"Interpolation" is a broad term for starting with two concrete bounds on a mathematical object, and producing a range of other bounds that continuously fill in "the abstract space of bounds" between the extreme points that you started with.
A very typical example is provided by linear operators on the set of functions on a measure space. If $T$ is a linear map defined on simple functions, then there are all sorts of continuity properties $T$ might have, depending on which $L^p$ norm I pick for its domain and which $L^p$ norm I pick for its image. We can talk about the norm $$\lVert T \rVert_{p \to q} = \sup_{\lVert f \rVert_p = 1} \lVert T f \rVert_q,$$ and ask whether this is finite for a given choice of $p$ and $q,$ or more concretely compute this norm as a function of $(p,q).$
Interpolation essentially says: supposing you know $\lVert T \rVert_{p_1, q_1}$ and $\lVert T \rVert_{p_2, q_2}$, can you deduce an upper bound for a family of norms $\lVert T \rVert_{p_t, q_t}$ that "draw a line," in the space of norms, between the specified endpoints?
This is really useful in practice, because often one has a family of norms where some are easier to compute than others, and one can get a lot of information about an object by computing in the "easy cases" then interpolating. A few examples of physics applications that come to mind are the proof of the data processing inequality for Sandwiched Renyi relative entropies and the characterization of "Sobolev class" for UV-regular states in free field theory. Both of these subjects are closely related to different parts of my research, so I've been trying to understand the technique of interpolation more robustly.
This post is a pedagogical explanation of the theory of complex interpolation. I'm also interested in the "real interpolation method," but I'll leave that to future investigations and maybe a future post. I'd also like to make a post explaining applications to e.g. quantum information theory, but given how rarely I actually update this blog, we'll leave that aspiration in the "maybe" category.
One thing to say up front is that we will develop complex interpolation in a very abstract setting before applying it to specific examples of families of norms. The main advantage of the abstract formulation is that it can be applied to norms that aren't already known to lie in some continuous family. For example, it makes intuitive sense that you might be able to interpolate between $L^p$ norms at different values of $p$; you expect that the "interpolation spaces" will lie between two "endpoint" values of $p.$ But the general theory of complex interpolation, which we'll develop here, actually works for any two Banach spaces $A$ and $B$ that are sufficiently compatible in a sense we'll spell out below; the theory produces a continuous family of norms, even if you didn't know what that family was to start with.
Here's how the post is laid out. Note that the first 4 sections are all abstract; we'll come up with general theorems that take pairs $A$ and $B$ and spit out "interpolating spaces" $(A,B)_x$ that are defined in abstract language. In the last two sections, we actually compute the spaces $(A,B)_x$ in two physically interesting settings, and we'll find out that the spaces $(A,B)_x$ that pop out are spaces we were already familiar with, except that now we have new bounds on those old spaces by virtue of the fact that they can be obtained by interpolation.
Table of contents
- Phragmen-Lindelof and the Hadamard three-lines theorem
- Interpolation of nested Banach norms
- Interpolation of compatible Banach norms
- Interpolation applied to maps
- Application: Lp spaces
- Application: Sobolev spaces
Prerequisites: Comfort with complex analysis and with the term "Banach space." For examples, it will be helpful to understand $L^p$ spaces at the level of my blog post on the subject.
1. Phragmen-Lindelof and the Hadamard three-lines theorem
The basic idea behind complex interpolation is to abuse the rigidity of holomorphic functions. Recall that any holomorphic function $f : \mathbb{C} \to \mathbb{C}$ obeys the maximum modulus principle: in any region $\Omega,$ the modulus $|f|$ cannot attain a local maximum; when $\Omega$ is relatively compact, the maximum value of $|f|$ within $\bar{\Omega}$ is realized on $\partial \Omega.$
A powerful technique known as the Phragmen-Lindelof principle lets us apply a version of the maximum modulus principle to certain non-compact domains with sufficient symmetry. The example we will be concerned with in this article is that of a vertical strip of width 1, sketched below.
Let $f$ be a function that is holomorphic in the interior of the strip, and that has a continuous extension to its boundary. We would like to deduce some bound on the interior values of $|f|$ in terms of the values it takes on the boundary of the strip.
The trick is to multiply $f$ by a well-chosen function. Let $S(z)$ be a function that is holomorphic in the strip and that is continuous on its boundary — here $S$ stands for "smearing." The function $f(z) S(z)$ is well defined in the strip, with appropriate holomorphicity/continuity properties.
Now, let's chop off the strip at vertical values $\pm Y$; this is sketched below. Within the rectangle we have created, the maximum modulus principle holds, and we have
$$ \sup_{|y| \leq Y} |f(x + i y) S(x + i y)| \leq \max\{ \sup_{|y| \leq Y} |f(i y) S(i y)| , \sup_{|y| \leq Y} |f(1+i y) S(1+ i y)|, \sup_{x} |f(x + i Y) S(x + i Y)|, \sup_{x} |f(x - i Y) S(x - i Y)|\}.$$
If we are able to choose $S$ so that the product $f S$ decays sufficiently quickly at large $y,$ then the last two options won't matter when we take $Y$ large, and we'll get
$$ \sup_{y} |f(x + i y) S(x + i y)| \leq \max\{ \sup_{y} |f(i y) S(i y)| , \sup_{y} |f(1+i y) S(1+ i y)|\}.$$
A common example is to assume that $|f(z)|$ is bounded above by one on the boundary of the strip, and that it grows at most exponentially in the interior of the strip, $|f(z)| \leq \alpha e^{\beta |y|}.$ Then one picks $S(z) = A e^{B z^2}$ and gets
$$|f(z) S(z)| = A |f(z)| e^{B(x^2 - y^2)},$$
which decays for large $y,$ so that we have
$$ e^{B x^2} \sup_{y} |f(x + i y)| e^{- B y^2} \leq \max\{ \sup_{y} |f(i y)| e^{- B y^2}, \sup_{y} |f(1+i y)| e^{B}e^{- B y^2} \}.$$
Taking $B$ to zero, one learns that $|f(z)|$ is bounded by one everywhere.
