I recently made a post about complex interpolation, which is a machine for taking two Banach spaces, $A$ and $B$, and spitting out a continuous family of "interpolating" Banach spaces $(A,B)_x.$ The properties of these spaces were:
- For any $v$ in the intersection $A \cap B$, one has a bound $$\lVert v \rVert_{(A,B)_x} \leq \lVert v \rVert_A^{1-x} \lVert v \rVert_B^x.$$
- The above characterization descends to maps; i.e., for continuous maps $T : A_1 \to B_1$ and $T : A_2 \to B_2$ that are "compatible with the sum $A_1 + A_2$" in some technical sense, one obtains an interpolating bound $$\lVert T \rVert_{(A_1, A_2)_x \to (B_1, B_2)_x} \leq \lVert T \rVert_{A_1 \to B_1}^{1-x} \lVert T \rVert_{A_2 \to B_2}^{x}.$$
There's a completely different technique for creating "interpolating bounds" that seems to be well known to a certain class of mathematicians, but that I haven't seen used in physics. This is the method of "real interpolation." I'm sure there are applications in physics, and maybe one reason more haven't been found is that real interpolation is conceptually thornier than complex interpolation. This blog post explains the theory of real interpolation; I wrote it to satisfy my own mathematical curiosity, with the hope that maybe I'll find a physics application later.
There are a few main differences between real interpolation and complex interpolation. The "upsides" are:
- Real interpolation allows $A$ and $B$ to be quasi-Banach spaces. These are complete spaces with respect to a "quasi-norm," meaning that the triangle inequality only holds up to some uniform constant: $$\lVert x + y \rVert_{A} \leq Z_A (\lVert x \rVert_{A} + \lVert y \rVert_{A}).$$ A nice example is provided by the $L^p$ spaces with $0 < p < 1$, which satisfy $$\lVert f + g\rVert_{p} \leq 2^{\frac{1}{p} - 1} (\lVert f \rVert_{p} + \lVert g \rVert_p).$$
- Real interpolation gives a two-parameter interpolating family $(A, B)_{x; q}.$ (We remind ourselves that we're doing real interpolation, and not complex interpolation, because we've put the $q$ there!) When $A$ and $B$ are Banach, so is $(A,B)_{x;q}$, and similar kinds of bounds-in-$x$ that hold in the complex interpolation case will also hold in the real case.
- Real interpolation norms apply to a broader class of maps than continuous, linear operators $T : A_1 + A_2 \to B_1 + B_2$; in particular, the maps can be taken to be "sublinear." I won't talk about this any further in the present post.
The "downsides" are:
- Real interpolation norms can be harder to compute in practice due to the way the theory is formulated (optimization over a splitting, rather than optimization over holomorphic functions).
- The real interpolation norms on operators are the same as in the complex interpolation case: $$\lVert T \rVert_{(A_1, A_2)_{x;q} \to (B_1, B_2)_{x;q}} \leq \lVert T \rVert_{A_1 \to B_1}^{1-x} \lVert T \rVert_{A_2 \to B_2}^{x},$$ but the bounds on vectors are different; we generically have $$\lVert v \rVert_{(A, B)_{x;q}} \leq \frac{1}{(qx(1-x))^{1/q}}\lVert v \rVert_{A}^{1-x} \lVert v \rVert_{B}^{x}.$$ Relatedly, the interpolation between a space and itself is not trivial except in the limit $q \to \infty$; we have
$$\lVert v \rVert_{(A,A)_x} = \frac{1}{(q x(1-x))^{1/q}} \lVert v \rVert_A.$$ - Because the complex interpolation spaces are obtained using the rigid structure of holomorphicity, they can often be nicer spaces, structurally speaking, than the ones obtained by real interpolation. Even if the complex interpolation spaces show up as a sub-family of the real interpolation spaces, the fact that they were obtained using holomorphic tools often means one has stronger control over their properties.
In this post, we'll start by explaining the abstract idea of real interpolation, introducing the "$K$-functional." We'll show that the spaces obtained by $K$-functional interpolation are Banach or quasi-Banach depending on the properties of the two spaces we start with. Finally, we'll compute an explicit example of the real interpolation spaces between the $L^1$ and $L^{\infty}$ spaces coming from a generic measure space.
Table of Contents
Prerequisites: It will be helpful to have read my post on complex interpolation, as the notion of a compatible pair of Banach spaces is explained more thoroughly there.
Sources: I learned most of this material by consulting the textbook Interpolation Spaces: An Introduction. Many more details, though with a terser presentation, can be found there.
1. The K-functional
The basic idea behind real interpolation is to start with two spaces $A$ and $B$, to look at a vector that splits as $v = a + b,$ then to optimize over the choice of splitting.
