In this final lecture we will give a proof of Pick’s interpolation theorem that is based on operator theory.
Theorem 1 (Pick’s interpolation theorem): Let $latex z_1, ldots, z_n in D$, and $latex w_1, ldots, w_n in mathbb{C}$ be given. There exists a function $latex f in H^infty(D)$ satisfying $latex |f|_infty leq 1$ and
$latex f(z_i) = w_i ,, ,, i=1, ldots, n$
if and only if the following matrix inequality holds:
$latex big(frac{1-w_i overline{w_j}}{1 – z_i overline{z_j}} big)_{i,j=1}^n geq 0 .$
Note that the matrix element $latex frac{1-w_ioverline{w_j}}{1-z_ioverline{z_j}}$ appearing in the theorem is equal to $latex (1-w_i overline{w_j})k(z_i,z_j)$, where $latex k(z,w) = frac{1}{1-z overline{w}}$ is the reproducing kernel for the Hardy space $latex H^2$ (this kernel is called the Szego kernel). Given $latex z_1, ldots, z_n, w_1, ldots, w_n$, the matrix
$latex big((1-w_i overline{w_j})k(z_i,z_j)big)_{i,j=1}^n$
is called the Pick matrix, and it plays a central role in various interpolation problems on various spaces.
I learned this material from Agler and McCarthy’s monograph [AM], so the following is my adaptation of that source.
(A very interesting article by John McCarthy on Pick’s theorem can be found here).
1. A necessary condition for interpolation
Lemma 1: Let $latex T in B(H)$. Then $latex |T|leq 1$ if and only if $latex I – T^*T geq 0$.
Proof: Let $latex |h|=1$ in $latex H$. Then
$latex |Th|^2 leq 1 = |h|^2 Leftrightarrow langle (I – T^*T)h,h rangle = langle h,h rangle – langle Th, Th rangle geq 0.$
Proposition 2: Let $latex H$ be a RKHS and a set $latex X$ with kernel $latex k$. A function $latex f : X mathbb{C}$ is a multiplier of norm $latex 1$ if and only if for every $latex n in mathbb{N}$ and every $latex n$ points $latex x_1, ldots, x_n in X$ the associated Pick matrix is positive semi-definite, meaning:
$latex big((1-f(x_i)overline{f(x_j)}) k(x_i, x_j) big) geq 0.$
Proof: Let us define the operator $latex T: span{k_x : x in X} rightarrow H$ by extending linearly the rule
$latex T k_x = overline{f(x)} k_x .$
Suppose first that $latex f$ is a multiplier. Then by Proposition 3 in the previous post $latex T$ extends to $latex H$ and is equal to $latex T = M_f^*$. Let $latex x_1, ldots, x_n in X$ and $latex c_1, ldots, c_n in mathbb{C}$. Then
$latex langle (I-T^*T) sum c_j k_{x_j}, sum c_i k_{x_i} rangle = $
$latex = sum_{i,j} c_joverline{c_i} (1-f(x_i)overline{f(x_j)}) langle k_{x_j}, k_{x_i} rangle = sum_{i,j} c_j overline{c_j}(1- f(x_i)overline{f(x_j)}) k(x_i,x_j).$
Since the span of $latex k_x$s is dense in $latex H$, the lemma implies that $latex |f| = |M_f^*|leq 1$ if and only if the Pick matrix is positive semi-definite. The other direction is similar.
Exercise A: Complete the above proof.
Corollary 3: Let $latex z_1, ldots, z_n in D$, and $latex w_1, ldots, w_n in mathbb{C}$ be given. A necessary condition for there to exist a function $latex f in H^infty(D)$ satisfying $latex |f|_infty leq 1$ and
$latex f(z_i) = w_i ,, ,, i=1, ldots, n$
is that the following matrix inequality holds:
$latex big(frac{1-w_i overline{w_j}}{1 – z_i overline{z_j}} big)_{i,j=1}^n geq 0 .$
Proof: This follows from Proposition 2 and from $latex Mult(H^2(D)) = H^infty(D)$.
Exercise B: Find the correct version of the above three results when the condition of having norm less than or eqaul to $latex 1$ is replaced by the condition of being bounded.
Remark: Note that it is important to put the bar on the right element. If $latex f$ is a norm one $latex H^infty$ function then there must be $latex z_1, ldots, w_n$ such that the matrix
$latex big(frac{1-w_j overline{w_i}}{1 – z_i overline{z_j}} big)_{i,j=1}^n $
is not positive semi-definite (why?).