A useful application is to consider a function $f(z)$ that is known to be bounded on the strip, with bound $Z_L$ on the left boundary of the strip $Z_R$ on the right boundary of the strip. The function
$$F(z) = f(z) Z_L^{z-1} Z_R^{-z}$$
grows at most exponentially in the interior of the strip, and is bounded by one on the boundaries of the strip. So applying Phragmen-Lindelof, one finds $|F(z)| \leq 1,$ or
$$|f(x + i y)| \leq Z_L^{1-x} Z_R^{x}.$$
This is known as the Hadamard three-lines theorem.
2. Interpolation of nested Banach norms
Now let's try to use the Phragmen-Lindelof principle to do the following: given two Banach spaces $A$ and $B$, let's try to make sense of a family of norms $(A,B)_{x}$ that "interpolate" between $A$ and $B$.
A priori it's a little hard to think about what we might mean by this, since $A$ and $B$ are completely different abstract vector spaces. So where is the norm $(A,B)_x$ supposed to live? Is it supposed to live on elements of $A$? On elements of $B$? On some auxiliary vector space we construct?
To get around this, we'll assume for the present section that there is a natural inclusion $A \subseteq B.$ In other words, we assume the existence of a norm $\lVert \cdot \rVert_{A}$ that can be defined on all elements of $B$, for which it may take finite or infinite values; $A$ will be identified as the space of finite-$\lVert \cdot \rVert_A$-norm objects of $B$. A simple example of this is $\ell_1 \subseteq \ell_2,$ the inclusion between summable and square-summable sequences. One can compute the sum of absolute values for any sequence, it just might end up being infinite. On the other hand, any summable sequence is square-summable. We'll also assume, as is the case for the example we've just given, that the inclusion $A \subseteq B$ is continuous. (This is true for the sequence example due to $\lVert u \rVert_{\ell_2} \leq \lVert u \rVert_{\ell_1}.$)
So now we want to pick a generic element $b \in B$ and assign a new norm to it, parametrized by some number $0 < x < 1.$ Well, we introduced the Phragmen-Lindelof principle above for a reason; let's try to exploit the rigidity of holomorphic functions! Let $f$ be a bounded function from the strip to $B$ that has appropriate holomorphicity properties. (While we formulated the Phragmen-Lindelof principle in terms of maps from $\mathbb{C}$ to $\mathbb{C}$, holomorphicity still makes perfect sense for maps from $\mathbb{C}$ to a generic Banach space; see for example my post on Bochner integration.) Suppose we have $f(x) = b.$ Then we automatically know, by the Phragmen-Lindelof principle, that we have
$$\lVert b \rVert_{B} \leq \max \left( \sup_{y} \lVert f(iy) \rVert_B , \sup_{y} \lVert f(1+i y)\rVert_B \right).$$
So far this isn't so interesting, since both sides only refer to $B$. But note that since we have a continuous inclusion $A \subseteq B,$ the $B$ norm is dominated by the $A$ norm, i.e., there is some constant $C > 1$ with $\lVert \cdot \rVert_{B} \leq C\lVert \cdot \rVert_A.$ So we could write, for example,
$$\lVert b \rVert_{B} \leq C \max\left( \sup_{y} \lVert f(iy) \rVert_A, \sup_{y} \lVert f(1+i y)\rVert_B \right),$$
at the expense of allowing for the right-hand side to become infinite if $f(iy)$ leaves the part of $B$ where $A$ is finite.