As in the setting of complex interpolation, one assumes the spaces $A$ and $B$ are continuously embedded into a topological vector space $W$. For the moment, we will assume that $A$ and $B$ are Banach spaces, but we will relax this later. For a vector $v$ in $A + B$, one defines the sum norm
$$\lVert v \rVert_{A + B} = \inf\{\lVert a \rVert_A + \lVert b \rVert_B \,|\, a + b = v\}.$$
One can also define the intersection norm
$$\lVert v \rVert_{A \cap B} = \max\{\lVert v \rVert_A, \lVert v \rVert_B\},$$
and this clearly takes finite values only if both the $A$ and $B$-norms are finite, i.e., if and only if we have $v \in A \cap B.$
We'll now take a sharper look at the $A+B$ norm and look at some different ways of optimizing over the splitting. In particular, we'll introduce a weight, $t,$ which can be adjusted to give the $A$ or the $B$ norm more importance; roughly we want to recover the topology of the $A$-norm in the limit $t \to 0,$ and the $B$-norm in the limit $t \to \infty.$ For now, let's not worry if this defines a norm, and let's just define the quantity
$$\lVert v \rVert_{A+B;t} = \inf\{\lVert a \rVert_A + t \lVert b \rVert_B \,|\, a + b = v\}.$$
This quantity has a name in the mathematics literature; it is called the $K$-functional, and we will henceforth use that notation, writing
$$K(t, v; A,B) = \inf\{\lVert a \rVert_A + t \lVert b \rVert_B \,|\, a + b = v\}.$$
Real interpolation spaces are defined by integrating over the $K$-functional in $t,$ with respect to an appropriate measure. When the measure puts a lot of weight near $t=0$ we expect to be close to the $A$ norm, and when it puts a lot of weight near $t=\infty$ we expect to be close to the $B$ norm.
Let's think about what kinds of measures we might consider. For the moment let $\mu(t)$ be a general function, and consider the integral
$$\int_{0}^{\infty} dt\, \mu(t) K(t, v; A, B).$$
What we'd like is to get the same kind of bound we had in the case of complex interpolation, i.e., for $v \in A \cap B,$ we'd like to get
$$\int_{0}^{\infty} dt\, \mu(t) K(t, v; A, B) \leq \lVert v \rVert_{A}^{1-x} \lVert v \rVert_{B}^{x}.$$
This turns out not to be possible for reasons that we'll explore later; the right-hand side above is actually going to have to pick up some $x$-and-$q$-dependent coefficient. Luckily, we can still choose $\mu(t)$ to get the usual interpolation inequality for maps like $T : A_1 + A_2 \to B_1 + B_2.$ In this case, we have, for $v \in A_1 + A_2,$ the equation
$$K(t, T v; B_1, B_2) = \inf\{\lVert b_1 \rVert_{B_1} + t \lVert b_2 \rVert_{B_2} | b_1 + b_2 = T v\}.$$
We can find an upper bound on this by restricting the infimum to pairs $b_1$ and $b_2$ that are in the image of $T$, so that we get
$$K(t, T v; B_1, B_2) \leq \inf\{\lVert T a_1 \rVert_{B_1} + t \lVert T a_2 \rVert_{B_2} | a_1 + a_2 = v\}.$$
Then using the boundedness of $T$ on $A_1 \to B_1$ and $A_2 \to B_2,$ we get
$$K(t, T v; B_1, B_2) \leq \lVert T \rVert_{A_1 \to B_1} \inf\{\lVert a_1 \rVert_{B_1} + t \frac{\lVert T \rVert_{A_2 \to B_2}}{\lVert T \rVert_{A_1 \to B_1}} \lVert a_2 \rVert_{A_2} | a_1 + a_2 = v\}.$$
In other words, we have
$$K(t, T v; B_1, B_2) \leq \lVert T \rVert_{A_1 \to B_1} K\left( \frac{\lVert T \rVert_{A_2 \to B_2}}{\lVert T \rVert_{A_1 \to B_1}} t, v; A_1, A_2\right).$$
This gives a certain kind of homogeneity of the $K$-functionals; when you shift from the "image pair" $(B_1, B_2)$ back to the "domain pair" $(A_1, A_2)$, you get a bound that involves rescaling $t$ and multiplying by a constant. Integrating with respect to the measure $\mu(t)$, then rescaling $t$ on the right-hand side, gives
$$\int_0^{\infty} dt\, \mu(t) K(t, T v; B_1, B_2) \leq \frac{(\lVert T \rVert_{A_1 \to B_1})^2}{\lVert T \rVert_{A_2 \to B_2}} \int_{0}^{\infty} dt \mu\left( \frac{\lVert T \rVert_{A_1 \to B_1}}{\lVert T \rVert_{A_2 \to B_2}} t \right) K\left( t, v; A_1, A_2\right).$$