Proposition 2 gives one way of proving Pick’s theorem: one shows that values of $latex f$ can be chosen at all points of the disc so that the Pick matrix (on $latex D times D$) is positive semi-definite. This approach is perhaps the purest RKHS approach. We will use a different approach, which will actually give a formula for a solution. Both approaches are treated in [AM].
2. The realization formula
A key to the approach we will see for the interpolation problem is the following characterization of functions in $latex (H^infty)_1$ — the closed unit ball of $latex H^infty(D)$.
Theorem 4: Let $latex f : D rightarrow mathbb{C}$. Then $latex f in (H^infty)_1$ if and only if there is a Hilbert space $latex K$ and an isometry $latex V: mathbb{C} oplus K rightarrow mathbb{C} oplus K$ with block structure
$latex V =left(begin{smallmatrix} a&X \ Y&Z end{smallmatrix}right),$
such that
(*) $latex f(w) = a + wX(I-wZ)^{-1}Y$
for all $latex w in D$.
Proof: For sufficiency, suppose that $latex V$ is an isometry as in the theorem. Then $latex Z$ is also a contraction, thus $latex I-wZ$ is invertible for all $latex w in D$ and $latex (I-wZ)^{-1}$ is an analytic operator valued function. Therefore $latex f$ defined by (*) is analytic in D. To see that $latex f in (H^infty)_1$ we make a long computation:
$latex 1 – overline{f(w)} f(w) = 1 – (a + wX(I-wZ)^{-1} Y)^* (a+wX(I-wZ)^{-1} Y) = ldots $
$latex ldots = Y^*((I-wZ)^{-1})^*[1-|w|^2](I-wZ)^{-1}Y geq 0 ,$
where to get from the first line to the second one we used the fact that $latex V$ is an isometry:
$latex V^*V = left(begin{smallmatrix} 1&0 \ 0& I end{smallmatrix}right)$
which gives $latex overline{a} a + Y^* Y = 1$, $latex overline{a} X + Y^* Z = 0$ and $latex X^* X + Z^* Z = I$.
We leave the converse as an exercise for two reasons. First, we need the sufficiency part of the above theorem for our proof of Pick’s theorem. Second, our proof of Pick’s theorem contains the same idea that is used to proof the necessity part.
Exercise C: Complete the proof of the above theorem (suggestion: first give it a try on your own, and if you get stuck you can look at the proof of Pick’s theorem below).
3. Proof of Pick’s theorem
We now complete the proof of Pick’s theorem. Corollary 3 takes care of one direction: positivity of the Pick matrix is a necessary condition for a norm one multiplier interpolant to exist. It remains to show that this condition is sufficient. For that, we require first a lemma, which interest of its own.
Lemma 5: Let $latex k : X times X rightarrow mathbb{C}$ be a positive semi-definite kernel. Then there exists a Hilbert space $latex K$ and a function $latex F: X rightarrow K$ such that $latex k(x,y) = langle F(x), F(y) rangle$ for all $latex x,y in X$. If $latex X$ is a set with $latex n$ points, then the dimension of $latex K$ can be chosen to be the rank of $latex k$ when considered as a positive semi-definite $latex n times n $ matrix.
Proof: Let $latex K$ be the RKHS associated with $latex k$ as in Theorem 1 in the previous post (in that theorem the space was denoted $latex H$). Define $latex F: X rightarrow K$ by $latex F(x) = k_x$. Then we have
$latex langle F(x), F(y) rangle = k(y,x) .$
This is not exactly what the theorem asserts; to obtain the assertion of the theorem apply what we just did to the kernel $latex tilde{k}(x,y) = k(y,x)$, which is also positive semi-definite.
To prove that theorem in the finite case, assume that $latex X = {1,2,ldots, n}$ and consider the positive semi-definite matrix $latex A = big(k(i,j) big)_{i,j=1}^n$. Let the rank of this matrix be $latex m leq n$. Then by the spectral theorem $latex A = sum_{l=1}^m v_l v_l^*$, where $latex v_1, ldots, v_m in mathbb{C}^n$ are the eigenvectors corresponding to non-zero eigenvalues. Denote $latex v_l = (v_l^1, v_l^2, ldots, v_l^n)$ for $latex l=1, ldots, m$. Now define $latex F: X rightarrow mathbb{C}^m$ by $latex F(i) = (v_1^i, v_2^i, ldots, v_m^i)$ for $latex i=1, ldots, n$. Then we have that
$latex k(i,j) = (A)_{i,j} = sum_{l=1}^m (v_l v_l^*)_{i,j} = sum_{l=1}^m v_l^i overline{v_l^j} = langle F(i), F(j) rangle ,$
as was to be proved.