This expression we've just created is pretty interesting, since it involves a $B$ on the left-hand side, and both an $A$ and a $B$ on the right-hand side. We could imagine creating some new norm by optimizing over choices of $f.$ Indeed, this is what we'll do. We now restrict to the set of bounded holomorphic functions $f$ that have $\sup_{y} \lVert f(iy) \rVert_A < \infty$ and $\sup_y \lVert f(1 + i y) \rVert_B < \infty.$ And of course we'll keep the requirement $f(x) = b.$ Call this class of functions $\mathcal{F}(A,B)_{b;x}.$ Then we get
$$\lVert b \rVert_{B} \leq C \inf_{f \in \mathcal{F}(A,B)_{b;x}} \max\left( \sup_{y} \lVert f(iy) \rVert_A, \sup_{y} \lVert f(1+i y)\rVert_B \right).$$
Then we just define
$$\lVert b \rVert_{(A,B)_x} \equiv \inf_{f \in \mathcal{F}(A,B)_{b;x}} \max\left( \sup_{y} \lVert f(iy) \rVert_A, \sup_{y} \lVert f(1+i y)\rVert_B \right).$$
To see that this is actually a norm, note first that it is positive definite, since we know that $\lVert b \rVert_B$ is dominated by $\lVert b \rVert_{(A,B)_x},$ so if the latter vanishes, then the former vanishes, and one has $b=0.$ To see the triangle inequality, fix $b_1$ and $b_2$ with finite $\lVert \cdot \rVert_{(A,B)_X},$ and let $f_{b_1}$ and $f_{b_2}$ be functions such that the infima are satisfied to within an $\epsilon$ threshold. Then $f_{b_1} + f_{b_2}$ is included in the optimization over $\lVert b_1 + b_2 \rVert_{(A,B)_x},$ and we have
$$\lVert b_1 + b_2 \rVert_{(A,B)_x} \leq \max\left( \sup_{y} \lVert f_{b_1}(iy) + f_{b_2}(i y) \rVert_A, \sup_{y} \lVert f_{b_1}(1+i y) + f_{b_2}(1 + i y)\rVert_B \right),$$
and using the triangle inequalities for $A$ and $B$ gives
$$\lVert b_1 + b_2 \rVert_{(A,B)_x} \leq \max\left( \sup_{y} \lVert f_{b_1}(iy)\rVert_A + \lVert f_{b_2}(i y) \rVert_A, \sup_{y} \lVert f_{b_1}(1+i y)\rVert_B + \lVert f_{b_2}(1 + i y)\rVert_B \right) \leq \lVert b_1 \rVert_{(A,B)_x} + \lVert b_2 \rVert_{(A,B)_x} + 2 \epsilon,$$
at which point you can take $\epsilon \to 0.$
It turns out that $\lVert \cdot \rVert_{(A,B)_x}$ actually gives a Banach norm, i.e., it is complete. We will show this later. First, let us understand more about the relation between the norm $\lVert \cdot \rVert_{(A,B)_x}$ and the norms $\lVert \cdot \rVert_A$ and $\lVert \cdot \rVert_B.$
We already know, by construction, that there is an inequality of the form $\lVert \cdot \rVert_B \leq C \lVert \cdot \rVert_{(A,B)_x}.$ But note that $f(z) = b$ is a legal function that appears in the optimization, which gives the bound
$$\lVert b \rVert_{(A,B)_x} \leq \max\left( \lVert b \rVert_A, \lVert b \rVert_B \right) \leq C \lVert b \rVert_A.$$
So the norm we have constructed actually sits topologically "between" $\lVert \cdot \rVert_B$ and $\lVert \cdot \rVert_A.$ We can say a little more about the relationship between these things by using a trick similar to the one in the three-lines thoerem. Assuming that $\lVert b \rVert_A$ and $\lVert b \rVert_B$ are both finite, one can consider the function
$$f(z) = (\lVert b \rVert_A)^{z-x} (\lVert b \rVert_B)^{x-z} b,$$
which is easily checked to be in the class $\mathcal{F}(A,B)_{b;x}.$ So one has
$$\lVert b \rVert_{(A,B)_x} \leq \max\left( \sup_{y} \lVert f(iy) \rVert_A, \sup_{y} \lVert f(1+i y)\rVert_B \right),$$
and plugging in the specific form of $f,$ this gives
$$\lVert b \rVert_{(A,B)_x} \leq (\lVert b \rVert_A)^{1-x} (\lVert b \rVert_B)^{x}.$$
This is our first interpolation inequality relating Banach spaces $A \subseteq B$ and an interpolating family of norms $(A,B)_x.$
To close this section, let us see how to convince ourselves that the norm $\lVert \cdot \rVert_{(A,B)_x}$ is actually complete within $B$. The idea is first to convince ourselves that $\mathcal{F}(A,B)$ (i.e., the space of functions with no particular $f(x) = b$ restriction) is complete with respect to the norm
$$\lVert f \rVert_{\mathcal{F}(A,B)} = \max(\sup_{y} \lVert f(i y)\rVert_{A}, \sup_{y} \lVert f(1+i y)\rVert_B).$$
Then we will get a natural bounded map from $\mathcal{F}(A,B)$ to $B$ defined by evaluation at $x$, $f \mapsto f(x)$. It is easily seen that $(A, B)_x$ is isomorphic to the quotient of the evaluation map by its kernel, and the quotient of a Banach space by a closed subspace is automatically a Banach space. So all we need to know is that $\mathcal{F}(A,B)$ is Banach.
To see this, suppose we have a sequence of functions $f_n \in \mathcal{F}(A,B)$ that is Cauchy with respect to $\lVert \cdot \rVert_{\mathcal{F}(A,B)}.$ In particular, this means that $f_n(iy)$ and $f_n(1+iy)$ are uniformly Cauchy in $y$; moreover, by Phragmen-Lindelof, $f_n(z)$ is uniformly Cauchy for any open set in the interior of the strip. Basic tools of complex analysis then tell us that $f_n$ converges pointwise to some nice holomorphic function $f(z)$ on the strip, and it is easily seen that one has $\lVert f_n - f \rVert_{\mathcal{F}(A,B)}$ goes to zero at large $n.$
3. Interpolation of compatible Banach norms
The structure we used in the previous section, which was a continuous embedding of Banach spaces $A \subseteq B,$ is way too strong in general. It holds for sequence spaces like $\ell_1 \subseteq \ell_2,$ but even for general measure spaces it isn't true — on the real line there are $L^1$ functions that are not $L^2,$ and there are $L^2$ functions that are not $L^1.$ So how can we do interpolation in a more general setting?