Seeking symmetry between the left- and right-hand sides, one wants $\mu(t)$ to satisfy $\mu(c t) \propto \mu(t),$ i.e., we need $\mu(t) = \alpha t^{\beta}$ for some coefficients $\alpha, \beta.$ In defining a norm, overall constants aren't really important, so we might as well set $\alpha=1,$ and write
$$\frac{\int_0^{\infty} dt\, t^{\beta} K(t, T v; B_1, B_2)}{\int_0^{\infty} dt\, t^{\beta} K\left( t, v; A_1, A_2\right)} \leq \lVert T \rVert_{A_1 \to B_1}^{2+\beta} \lVert T \rVert_{A_2 \to B_2}^{-1-\beta}.$$
Matching to the usual interpolation inequality, we may as well parametrize $\beta=-1-x;$ then we get
$$\frac{\int_0^{\infty} \frac{dt}{t}\, t^{-x} K(t, T v; B_1, B_2)}{\int_0^{\infty} \frac{dt}{t}\, t^{-x} K\left( t, v; A_1, A_2\right)} \leq \lVert T \rVert_{A_1 \to B_1}^{1-x} \lVert T \rVert_{A_2 \to B_2}^{x}.$$
This defines the so-called $q=1$ real interpolation norm,
$$\lVert v \rVert_{(A, B)_{x;1}} \equiv \int_{0}^{\infty} \frac{dt}{t}\, t^{-x} K(t, v;A,B).$$
We will explore its properties below, in particular we will show norm properties and completeness that are inherited from $\lVert \cdot \rVert_A$ and $\lVert \cdot \rVert_B.$ But first we will embed this story within a larger class of norms $(A,B)_{x;q}.$
The idea is to note that what really mattered was the homogeneity of $K$ as one moves from the space $A_1 + A_2$ to $B_1 + B_2.$ If we raised $K$ to a power $q$ before integrating, we could compensate for "changing the homogeneity" by adding additional factors of $t$ to $\mu(t).$ It is easy to see that one has
$$\frac{\left(\int_0^{\infty} \frac{dt}{t}\, t^{-q x} K(t, T v; B_1, B_2)^q\right)^{1/q}}{\left(\int_0^{\infty} \frac{dt}{t}\, t^{-qx} K\left( t, v; A_1, A_2\right)^q\right)^{1/q}} \leq \lVert T \rVert_{A_1 \to B_1}^{1-x} \lVert T \rVert_{A_2 \to B_2}^{x}.$$
This inspires the family of real-interpolation quantities
$$\lVert v \rVert_{(A,B)_{x;q}} = \left(\int_0^{\infty} \frac{dt}{t}\, t^{-q x} K(t, v; A, B)^q\right)^{1/q}.$$
We will explore the properties of these quantities in the next section. Note that in the limit $q \to \infty,$ we get
$$\lVert v \rVert_{(A,B)_{x;\infty}} = \sup_{t > 0} t^{-x} K(t, v; A,B) = \sup_{t > 0} t^{-x} \inf\{\lVert a \rVert_{A} + t \lVert b \rVert_{B} \,|\, a + b = v\}.$$
2. Properties of real interpolation spaces
Now let's take a closer look at the quantity we have defined,
$$\lVert v \rVert_{(A,B)_{x;q}} = \left(\int_0^{\infty} \frac{dt}{t}\, t^{-q x} K(t, v; A, B)^q\right)^{1/q}.$$
We will assume that $A$ and $B$ are quasi-Banach spaces, meaning that we have constants $Z_A$ and $Z_B$ with
$$\lVert a_1 + a_2 \rVert_{A} \leq Z_A (\lVert a_1\rVert_{A} + \lVert a_2 \rVert_{A})$$
and
$$\lVert b_1 + b_2 \rVert_{B} \leq Z_B (\lVert b_1\rVert_{B} + \lVert b_2 \rVert_{B}),$$
and that $A$ and $B$ are complete with respect to the associated topologies. We will now show a few facts:
- $(A,B)_{x;q}$ is a quasi-Banach space for every $x \in (0,1)$ and every $q > 0.$ If $A$ and $B$ started a Banach spaces, then $(A,B)_{x;q}$ is in fact Banach for $q \geq 1.$
- One has continuous embeddings $$A \cap B \subseteq (A,B)_{x;q} \subseteq A + B.$$ In fact, one additionally has the continuous embedding $$(A,B)_{x;q_1} \subseteq (A,B)_{x; q_2}$$ for $q_1 \leq q_2,$ with concrete inequality $$\lVert v \rVert_{(A,B)_{x;q_2}} \leq (q_1 x(1-x))^{(q_2 - q_1)/q_1 q_2} \lVert v \rVert_{(A,B)_{x;q_1}}.$$
- One generically has $$\lVert v \rVert_{(A,B)_{x;q}} \leq \frac{1}{(q x(1-x))^{1/q}} \lVert v \rVert_{A}^{1-x} \lVert v \rVert_{B}^x$$ and $$\lVert T \rVert_{(A_1,A_2)_{x;q} \to (B_1, B_2)_{x;q}} \leq \lVert T \rVert_{A_1 \to B_1}^{1-x} \lVert T \rVert_{A_2 \to B_2}^{x}.$$
Let us start by examining what version of a triangle inequality might hold on $(A,B)_{x;q}.$ We have
$$K(t, v_1 + v_2; A, B) = \inf\{\lVert a \rVert_{A} + t \lVert b \rVert_{B} \,|\, a + b = v_1 + v_2\},$$
and by restricting in the infimum to configurations with $v_1 = a_1 + b_1$ and $v_2 = a_2 + v_2,$ we get