The following theorem completes the proof of Pick’s theorem. In fact, we obtain a little bit more information then claimed in Theorem 1.
Theorem 6: Let $latex z_1, ldots, z_n in D$ and $latex w_1, ldots, w_n in mathbb{C}$ be given. If the Pick matrix
$latex left ( frac{1-w_i overline{w_j}}{1-z_i overline{z_j}} right)_{i,j=1}^n $
is positive semi-definite and has rank $latex m$, then there exists a rational function $latex f$ of degree at most $latex m$ such that $latex |f|_infty leq 1$ and $latex f(z_i) = w_i$, for all $latex i=1, ldots, n$.
Remark: By degree of a rational function we mean either the degree of the numerator or the degree of the denominator — the bigger one — appearing in the rational function in reduced form. A consequence of the boundedness of $latex f$ is that the poles of $latex f$ lie away from the closed unit disc.
Proof: By Lemma 5 we know that there are $latex F_1, ldots, F_n in mathbb{C}^m$ such that $latex frac{1 – w_i overline{w}_j}{1 – z_i overline{z}_j} = langle F_i, F_j rangle$ for all $latex i,j$. We can rewrite this identity in the following form:
$latex 1 + langle z_i F_i, z_j F_j rangle = w_i overline{w_j} + langle F_i, F_j rangle .$
This means that we define a linear map $latex V$ from $latex span{(1, z_i F_i) : i=1, ldots, n} subseteq mathbb{C} oplus mathbb{C}^m$ into $latex span {(w_i, F_i) : i = 1, ldots, n} subseteq mathbb{C} oplus mathbb{C}^m$, by sending $latex (1, z_i F_i)$ to $latex (w_i, F_i)$ and extending linearly, then this $latex V$ is an isometry. Since isometric subspaces have equal dimension, we can extend $latex V$ to an isometry $latex V : mathbb{C} oplus mathbb{C}^m rightarrow mathbb{C} oplus mathbb{C}^m$.
Recall that the realization formula gives us a way to write down a function $latex f in (H^infty)_1$ given an isometric matrix in block form. So let us write
$latex V = left(begin{smallmatrix} a&X \ Y&Z end{smallmatrix} right) , $
where the decomposition is according to $latex mathbb{C} oplus mathbb{C}^m$, and define $latex f(w) = a + wX(I-wZ)^{-1}Y$. By Theorem 4, $latex f in (H^infty)_1$, and it is rational because there are rational formulas for the computation of the inverse of a matrix. The degree of $latex f$ is evidently not greater than $latex m$. We have to show that $latex f$ interpolates that data.
Fix $latex i in {1, ldots, n}$. From the definition of $latex V$ we have
$latex left(begin{smallmatrix} a&X \ Y&Z end{smallmatrix} right) left(begin{smallmatrix} 1\z_i F_i end{smallmatrix} right) = left(begin{smallmatrix} w_i\F_i end{smallmatrix} right).$
The first row gives $latex a + z_i X F_i = w_i$. The second row gives $latex Y + z_i Z F_i = F_i$, so solving for $latex F_i$ we obtain $latex F_i = (I – z_i Z)^{-1}Y$ (where the inverse is legal). Plugging this in the first row gives
$latex w_i = a + z_i X F_i = a+ z_i X (I-z_iZ)^{-1}Y = f(z_i)$
thus $latex f$ interpolates the data, and the proof is complete.
Exercise D: Go over the proof and make sure you understand how it gives a formula for the solution.
Remark: The argument that we used to prove Pick’s theorem has a cute name: the lurking isometry argument. It generalizes to broader contexts, and can be used to prove the necessity part of Theorem 4.
4. Brief on the commutant lifting approach
There is another approach for the interpolation problem that is “even more” operator theoretic. It is now known as the commutant lifting approach. The interested student can find a more detailed treatment in [AM].
We need the following definition. Let $latex T$ be a contraction on a Hilbert space $latex H$. Suppose that $latex K$ is a Hilbert space which contains $latex H$ as a subspace and that $latex V$ is an isometry on $latex K$. Denote by $latex P_H$ the orthogonal projection of $latex K$ onto $latex H$. If
$latex T^n = P_H V^n P_H $
for all $latex nin mathbb{N}$ then we say that $latex V$ is an isometric dilation of $latex T$, and that $latex T$ is a compression of $latex V$.
