)
\bigl(A_0^{1-\theta}A_1^\theta\bigr).
\]

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.