$$K(t, v_1 + v_2; A, B) \leq \inf\{\lVert a_1 + a_2 \rVert_{A} + t \lVert b_1 + b_2 \rVert_{B} \,|\, a_1 + b_1 = v_1, a_2 + b_ 2 = v_2\}.$$
Applying the quasi-triangle inequalities gives
$$K(t, v_1 + v_2; A, B) \leq \inf\{Z_A (\lVert a_1 \rVert_{A} + \lVert a_2 \rVert_{A}) + t Z_B (\lVert b_1 \rVert_{B} + \lVert b_2 \rVert_{B}) \,|\, a_1 + b_1 = v_1, a_2 + b_ 2 = v_2\},$$
then we can split up the infimum to get
$$K(t, v_1 + v_2; A, B) \leq \inf\{Z_A \lVert a_1 \rVert_{A} + t Z_B \lVert b_1 \rVert_B \,|\, a_1 + b_1 = v_1\} + \inf\{Z_A \lVert a_2 \rVert_{A} + t Z_B \lVert b_2 \rVert_{B}) \,|\, a_2 + b_ 2 = v_2\},$$
which we can rewrite as
$$K(t, v_1 + v_2; A, B) \leq Z_A K\left( \frac{Z_B}{Z_A} t, v_1; A, B \right) + Z_A K\left( \frac{Z_B}{Z_A} t, v_2; A, B\right).$$
Integrating, we get
$$\lVert v_1 + v_2 \rVert_{(A,B)_{x;q}} \leq \left( \int_0^{\infty} \frac{dt}{t}\, t^{-q x} \left( Z_A K\left( \frac{Z_B}{Z_A} t, v_1; A, B \right) + Z_A \left( \frac{Z_B}{Z_A} t, v_2; A, B\right) \right)^q \right)^{1/q},$$
and rescaling $t$ gives
$$\lVert v_1 + v_2 \rVert_{x;q} \leq Z_A^{1-x} Z_B^{x} \left( \int_0^{\infty} \frac{dt}{t}\, t^{-q x} \left( K\left( t, v_1; A, B \right) + \left(t, v_2; A, B\right) \right)^q \right)^{1/q}.$$
We can think of the integral as representing the $L^q$ (quasi)norm with respect to the measure $dt/t.$ So for $q \geq 1$ we get
$$\lVert v_1 + v_2 \rVert_{(A,B)_{x;q}} \leq Z_A^{1-x} Z_B^{x} (\lVert v_1 \rVert_{(A,B)_{x;q}} + \lVert v_2 \rVert_{(A,B)_{x;q}}),$$
and for $0 < q < 1$ we get
$$\lVert v_1 + v_2 \rVert_{(A,B)_{x;q}} \leq 2^{\frac{1}{q} - 1} Z_A^{1-x} Z_B^{x} (\lVert v_1 \rVert_{(A,B)_{x;q}} + \lVert v_2 \rVert_{(A,B)_{x;q}}).$$
So for any $q,$ we always get a quasi-triangle inequality. Note that if $A$ and $B$ had started as normed spaces, so that we had $Z_A = Z_B = 1,$ then we would get an exact triangle inequality for $q \geq 1,$ but not for $0 < q < 1.$
Now let us address the question of completeness for $(A,B)_{x;q}.$ To do this, we will first show the continuous embeddings
$$A \cap B \subseteq (A, B)_{x;q} \subseteq A + B.$$
For $v \in A + B,$ we clearly have (from the definition of $K$)
$$t \leq 1: K(t,v;A,B) \geq t \lVert v \rVert_{A + B}$$
and
$$t \geq 1: K(t,v;A,B) \geq \lVert v \rVert_{A+B}.$$
Adding these two inequalities, we get
$$K(t, v; A,B) \geq \Theta(1-t) t \lVert v \rVert_{A+B} + \Theta(t-1) \lVert v \rVert_{A+B}.$$
Multiplying by $t^{-x}$ and taking the $L^q$ norm with respect to $dt/t,$ we get
$$\lVert v \rVert_{A+B} \leq (q x(1-x))^{1/q} \lVert v \rVert_{(A,B)_{x;q}},$$
which proves the continuous embedding $(A,B)_{x;q} \subseteq A + B.$
On the other hand, for $v \in A \cap B,$ the infimum in the definition of $K$ can be upper bounded by putting all of the "weight" of $v$ in either $A$ or $B$, leading to the inequality
$$K(t,v;A,B) \leq \min(\lVert v \rVert_A, t \lVert v \rVert_B).$$
Multiplying by $t^{-x}$ and taking $L^q$ norms with respect to $dt/t,$ we get
$$\lVert v \rVert_{(A,B)_{x;q}} \leq \frac{1}{(q x(1-x))^{1/q}} \lVert v \rVert_{A}^{1-x} \lVert v \rVert_{B}^{x}.$$
Not only does this show the continuous inclusion $(A,B)_{x;q} \subseteq A \cap B,$ it also gives us a version of the norm-interpolation equality that held for complex interpolation norms, with the caveat that we now have this funny prefactor of $(q x(1-x))^{-1/q}.$ Depending on the specific values of $q$ and $x,$ the prefactor $1/(q x(1-x))^{1/q}$ may be bigger than or less than one; in the case that it is less than one, the norm-interpolation bound on vectors is stronger than in the case of complex interpolation; in the other case, it is weaker. Note that this is in contrast to the map-interpolation bound we used earlier to motivate our definitions of real interpolation; that bound had no prefactor. In other words, the bounds on interpolation of vectors may be qualitatively different than the bounds on interpolation of maps; we will have more to say about this later.