The trick is to realize that when one has a continuous embedding $A \subseteq B,$ one can identify $B$ with the "sum" $A + B,$ And can identify $A$ with the "intersection" $A \cap B.$ More generally, if $A$ and $B$ can be realized as subspaces of some shared vector space $W$, then one says $A$ and $B$ are compatible, and one can define
$$A + B = \{a + b \,|\, a \in A, b \in B\},$$
and endow it with the "smallest optimization norm"
$$\lVert v \rVert_{A + B} = \inf \{\lVert a \rVert_A + \lVert b \rVert_B \,|\, a \in A, b \in B, a + b = v\}.$$
One can also define
$$A \cap B = \{v | v \in A, v \in B\}$$
with the "largest optimization norm"
$$\lVert v \rVert_{A \cap B} = \max(\lVert v\rVert_A, \lVert v \rVert_B).$$
Obviously in the setting $A \subseteq B,$ one has $A + B = B$ as sets, and the inequality $\lVert v \rVert_B \leq C \lVert v \rVert_A$ (taking $C \geq 1$ WLOG) gives
$$\lVert b \rVert_{A + B} = \inf \{\lVert a \rVert_A + \lVert b - a \rVert_B \,|\, a \in A\} \geq \inf \{ \frac{1}{C} \lVert a \rVert_B + \lVert b - a \rVert_B \,|\, a \in A\} \geq \frac{1}{C} \inf \{\lVert a \rVert_A + \lVert b - a \rVert_B \,|\, a \in A\},$$
and now applying the triangle inequality gives
$$\lVert b \rVert_{A + B} \geq \frac{1}{C} \lVert b \rVert_B.$$
On the other hand, taking $a=0$ in the infimization gives the inequality
$$\lVert b \rVert_{A + B} \leq \lVert b \rVert_B.$$
So the $(A+B)$-norm is mutually bounded by the $B$-norm, and they induce the same topology on the set $B = A + B.$
In the same setting, i.e. the setting $A \subseteq B,$ one has $A \cap B = A$ as sets, and from the inequality $\lVert v \rVert_B \leq C \lVert v \rVert_A$ (taking $C \geq 1$ WLOG) one gets
$$\lVert v \rVert_{A \cap B} \leq C \lVert v \rVert_A,$$
but also
$$\lVert v \rVert_{A \cap B} \geq \lVert v \rVert_B,$$
so $A \cap B$ and $A$ induce compatible norms on the same set.
In the more general setting, where we do not have $A \subseteq B,$ we can still play our usual games with respect to the inclusion $A \cap B \subseteq A + B.$ The idea is that $A + B$ is a natural candidate for "the smallest Banach space containing both $A$ and $B$" while $A \cap B$ is a natural candidate for "the largest Banach space contained in both $A$ and $B.$"
Actually, we haven't shown that our new spaces are Banach spaces. We also haven't demonstrated that the set inclusions $A \cap B \subseteq A \subseteq A + B$ and $A \cap B \subseteq B \subseteq A + B$ are actually continuous embeddings. For both of these points, it is important to think a little bit carefully about the role of the "embedding space" $W$ in which $A$ and $B$ are assumed to live. Up above, we said "$A$ and $B$ are compatible if they can be realized as subspaces of a shared vector space $W$." But this is a bit misleading; any two Banach spaces can be realized as subspaces of the direct sum $A \oplus B.$ But there may be many different ways of embedding $A$ and $B$ into a common vector space, and in practice, one has a canonical choice. For example, the Banach spaces $L^p$ and $L^q$ both naturally live within the vector space $L^0$ of measurable functions; this has nothing to do with $L^p \oplus L^q,$ and one cannot embed $L^p \oplus L^q$ into $L^0.$ The choice of common vector space $W$ is actually an input into the theory of complex interpolation; different choices will yield different interpolation spaces.
So in practice, one assumes the existence of a vector space $W$ with $A, B \subseteq W.$ There are a few technical assumptions that needs to be made, namely that $W$ is a topological vector space — i.e., it carries a topology with respect to which addition and scalar multiplication are continuous — and that the embeddings $A, B \subseteq W$ are continuous. One also assumes that the topology on $W$ is Hausdorff, but as a physicist I've never thought about non-Hausdorff spaces, so I don't feel the need to keep this assumption in mind.
Now that we understand that we have some nice vector space $W$ containing both $A$ and $B$, we can show that one has continuous embeddings $A \cap B \subseteq A, B \subseteq A + B,$ and that $A+B$ and $A \cap B$ are Banach spaces. Continuity of the embeddings is easy to show directly from the definitions, so I'll leave it as an exercise. To see that $A \cap B$ is complete, one assumes the existence of a Cauchy sequence $v_n \in A \cap B$, and notes that because $A \cap B \subseteq A, B$ are continuous inclusions, the sequence $v_n$ must converge with respect to both $A$ and $B$. Because both of these are embedded in the Hausdorff space $A + B,$ and limits in Hausdorff spaces are unique, one learns that $v_n$ has a limit that is in both $A$ and $B$, hence in $A \cap B$; so $A \cap B$ is complete. As for showing completeness of $A + B,$ one uses a silly little trick, which is that the map $A \oplus B \to A + B$ given by $a \oplus b \to a + b$ is continuous, and its kernel is exactly $v \oplus (-v)$ with $v \in A \cap B.$ So one has $A + B \cong A \oplus B / A \cap B,$ and the quotient of a Banach space by a closed subspace is a Banach space.
Now that we're happy (hopefully) with $A + B$ and $A \cap B$ as Banach spaces, let's beef up our complex interpolation arguments from the previous section. What we want now is to define a norm $\lVert \cdot \rVert_{(A,B)_x}$ on a generic element of $A + B.$ Let's call this generic element $v.$ In analogy to the previous section, we'll define $\mathcal{F}(A,B)_{v;x}$ as the set of all bounded holomorphic functions from the strip to $A + B$ satisfying:
- $\sup_{y} \lVert f(i y)\rVert_A < \infty$,
- $\sup_{y} \lVert f(1+ i y)\rVert_B < \infty$,
- $f(x) = v.$
Then we'll define the norm
$$\lVert v \rVert_{(A,B)_x} \equiv \inf_{f \in \mathcal{F}(A,B)_{v;x}} \max(\sup_{y} \lVert f(i y)\rVert_A, \sup_y \lVert f(1 + i y)\rVert_B).$$
Tada! We've defined a norm. What can we do with it?