An isometric dilation $latex (V,K)$ of $latex (T,H)$ is said to be minimal if the only subspace of $latex K$ which contains $latex H$ and is invariant for $latex V$ is $latex K$ itself.
Remark: Note that if $latex (V, K)$ is a minimal isometric dilation for $latex T$ then $latex K = overline{span}{V^n h : n in mathbb{N}, h in H}$. Also, if $latex mleq n$ then
$latex langle V^n h, V^m g rangle = langle V^{n-m} h, g rangle = T^{n-m} h, g$
which depends only on $latex m,n,h,g$.
Theorem 7 (Sz.-Nagy’s isometric dilation): Every contraction $latex T$ on a Hilbert space $latex H$ has a minimal isometric dilation. The minimal isometric dilation is unique in the following sense: if $latex V_i$ are both minimal isometric dilations of $latex T$ on Hilbert spaces $latex K_i$, $latex i=1, 2$ then there exists a unitary $latex U : V_1 rightarrow V_2$ such that
$latex Uh = h$
for all $latex h in H$ and such that $latex V_2 = U V_1 U^*$.
Proof: Define $latex d_T = sqrt{I-T^*T}$, and let $latex D = overline{d_T(H)}$. Define $latex K = H oplus D oplus D oplus D oplus ldots $ and
$latex V(h, f_1, f_2, f_3, ldots) = (Th, d_T h, f_1, f_2, ldots) .$
We can check that $latex V$ is indeed a minimal isometric dilation for $latex T$ (exercise). As for minimality, if $latex (V_i, K_i)$ are two minimal isometric dilations we define $latex U : V_1 rightarrow V_2$ by
$latex U sum_n V_1^n h_n = sum_n V_2^n h_n.$
By the remark before the theorem we have that
$latex |sum_n V_2^n h_n |^2 = sum_{m,n} langle V_2^m h_m , V_2^n h_n rangle = $
$latex = sum_{m,n} langle V_1^m h_m , V_1^n h_n rangle = | sum_n V_1^n h_n |^2.$
so $latex U$ extends to a unitary between $latex K_1$ and $latex K_2$. Again, we may check that $latex U V_1 = V_2 U$ and that $latex Uh = h$ for $latex h in H$.
Example: Let $latex S$ be the multiplication operator $latex M_z$ on $latex H^2$. Then $latex M_z$ is an isometry. If $latex z_1, ldots, z_n in D$ and $latex G = span{k_{z_i} : i = 1, ldots, n}$, then $latex G$ is an invariant subspace for $latex S^*$ (why?) therefore $latex T = P_G Sbig|_G$ is a compression of $latex S$ and $latex S$ is an isometric dilation for $latex T$. $latex S$ is in fact the minimal isometric dilation for $latex T$ (exercise).
The key operator theoretic result we shall need is the following:
Theorem 8 (Foias–Sz.-Nagy commutant lifting theorem): Let $latex V in B(K)$ be an isometric dilation of $latex T in B(H)$. Suppose that $latex X in B(H)$ satisfies $latex TX = XT$. Then there exists $latex Z in B(K)$ such that $latex Z^* big|_H = T^*$ which satisfies $latex VZ = ZV$ and $latex |Z| = |X|$.
I will not supply a proof of this theorem here. We will also require the following fact:
Proposition 9: Let $latex S$ be as in the example above and let $latex Z in B(H^2)$ satisfy $latex SZ = ZS$. Then there exists $latex f in H^infty$ such that $latex Z = M_f$.
Proof: Let $latex f = Z 1 in H^2$. For every polynomial $latex p$, $latex Z p = Z M_p 1 = Z p(M_z) 1 = p(M_z) Z 1 = pf = M_f p$. From the density of polynomials in $latex H^2$ and the boundedness of $latex Z$ it follows that $latex f$ is a multiplier and that $latex Z = M_f$.
Alternative proof of Pick’s theorem: Suppose that the Pick matrix is positive semi-definite. Define $latex X$ on $latex G = span {k_{z_1}, ldots, k_{z_n}}$ to be the adjoint of $latex k_{z_i} mapsto overline{w_i} k_{z_i}$. Positivity of the Pick matrix implies that $latex X$ is a contraction. Now $latex X$ commutes with $latex T = P_G S big|_G$ (their adjoints are both diagonal with respect to the basis $latex {k_{z_1}, ldots, k_{z_n}}$). So by the commutant lifting theorem and the above proposition there exists some $latex f in H^infty$ such that $latex M_f^* big|_{G} = T^*$ and $latex |f|_infty = |M_f| leq 1$. This $latex f$ solves the interpolation problem.