Let's return to completeness of $(A,B)_{x;q}.$ Because of the continuous embedding $(A,B)_{x;q} \subseteq A + B,$ any Cauchy sequence in $(A,B)_{x;q}$ is also Cauchy in $A+B.$ So given a Cauchy sequence $v_n \in (A,B)_{x;q},$ we know there is a unique vector $v \in A + B$ with
$$\lVert v_n - v \rVert_{A+B} \to 0.$$
We need to show that the convergence $v_n \to v$ also happens with respect to the norm $(A,B)_{x;q}.$ We have
$$(\lVert v_n - v \rVert_{(A,B)_{x;q}})^q = \int_0^{\infty} \frac{dt}{t}\, (t^{-x} K(t,v_n - v;A,B))^{q}.$$
Pointwise in $t$, we have the convergence
$$K(t,v_n - v;A,B) = \lim_{m \to \infty} K(t,v_n - v_m)$$ which follows from the reverse quasi-triangle inequality for the $K$-functional, together with the bound $K(t, v; A, B) \leq \max(1,t) \lVert v \rVert_{A+B}.$
We can now use a standard real-analytic trick and apply Fatou's lemma to get
$$(\lVert v_n - v \rVert_{(A,B)_{x;q}})^q \leq \liminf_{m \to \infty} \int_0^{\infty} \frac{dt}{t}\, (t^{-x} K(t,v_n - v_m;A,B))^{q} = \liminf_{m \to \infty} (\lVert v_n - v_m\rVert_{(A,B)_{x;q}})^q.$$
The Cauchy property of the $v_n$ sequence with respect to $(A,B)_{x;q}$ then gives convergence of the left-hand side at large $n.$
Now we know that the $(A,B)_{x;q}$ spaces are (quasi)-Banach spaces obeying a norm-interpolation inequality
$$\lVert v \rVert_{(A,B)_{x;q}} \leq \frac{1}{(q(1-x)x)^{1/q}} \lVert v \rVert_{A}^{1-x} \lVert v \rVert_{B}^{x}.$$
Because of the way we designed these spaces in the previous section, we also expect that they should give an interpolation inequality for maps. This is easy to show, just by following the same logic we did when we constructed these spaces in the first place. Given a linear, bounded $T : A_1 + A_2 \to B_1 + B_2,$ with bounded restrictions $T : A_1 \to B_1$ and $T : A_2 \to B_2,$ one has
$$K(t, T v;B_1, B_2) = \inf\{\lVert b_1 \rVert_{B_1} + \lVert b_2 \rVert_{B_2} \,|\, b_1 + b_2 = T v\},$$
and by restricting the optimization to $b_1$ and $b_2$ to be of the form $T a_1$ and $T a_2$ with $a_1 + a_2 = v,$ we get
$$K(t, T v;B_1, B_2) \leq \inf\{\lVert T \rVert_{A_1 \to B_1} \lVert a_1 \rVert_{A_1} + \lVert T \rVert_{A_2 \to B_2} \lVert a_2 \rVert_{A_2} \,|\, a_1 + a_2 = v\},$$
and rearranging a bit gives
$$K(t, T v;B_1, B_2) \leq \lVert T \rVert_{A_1 \to B_1} K\left( t \frac{\lVert T \rVert_{A_2 \to B_2}}{\lVert T \rVert_{A_1 \to B_1}}, v, A_1, A_2\right).$$
Multiplying by $t^{-x}$ and taking an $L^q$ norm with respect to $dt/t$ gives
$$\lVert T \rVert_{(A_1, A_2)_x \to (B_1, B_2)_x} \leq \lVert T \rVert_{A_1 \to B_1}^{1-x} \lVert T \rVert_{A_2 \to B_2}^x,$$
with no additional subtleties when the spaces are assumed only to be quasi-Banach instead of Banach.
Now let us address the asymmetry between the interpolation bounds for vectors in $(A,B)_x$ and the interpolation bounds for maps from $(A_1, A_2)_x$ to $(B_1, B_2)_x.$ The norm bounds have an additional $q$-and-$x$-dependent prefactor that is not there for maps (though it does go away in the limit $q \to \infty$). One reason this might seem a little strange is that in the complex interpolation case, there is an absolute symmetry between the vector-interpolation and map-interpolation bounds. Indeed, in that post, I explained that the vector-interpolation bounds could be thought of as a special case of map-interpolation bounds, by thinking of a vector in a space $A$ as a bounded map from $\mathbb{C}$ to $A$. What goes wrong with that logic in the setting of real interpolation?
The issue is that for real interpolation, interpolation between a space and itself can be a nontrivial task. In particular, $(A,A)_{x;q}$ is not simply $A$, except in the case $q \to \infty.$ The same basic manipulation we have already done several times in fact show
$$\lVert v \rVert_{(A,A)_{x;q}} = \frac{1}{(q x (1-x))^{1/q}} \lVert v \rVert_{A}.$$
So the map interpolation bound applied to a vector $v \in A + B,$ thought of as a map $v : \mathbb{C} \to A + B,$ gives
$$\lVert v \rVert_{(\mathbb{C}, \mathbb{C})_{x,q} \to (A,B)_{x;q}} \leq \lVert v \rVert_A^{1-x} \lVert v \rVert_{B}^x,$$
i.e.,
$$\frac{\lVert v \rVert_{(A,B)_{x;q}}}{\lVert 1 \rVert_{(\mathbb{C}, \mathbb{C})_{x,q}}} \leq \lVert v \rVert_A^{1-x} \lVert v \rVert_{B}^x.$$
Finally, plugging in our formula for $(\mathbb{C}, \mathbb{C})_{x,q},$ we get
$$\lVert v \rVert_{(A,B)_{x;q}} \leq \frac{1}{(q x(1-x))^{1/q}} \lVert v \rVert_A^{1-x} \lVert v \rVert_{B}^x,$$
which is the usual vector-interpolation bound.