Well, in the setting $A \subseteq B,$ we had a continuous inclusion $A \subseteq (A,B)_{x} \subseteq B.$ Now that $A$ and $B$ have no natural nesting properties, we can't hope to have any such nesting. But we can hope to have a continuous nesting
$$A \cap B \subseteq (A,B)_x \subseteq A + B,$$
and we can also hope that for anything in $A \cap B,$ we are still able to recover the bound
$$\lVert v \rVert_{(A,B)_x} \leq (\lVert v \rVert_{A})^{1-x} (\lVert v \rVert_B)^{x}.$$
Of course we also want $(A,B)_x$ to define a Banach subspace of $A + B$, i.e., we want it to be complete.
All of the arguments are basically the same as in the preceding section. In particular, the triangle norm proofs are entirely unchanged, and the completeness argument is basically the same once you replace $B$ with $A+B.$ As for the continuous inclusions, for $v \in A \cap B,$ one can pick the constant function $f(z) = v$ and get
$$\lVert v \rVert_{(A,B)_x} \leq \max(\lVert f(v) \rVert_A, \lVert f(v) \rVert_B) = \lVert v \rVert_{A \cap B},$$
while on the other hand for any $f \in \mathcal{F}(A,B)_{v;x}$ one gets
$$\lVert v \rVert_{A+B} \leq \max \left( \lVert f(i y) \rVert_{A + B}, \lVert (f1+i y)\rVert_{A + B}\right) \leq \max \left( \lVert f(i y) \rVert_{A}, \lVert f(1+i y)\rVert_{B}\right),$$
and taking the infimum gives $\lVert v \rVert_{A + B} \leq \lVert v \rVert_{(A,B)_x}.$
As for our famous "interpolation bound," nothing at all is changed; we just define the function
$$f(z) = (\lVert v \rVert_A)^{z-x} (\lVert v \rVert_B)^{x-z} v$$
and see that this is in our class $\mathcal{F}(A,B)_{v;x},$ which gives the bound
$$\lVert v \rVert_{(A,B)_x} \leq (\lVert v \rVert_A)^{1-x} (\lVert v \rVert_B)^{x}.$$
4. Interpolation applied to maps
So far we've just interpolated between Banach spaces. But what one often wants to do in practice — which is the motivation I gave in the introduction — is to interpolate maps between Banach spaces. It's tempting to think of this as a direct extension of what we've already done, since maps between Banach spaces themselves naturally live in Banach spaces.
Concretely, if $A$ and $B$ are Banach spaces and $T : A \to B$ is linear, then there is a natural norm on $T$ given by
$$\lVert T \rVert_{A \to B} \equiv \sup_{\lVert v \rVert_{A} = 1} \lVert T v \rVert_B.$$
It's a pretty easy exercise to show that with respect to this norm, the space of bounded linear maps from $A$ to $B$ is a Banach space. (In fact this only requires $B$ to be a Banach space, $A$ just needs to be normed). Let's call this new Banach space
$$\mathcal{B}(A, B).$$
Now suppose we have two pairs of Banach spaces, $A_1$ and $A_2,$ and $B_1$ and $B_2.$ This gives us a pair of Banach spaces of maps,
$$\mathcal{B}(A_1, B_1) \quad \text{and} \quad \mathcal{B}(A_2, B_2).$$
What we'd like to do is interpolate between these spaces, and relate this to the interpolations $(A_1,A_2)_x$ and $(B_1, B_2)_x.$ We're gunning for a statement like the following:
- When $(A_1, A_2)$ and $(B_1, B_2)$ are compatible, then the mapping spaces are also compatible, and one has
$$\left( \mathcal{B}(A_1, B_1), \mathcal{B}(A_2, B_2)\right)_x \hookrightarrow \mathcal{B}((A_1, A_2)_x, (B_1, B_2)_x).$$
If this were true, then we could just apply the results of the preceding section to obtain the bound
$$\lVert T \rVert_{(A_1, A_2)_x \to (B_1, B_2)_X} \leq \lVert T \rVert_{A_1 \to B_1}^{1-x} \lVert T \rVert_{A_2 \to B_2}^{x}.$$
The trouble is, the reasoning we've described above requires a pretty special set of assumptions. The fact that $A_1$ and $A_2$ live in some common space $W^{(A)},$ and that $B_1$ and $B_2$ live in some common space $W^{(B)},$ does not actually let us find a common space containing $\mathcal{B}(A_1, B_1)$ and $\mathcal{B}(A_2, B_2).$ I'm pretty sure that in the special case where $A_1 \cap A_2$ is dense in both $A_1$ and $A_2,$ one can actually treat these spaces as compatible with an embedding into $\mathcal{B}(A_1 \cap A_2, B_1 + B_2)$; but this requires assumptions that are too strong in general.