It seems conceptually interesting that in the case of real interpolation, vector bounds are qualitatively different from map bounds, and that interpolation of a space to itself introduces a coefficient into the norm. We can't fix this by rescaling the norms on $(A,B)_{x;q}$ by $(q x(1-x))^{1/q},$ because then we'd mess up the interpolation bounds on maps. This seems like a genuinely important difference between real and complex interpolation; real interpolation gives you a two-parameter family of norms instead of a one-parameter family, but the norms have funny artifacts in the form of additional coefficients.
To close this section, let's treat the issue of how the $(A,B)_{x;q}$ norms are related to one another. There will generally not be any relation between norms at different values of $x$, but there will be a nesting in $q.$ For $q_1 < q_2,$ we will have the continuous embedding
$$(A,B)_{x;q_1} \subseteq (A,B)_{x; q_2}.$$
Let's prove this.
The idea is to note that there is an inequality relating the $K$-functional to the $(A,B)_{x;q}$ norm, which we obtain by starting with the observation
$$K(t,v;A,B) = \inf\{\lVert a \rVert_{A} + t \lVert b \rVert_B\} = \inf\{\lVert a \rVert_{A} + \frac{t}{s} s \lVert b \rVert_B\} \leq \max(1,t/s) K(s,v;A,B).$$
We can rearrange this as
$$\frac{1}{\max(1,t/s)} K(t,v;A,B) \leq K(s,v;A,B).$$
Multiplying by $s^{-x}$ and taking $L^q(ds/s)$ norms on both sides, we get
$$\frac{t^{-x}}{(q x(1-x))^{1/q}} K(t,v;A,B) \leq \lVert v \rVert_{(A,B)_{x;q}},$$
and rearranging gives
$$K(t,v;A,B) \leq (q x(1-x))^{1/q} t^{x} \lVert v \rVert_{(A,B)_{x;q}}.$$
Now if we have $q_2 \geq q_1,$ we start by writing $(A,B)_{x;q_2}$ explicitly as
$$\lVert v \rVert_{(A,B)_{x;q_2}}^{q_2} = \int_0^{\infty} \frac{dt}{t}\, t^{-q_2 x} K(t,v;A,B)^{q_2}.$$
Then we split up $K(t,v;A,B)^{q_2}$ as $K(t,v;A,B)^{q_2 - q_1} K(t,v;A,B)^{q_1},$ and apply the inequality we have just derived to obtain
$$\lVert v \rVert_{(A,B)_{x;q_2}}^{q_2} \leq \int_0^{\infty} \frac{dt}{t}\, t^{-q_2 x} t^{x(q_2 - q_1)} \left( (q_1 x(1-x))^{1/q_1} \right)^{q_2 - q_1} \lVert v \rVert_{(A,B)_{x;q_1}}^{q_2 - q_1} K(t,v;A,B)^{q_1},$$
which simplifies to
$$\lVert v \rVert_{(A,B)_{x;q_2}}^{q_2} \leq (q_1 x(1-x))^{(q_2 - q_1)/q_1} \lVert v \rVert_{(A,B)_{x;q_1}}^{q_2 },$$
hence
$$\lVert v \rVert_{(A,B)_{x;q_2}} \leq (q_1 x(1-x))^{(q_2 - q_1)/q_1 q_2} \lVert v \rVert_{(A,B)_{x;q_1}},$$
which is the continuous embedding we claimed.
3. Example: L1 and L-infinity
Because real interpolation is pretty complicated, we won't do the full-on real interpolation of $L^p$ spaces; in this post, we'll content ourselves by looking at $L^1$ and $L^{\infty}.$ These are both topologically embedded in the vector space $L^0$ of measurable functions, so they form a compatible pair. Our goal is to compute the $K$-functional $K(t,f;L^1(\mu), L^{\infty}(\mu))$ where $f$ is in $L^1(\mu) + L^{\infty}(\mu).$
Since simple functions of finite support are dense in $L^1,$ and simple functions of arbitrary support are dense in $L^{\infty},$ we know that general simple functions are dense in $L^1(\mu) + L^{\infty}(\mu).$ For a generic simple function $s,$ it is natural to split it as a sum of simple functions $s = s_1 + s_{\infty},$ where $s_1$ has finite support and $s_{\infty}$ need not.
Let's order the magnitudes of the complex numbers in the image of $s$ as $y_1, \dots, y_N,$ where we choose a decreasing order. If we start with $s_1 = 0, s_{\infty} = s,$ then we have
$$\lVert s_1 \rVert_{1} + t \lVert s_{\infty} \rVert_{\infty} = t y_1.$$
The only way to decrease $\lVert s_{\infty} \rVert_{\infty}$ is to take the largest magnitude in the image of $s$ and start shrinking it, pushing those values into $s_1$. What this looks like is taking some continuous parameter $\lambda \geq 0$ and writing
$$s_1(x) = \lambda\delta_{s(x)=y_1}$$
and
$$s_\infty(x) = s(x) - \lambda \delta_{s(x)=y_1}.$$
For sufficiently small $\lambda,$ such that we don't actually hit the next value $y_2$, we get
$$\lVert s_1 \rVert_1 + t \lVert s_{\infty} \rVert_{\infty} = \lambda \mu(|s|^{-1}(y_1)) + t (y_1 - \lambda).$$
Once $\lambda$ hits $y_1 - y_2,$ we need to start shrinking the second-largest value in the image of $s$ as well; at that point, we get
$$s_1(x) = \lambda \delta_{s(x) = y_1} + (\lambda - (y_1 - y_2)) \delta_{s(x) = y_2},$$
and $s_{\infty} = s - s_1,$ with
$$\lVert s_1 \rVert_{1} + t \lVert s_{\infty} \rVert_{\infty} = \lambda \mu(|s|^{-1}(y_1)) + (\lambda - (y_1 - y_2)) \mu(|s|^{-1}(y_2)) + t (y_1 - \lambda).$$
This kind of pattern repeats until we hit $\lambda = y_1,$ in which case all of the "weight" of $s$ has been transferred into $y_1.$ A universal-in-$\lambda$ formula is
$$\lVert s_1 \rVert_{1} + t \lVert s_{\infty} \rVert_{\infty} = \sum_{j=1}^{N} \max(0, \lambda - (y_1 - y_j)) \mu(|s|^{-1}(y_j)) + t (y_1 - \lambda),$$
and to get the $K$-functional we need to infimize over $\lambda \leq y_1.$ It's actually convenient to relabel $\lambda \to y_1 - \lambda,$ so that what we are trying to compute is
$$K(t, s; L^{1}(\mu), L^{\infty}(\mu)) = \inf_{\lambda \geq 0} \left[ \sum_{j=1}^{N} \max(0, y_j - \lambda) \mu(|s|^{-1}(y_j)) + t \lambda \right],$$
even more nicely expressed as
$$K(t, s; L^{1}(\mu), L^{\infty}(\mu)) = \inf_{\lambda \geq 0} \left[ \sum_{y_j \geq \lambda} (y_j - \lambda) \mu(|s|^{-1}(y_j)) + t \lambda \right].$$
There is a productive way to rewrite the above expression in terms of a different simple function $s^*$, which is called the "decreasing rearrangement" of $s.$ While $s$ may have lived on a general measure space, $s^*$ is a simple function on the positive real line $(0, \infty).$ The idea is that $s^*(x)$ takes all of the values in the image of $|s|$ and places them above the positive real axis, in decreasing order, with the length of each segment being given by the measure of the corresponding set in the support of $s.$ In other words, we have $$s^* : (0, \mu(|s|^{-1}(y_1))] \to y_1,$$ then $$s^*\Big(\big(\mu(|s|^{-1}(y_1),) \mu(|s|^{-1}(y_1)) + \mu(|s|^{-1}(y_2))\big]\Big) \to y_2,$$ and so on. Formally, this is
$$s^*(x) = \inf\{\lambda \,|\, \mu\{|s| \geq \lambda\} \leq x\}.$$
In terms of this function, we can rewrite the $K$-functional as
$$K(t, s; L^{1}(\mu), L^{\infty}(\mu)) = \inf_{\lambda \geq 0} \left[ \int_{s^*(x) \geq \lambda} dx\, (s^*(x) - \lambda) + t \lambda \right].$$
How do we find the infimum of this expression? The first thing to note is that the expression in brackets is convex in $\lambda,$ so if the derivative with respect to $\lambda$ is zero anywhere, then that is automatically a global minimum; otherwise, the global minimum will occur at a place where the derivative jumps discontinuously. If $\lambda$ is not in the image of $|s|,$ then an infinitesimal change in $\lambda$ does not change the range of integration, and the change in the bracketed expression is
$$(t - \mu(s^*(x) \geq \lambda)) \delta \lambda.$$
For this to be zero, there must be a value of $\lambda$ for which the measure of the set with $s^*(x) \geq \lambda$ is exactly $t$. In this case the infimum is achieved as
$$K(t, s; L^{1}(\mu), L^{\infty}(\mu)) = \int_{0}^{t} dx\, (s^*(x) - \lambda) + t \lambda = \int_0^{t} dx\, s^*(x).$$
If this does not occur, we must look for minima at values of $\lambda$ that coincide with something in the image of $s^*.$ When one increases $\lambda$ slightly, the integration range $s^*(x) \geq \lambda$ shrinks by $\mu(s^*(x) = \lambda).$ So for small positive $\delta \lambda,$ we get the change