So instead, we will try to prove the inequality
$$\lVert T \rVert_{(A_1, A_2)_x \to (B_1, B_2)_x} \leq \lVert T \rVert_{A_1 \to B_1}^{1-x} \lVert T \rVert_{A_2 \to B_2}^{x}$$
by a direct route that uses our existing knowledge about the spaces $(A_1, A_2)_x$ and $(B_1, B_2)_x$, but without trying to interpolate the operator norms directly. This turns out to be pretty easy. The point is that the norms $\lVert \cdot \rVert_{(A_1, A_2)_x}$ and $\lVert \cdot \rVert_{(B_1, B_2)_x}$ are both defined with respect to certain choices of analytic functions on the strip, and $T$ will map analytic functions showing up in the $(A_1, A_2)_x$ norm to analytic functions showing up in the $(B_1, B_2)_x$ norm. Concretely, pick a vector $v$ in $(A_1, A_2)_x.$ Then for any $\epsilon > 0,$ there is some bounded function $v_{\epsilon}(z)$ on the strip, with $v_{\epsilon}(x) = v,$ and satisfying
$$\lVert v \rVert_{(A_1, A_2)_{x}} + \epsilon \geq \max\{ \sup_{y} \lVert v_{\epsilon}(i y) \rVert_{A_1}, \sup_{y} \lVert v_{\epsilon}(1 + i y) \rVert_{A_2}\}.$$
Now we assume that $T$ is a continuous map from $A_1 + A_2$ to $B_1 + B_2,$ and that it restricts to continuous maps $T : A_1 \to B_1$ and $T : A_2 \to B_2.$ From these assumptions, it is easy to see that $T v_{\epsilon}(z)$ is a bounded, holomorphic function on the strip that maps the left boundary into $B_1$ and the right boundary into $B_2,$ and that satisfies $T v_{\epsilon}(x) = T v.$ Consequently, we get the bound
$$\lVert T v \rVert_{(B_1 B_2)_{x}} \leq \max\{ \sup_{y} \lVert T v_{\epsilon}(i y) \rVert_{B_1}, \sup_{y} \lVert T v_{\epsilon}(1 + i y) \rVert_{B_2}\},$$
or
$$\lVert T v \rVert_{(B_1 B_2)_{x}} \leq \max\{ \sup_{y} \lVert T \rVert_{A_1 \to B_1} \lVert v_{\epsilon}(i y) \rVert_{A_1}, \sup_{y} \lVert T \rVert_{A_2 \to B_2} \lVert v_{\epsilon}(1 + i y) \rVert_{A_2}\}.$$
This is pretty close to giving us something useful, except that the left and right boundaries are treated asymetrically, so we can't productively simplify this expression. To get around this, we will repeat the same argument with the function $\lVert T \rVert_{A_1 \to B_1}^{z - x} \lVert T \rVert_{A_2 \to B_2}^{x - z} T v_{\epsilon}(z),$ which will give us the nicer bound
$$\lVert T v \rVert_{(B_1, B_2)_{x}} \leq \lVert T \rVert_{A_1 \to B_1}^{1-x} \lVert T \rVert_{A_2 \to B_2}^{x} \max\{ \sup_{y} \lVert v_{\epsilon}(i y) \rVert_{A_1}, \sup_{y} \lVert v_{\epsilon}(1 + i y) \rVert_{A_2}\},$$
and our choice of $v_{\epsilon}$ gives
$$\lVert T v \rVert_{(B_1, B_2)_{x}} \leq \lVert T \rVert_{A_1 \to B_1}^{1-x} \lVert T \rVert_{A_2 \to B_2}^{x} \left( \lVert v \rVert_{(A_1, A_2)_x} + \epsilon\right),$$
and taking $\epsilon \to 0$ gives the desired result.
Before going to examples, let's pause to think about an interesting conceptual point. We first proved a vector-interpolation bound
$$\lVert v \rVert_{(A_1, A_2)_x} \leq \lVert v \rVert_{A_1}^{1-x} \lVert v \rVert_{A_2}^{x},$$
then a map-interpolation bound
$$\lVert T \rVert_{(A_1, A_2)_x \to (B_1, B_2)_x} \leq \lVert T \rVert_{A_1 \to B_1}^{1-x} \lVert T \rVert_{A_2 \to B_2}^{x}.$$
The techniques we used to prove these two bounds were very similar, but as emphasized above, the second inequality is not just a special case of the first. However, the first equality is just a special case of the second! This is because any vector $b$ in a Banach space $B$ can be thought of as a bounded map $\mathbb{C} \to B,$ just defined by mapping $1 \in \mathbb{C}$ to $b \in B.$ So in our second inequality, let's take $A_1 = A_2 = \mathbb{C}.$ Then any vector $v$ in $B_1 + B_2$ forms a bounded map from $\mathbb{C} = A_1 + A_2$ to $B_1 + B_2,$ and the second inequality becomes
$$\lVert v \rVert_{\mathbb{C} \to (B_1, B_2)_x} \leq \lVert v \rVert_{\mathbb{C} \to B_1}^{1-x} \lVert v \rVert_{\mathbb{C} \to B_2}^{x},$$ which can be rewritten in terms of vector norms as
$$\lVert v \rVert_{(B_1, B_2)_x} \leq \lVert v \rVert_{B_1}^{1-x} \lVert v \rVert_{B_2}^{x}.$$
This is the vector-interpolation inequality.
5. Application: Lp spaces
Let's apply complex interpolation to $L^p$ spaces. Consider an arbitrary measure space, with $L^p$ the Banach space of measurable functions with norm
$$\lVert f \rVert_{p} \equiv \left( \int d\mu\, |f|^{p} \right)^{1/p}.$$
And of course we'll take the usual convention that $L^{\infty}$ is endowed with the "essential supremum" norm.