$$(t - \mu(s^*(x) \gneq \lambda)) \delta \lambda.$$
On the other hand, shrinking $\lambda$ slightly makes no change in the measure, and we get
$$(t - \mu(s^*(x) \geq \lambda)) \delta \lambda.$$
If we are at a global minimum, then the first term is nonnegative while the second is nonpositive; this gives the constraints
$$t \geq \mu(s^*(x) \gneq \lambda)$$
and
$$t \leq \mu(s^*(x) \geq \lambda).$$
If we have equality in either expression we are in the case we have already analyzed, so we might as well assume strict inequality, in which case we learn that $t$ lies somewhere in the domain with $s^*(t) = \lambda.$ The integrand $s^*(x) - \lambda$ is zero in this whole range, so we might as well take the integral over the region $s^*(x) \geq \lambda$ to be in the range $[0, t],$ in which case we get the infimum as
$$\int_0^t dx\, s^*(x) - \lambda + t \lambda = \int_0^t dx\, s^*(x),$$
which is the same expression as in the other case. So no matter which set of assumptions is satisfied, we end up with the $K$-functional expression
$$K(t, s; L^{1}(\mu), L^{\infty}(\mu)) = \int_{0}^{t} dx\, s^*(x).$$
This lets us write the interpolation norm of simple functions as
$$\lVert s \rVert_{(L^1, L^{\infty})_{x;q}} = \left[ \int_0^{\infty} \frac{dt}{t}\, \left( t^{-x}\int_{0}^{t} du\, s^*(u) \right)^q \right]^{1/q}.$$
In the special case $q=1,$ we can exchange the order of integration to obtain
$$\lVert s \rVert_{(L^1, L^{\infty})_{x;1}} = \int_{0}^{\infty} du\, \int_u^{\infty} \frac{dt}{t}\, t^{-x} s^*(u) = \frac{1}{x} \int_0^{\infty} du\, u^{-x} s^*(u).$$
The integral here has a name. For $p$ and $q$, the Lorentz norm $L^{p,q}$ is defined by
$$\lVert s \rVert_{p,q}= \left( \int_0^{\infty} \left(\frac{du}{u}\, u^{1/p} s^*(u)\right)^{q} \right)^{1/q}.$$
Matching terms, we see that we have
$$\lVert s \rVert_{(L^1, L^{\infty})_{x;1}} = \frac{1}{x} \lVert s \rVert_{L^{1/(1-x),1}}.$$
So the interpolation of $L^1$ and $L^{\infty}$, in the special case $q=1,$ naturally led us to the Lorentz space $L^{1/(1-x),1}.$ This is a nice thing we got out of real interpolation; even if we didn't know anything about the Lorentz spaces, we would have produced the relevant Banach norm!
There's more that can be said away from $q=1.$ In fact, one does find that the $(L^1, L^{\infty})_{x;q}$ space is the same as the Lorentz space $L^{1/(1-x),q}.$ But proving this turns out to be a little subtle, because one doesn't actually end up with an exact proportionality between the two norms; instead, the two norms end up being mutually bounded, meaning that they induce the same topology.
To see how this works, let's examine more closely the expression
$$\lVert s \rVert_{(L^1, L^{\infty})_{x;q}} = \left[ \int_0^{\infty} \frac{dt}{t}\, \left( t^{-x}\int_{0}^{t} du\, s^*(u) \right)^q \right]^{1/q}.$$
First let's rescale $u$ to move $t$ out of the bound of integration, so we get
$$\lVert s \rVert_{(L^1, L^{\infty})_{x;q}} = \left[ \int_0^{\infty} \frac{dt}{t}\, \left(\int_{0}^{1} du\, t^{1-x} s^*(t u) \right)^q \right]^{1/q}.$$
We can think of this as the $L^q$ norm, with respect to $dt/t,$ of some integral in $u.$ Using the triangle inequality for the $L^q$ norm, we get
$$\lVert s \rVert_{(L^1, L^{\infty})_{x;q}} \leq \int_0^1 du\, \left[ \int_0^{\infty} \frac{dt}{t}\, \left( t^{1-x} s^*(t u) \right)^q \right]^{1/q}.$$
Rescaling $t \to t/u$ gives
$$\lVert s \rVert_{(L^1, L^{\infty})_{x;q}} \leq \int_0^1 du\, u^{x-1} \left[ \int_0^{\infty} \frac{dt}{t}\, \left( t^{1-x} s^*(t) \right)^q \right]^{1/q},$$
and we can just evaluate the $u$ integral to get
$$\lVert s \rVert_{(L^1, L^{\infty})_{x;q}} \leq \frac{1}{x} \lVert s \rVert_{L^{1/(1-x),q}}.$$
To get a converse inequality, let's look back at the expression
$$\lVert s \rVert_{(L^1, L^{\infty})_{x;q}} = \left[ \int_0^{\infty} \frac{dt}{t}\, \left(\int_{0}^{1} du\, t^{1-x} s^*(t u) \right)^q \right]^{1/q}.$$
We have $tu\leq t$, and $s^*$ is decreasing, so we have $s^*(t u) \geq s^*(t).$ This gives
$$\lVert s \rVert_{(L^1, L^{\infty})_{x;q}} \geq \left[ \int_0^{\infty} \frac{dt}{t}\, \left(\int_{0}^{1} du\, t^{1-x} s^*(t) \right)^q \right]^{1/q},$$
which — as the $u$ integral is now trivial — is exactly the norm $L^{1/(1-x),q}.$
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