An important point is that on an arbitrary measure space, the $L^p$ spaces do not nest with one another in any particular way. However, there is a natural vector space containing all $L^p$ spaces, which is the space $L^0$ of all measurable functions. There is a natural topology on this space which can be framed in many different ways; a convenient one is to say that one has the $L^0$ convergence $f_n \to f$ if, within any finite-measure set, the measure of the subset with $|f_n - f| > \epsilon$ goes to zero for any $\epsilon$ at large $n.$ This is very much like a topology of uniform convergence, with "finite measure" playing the role of "compact." All embeddings $L^p \subseteq L^0$ are continuous, for if one has
$$\lVert f_n - f \rVert_{p} \to 0,$$
then there can't be any finite-measure set in which $f_n$ stays uniformly bounded away from $f.$
Now let's try to compute $(L^p, L^q)_x$. We start by considering functions $f^{(p)} \in L^p$ and $f^{(q)} \in L^q.$ Our goal is to compute $\lVert f^{(p)} + f^{(q)} \rVert_{(L^p, L^q)_x}.$ So let $\chi(z)$ be a bounded function on the strip with $\chi(x) = f^{(p)} + f^{(q)}$, such that the left boundary maps into $L^p$ And the right maps into $L^q.$ For any such function, we get an upper bound on the interpolation norm; the game is to pick a nice function that gives us a lot of data about the interpolation norm.
Let's simplify our task by starting in the case $p, q < \infty.$ Then it suffices to take $f^{(p)} + f^{(q)}$ to be a simple function $s$ with finite-measure support, since such functions are dense in $L^p$ and in $L^q,$ hence in $L^p + L^q.$ In this setting, a natural choice of $\chi$ is to fix some real numbers $a$ and $b$ and look at the function
$$\chi(z) = s |s|^{a z + b}.$$
We need a few constraints. The first is
$$\chi(x) = s, \quad \text{hence} \quad b = - a x.$$
The second is that on the left boundary of the strip we want to get some compatibility with the $L^p$ norm of $s$, and on the right with the $L^q$ norm of $s.$ A natural choice is to impose
$$(b + 1) p = (a + b + 1) q.$$
Together with $b = - a x,$ this gives
$$(b + 1)p = \frac{p q}{q + (p-q)x},$$
or
$$\frac{1}{(b + 1)p} = \frac{1-x}{p} + \frac{x}{q}.$$
So if we define a new exponent $r$ by
$$\frac{1}{r} = \frac{1-x}{p} + \frac{x}{q},$$
then we have
$$\max\left( \sup_{y} \lVert \chi(i y) \rVert_{p}, \sup_{y} \lVert \chi(1 + i y) \rVert_{q}\right) = (\lVert s \rVert_{r})^r,$$
hence
$$\lVert s \rVert_{(L^p, L^q)_x} \leq (\lVert s \rVert_r)^r.$$
But we can always divide $s$ by $\lVert s \rVert_{r}$ and use linearity to simplify this expression to
$$\lVert s \rVert_{(L^p, L^q)_x} \leq \lVert s \rVert_r.$$
This gives a one-way bound between $(L^p, L^q)_x$ and the $L^r$ norm with $1/r = (1-x)/p + x/q.$ To prove equality, we need to show the converse inequality. This is accomplished via the duality relation
$$\lVert s \rVert_{r} = \sup_{\lVert s' \rVert_{r'} = 1} |\int d\mu\, s' s|.$$
For any $\epsilon,$ there exists some bounded $\chi(z)$ with $\chi(x) = s$ and
$$\lVert s \rVert_{(L^p, L^q)_x} \geq \max(\sup_{y} \lVert \chi(i y)\rVert_p, \sup_{y} \lVert \chi(1 + i y)\rVert_q) - \epsilon.$$
With this choice of $\chi(z)$, and with $c$ being a real number we will fix later, we have
$$|\int d\mu\, s' s| = \left| \int d\mu\, s' |s'|^{c(z-x)} \chi(z) \right|_{z=x}.$$
It is easy to check that with the choice $c = (p' - r')/p' x,$ the function $s' |s'|^{c(z-x)}$ has unit $L^{p'}$ norm on the left boundary of the strip, and unit $L^{q'}$ norm on the right boundary of the strip. So using Holder's inequality and the Phragmen-Lindelof principle, we have
$$|\int d\mu\, s' s| \leq \max\left( \sup_{y} \lVert \chi(i y)\rVert_{p}, \sup_{y} \lVert \chi(1+iy)_q\rVert\right) \leq \lVert s \rVert_{(L^p, L^q)_x},$$
and taking $\epsilon \to 0$, then optimizing over $s'$, gives the desired statement
$$\lVert s \rVert_{(L^p, L^q)_x} = \lVert s \rVert_{L^r}$$
with
$$\frac{1}{r} = \frac{1-x}{p} + \frac{x}{q}.$$
To conclude, let's remember that above, we made an assumption $p, q < \infty$ in order to let ourselves work with simple functions of finite-measure support. In the setting where $p<\infty$ but $q=\infty,$ we have to be a little more careful with our logical reasoning. The functions that are dense in $L^p + L^{\infty}$ are the full set of simple functions, not just those of finite-measure support. This introduces some subtleties in the above arguments, because for a simple function $s$ with infinite-measure support, one can't cavalierly integrate $s$ against any exponent and expect to get a reasonable answer.
Now, by the arguments given in the $p,q < \infty$ case, one can still conclude that for any simple function $s$ of finite-measure support, we do have
$$\lVert s \rVert_{(L^p, L^\infty)_x} = \lVert s \rVert_{L^{p/(1-x)}}.$$
And we can get a Banach space by taking the closure of such functions, which is just the Banach space $L^{p/(1-x)}.$ The only concern is whether this actually captures the full space $(L^p, L^\infty)_x$; might we secretly have
$$L^{p/(1-x)} \subsetneq (L^p, L^{\infty})_x?$$
The answer is no, and the key is that the argument giving $(L^p, L^q)_x \subseteq L^{r}$ only used our ability to approximate $L^{r'}$ by simple functions, which is certainly the case for $r=p/(1-x).$ For if $f$ is any function in $L^p + L^{\infty},$ we can express its $r$-norm by optimizing over simple functions of finite-measure support:
$$\lVert f \rVert_{r} = \sup_{\lVert s' \rVert_{r'} = 1} |\int d\mu\, s' f|.$$
And if $f$ is in $(L^p, L^\infty)_x,$ then there's some function $\chi(z)$ that maps $i y$ uniformly into $L^p$ and $1 + i y$ uniformly into $L^{\infty}$, and with $\chi(x) = f.$ We then have no problem writing
$$\int d\mu\, s' f = \int d\mu\, s' |s'|^{(z-x)/(p-(1-x))} \chi(z) \quad \text{at $z=0$},$$
and getting that this is upper bounded by $\lVert \chi \rVert_{p}$ on the strip's left boundary and $\lVert \chi \rVert_{\infty}$ on the right boundary; this gives, by Phragmen-Lindelof, an $s'$-independent bound on $\int s' s,$ which tells us that $\lVert f \rVert_{r}$ is finite.
Putting all this together, we learn that we genuinely have
$$(L^p, L^q)_x = L^r$$
with
$$\frac{1}{r} = \frac{1-x}{p} + \frac{x}{q},$$
and this result holds even for $q=\infty.$
6. Application: Sobolev spaces
Now we'll prove a simple interpolation result for Sobolev spaces. These spaces seemed pretty scary to me when I first learned about them, but after spending some time with them in my research, I've decided they're not so bad. The idea is that on $\mathbb{R}^{D},$ there is a classic and easy-to-prove statement that a compactly supported function $f$ is smooth if and only if its Fourier transform decays at infinity faster than any rational function. On the "position-space" side of things, it makes sense to relax the regularity of $f$ by putting it in a class $C^k$ with finitely many derivatives. On the "momentum-space" side of things, this corresponds to restricting to Fourier transforms that satisfy decay estimates like
$$(1+|p|)^k |\hat{f}(p)| \leq Z.$$
In practice, it is often very useful to consider a different regularity condition, which has to do with integrability in momentum space. In particular, one might look at the space of functions for which we have
$$\int dp\, \left( (1 + |p|)^s |\hat{f}(p)| \right)^2 < \infty.$$
Functions of this kind are said to be in the Sobolev space $H^{s}$; nontrivially, one can show that this is a Banach space with respect to the norm
$$\lVert f \rVert_{H^s} = \sqrt{\int dp\, \left| (1 + |p|)^s |\hat{f}(p)| \right|^2 }.$$
There's a natural duality pairing between $H^s$ and $H^{-s}$ effected just by the $L^2$ inner product, since if $f^{(s)}$ is in $H^s$ while $g^{(-s)}$ is in $H^{-s},$ then we have (by Holder's inequality)
$$ |\int dp\, \hat{f}^{(s)}(p)^* \hat{g}^{(-s)}(p)| = |\int dp\, (1 + |p|)^s \hat{f}^{(s)}(p)^* (1 + |p|)^{-s} \hat{g}^{(-s)}(p)| \leq \lVert \hat{f}^{(s)} \rVert_{H^s} \lVert \hat{g}^{(-s)} \rVert_{H^{-s}}.$$
Moreover, one can obtain the usual duality relation
$$\lVert f \rVert_{H^s} = \sup_{\lVert g \rVert_{H^{-s}} = 1} |\int dp\, \hat{f}(p)^* \hat{g}(p)|,$$ since we have already shown the $\geq$ inequality, and saturation is achieved by taking $\hat{g} = (1 + |p|)^{2 s} f / \lVert f \rVert_{H^s}.$
The key claim about complex interpolation of Sobolev spaces is that for $s_1, s_2 \in \mathbb{R},$ we have
$$(H^{s_1}, H^{s_2})_x = H^{(1-x) s_1 + x s_2}.$$
In other words, complex interpolation of Sobolev spaces is linear.
The proof is basically the same as the proof of $L^p$ interpolation once you figure out exactly what your tools are. The important point is that the space of Schwartz functions is dense in every $H^s.$ So we can take $f$ to be any such function, and we just need to compute
$$\lVert f \rVert_{(H^{s_1}, H^{s_2})_x} = \lVert f \rVert_{H^{(1-x) s_1 + x s_2}}.$$
For the $\leq$ direction, we let consider the function $\chi(z)$ defined by
$$[\hat{\chi}(z)](p) = \hat{f}(p) (1 + |p|)^{(s_1 - s_2) (z-x)}.$$
At $z=iy,$ one has
$$\lVert \hat{\chi}(z)\rVert_{s_1} = \lVert f \rVert_{(1-x) s_1 + x s_2},$$
and the same bound works at $z=1+i y.$ So we end up with
$$\lVert f \rVert_{(H^{s_1}, H^{s_2})_x} \leq \lVert f \rVert_{(1-x) s_1 + x s_2}.$$
For the $\geq$ direction, we proceed by duality exactly as in the $L^p$ setting, but now using that $H^{-s}$ is the dual of $H^{s}.$ I leave this last step as an exercise to the dedicated reader.
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