NEVANLINNA-PICK INTERPOLATION: THE DEGENERATED CASE

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1 NEVANLINNA-PICK INTERPOLATION: THE DEGENERATED CASE Harald Woracek Institut für Technische Mathematik Technische Universität Wien A-040 Wien, Austria Introduction Let f be a complex valued function which is meromorphic in the open upper half plane C +. We consider the kernel N f (z, z ) = f(z) f (z ) z z (z, z D f ). () Here and in the following D f always denotes the set of all points of C + where f is holomorphic. Recall that by definition the kernel N f has κ ( IN 0 ) positive (negative, respectively) squares on D f if the following two conditions are satisfied:. For each k IN and points ζ,...,ζ k D f the hermitian form k i,j= N f (ζ i, ζ j )ξ i ξ j (2) has at most κ positive (negative, respectively) squares. 2. For some k IN and points ζ,...,ζ k D f the hermitian form (2) has exactly κ positive (negative, respectively) squares. Definition Let π, ν IN 0. Denote by N π ν the set of all functions f which are meromorphic in C + and such that the kernel N f has π positive and ν negative squares on D f. Further let N π and N ν be the sets of those functions where only the number of positive (negative, respectively) squares is prescribed and equal to π (ν, respectively). The sets N ν were introduced e.g. in [5]. In particular, N 0 is the Nevanlinna class consisting of all functions f holomorphic in C +, such that If(z) 0 for z C +.

2 2 It was shown essentially by G.Pick (see [6] and [7]) that for given distinct points z,...,z n C + and points w,...,w n C a function f N 0 satisfying exists if and only if the so called Pick matrix f(z i ) = w i for i =,...,n (3) ( wi w j IP = z i z j ) n j,i= (4) is nonnegative definite, in this case f is uniquely determined if and only if IP is singular. We will refer to a function f satisfying (3) as a solution of the interpolation problem with data (z, w ),...,(z n, w n ) C + C. In this context the points z,...,z n will always be distinct. In this note we are interested in the case that IP is singular. If, e.g., IP has rank m (< n) and is nonnegative definite it is shown that besides the unique solution of the interpolation problem in the set N 0, which is a rational function, there is no solution in any set N ν with 0 < ν < n m. In fact this result is proved in greater generality: Let π 0 (, respectively) be the number of positive (negative, respectively) squares of the Pick matrix IP and let rankip = m with m = π 0 + < n. Then the interpolation problem (3) has a unique solution in the set N π 0, if and only if the data points do not belong to some exactly described small exceptional set. This solution is given explicitly as a rational function. If either π 0 < π < π 0 + (n m) or < ν < + (m n) then (3) has no solution in the set N π N ν. In a following note it will be shown that solutions of the problem (3) in the set N π ν with π π 0 + (m n) and ν + (m n) always exist. These results are proved in Section 3, using in fact ideas of G.Pick (see [6]), and in Section 4, using some geometric properties of certain inner product spaces and their relation to interpolation problems of the considered type. In Sections 2 and 5 some results on rational functions and the sets N π ν are proved. Some of them are known, however, they are either used later or are immediate consequences of our results. There are a lot of different approaches to questions of the considered type. In particular we refer to [2] and [3] where rational interpolation problems are studied. There the solution (solutions, respectively) with minimal McMillan degree also are of particular interest. Within the framework of rational interpolation not only values of the function itself, but also values for its derivatives can be prescribed (see e.g. [] or [2]. Some generalizations to matrix valued interpolation problems can be found e.g. in [4], [6], [7] or [2]. Another approach to rational interpolation

3 3 problems using the so called Iohvidov laws can be found e.g. in [0]. This method is not applicable in our case. We refer to [5] for further references. Interpolation problems of Nevanlinna-Pick type have various applications. A whole part of the book [5] is devoted to connections with control and system theory (see also [9]). 2 A result on rational functions Before stating the first theorem we recall the notion of degree and signature. Let f be a rational function. The degree of f is defined as the number deg f of poles of f in C counted according to their multiplicities. If f(z) = p(z) q(z) with relatively prime polynomials p and q the degree of f equals max(deg p, deg q), where the degree of a polynomial is defined as usual. Let P be a hermitian n n-matrix. The signature sign P of P is the pair (π, ν) where π (ν, respectively) is the number of positive (negative, respectively) eigenvalues of P, counted according to their multiplicities. Obviously, the defect δ of P is n π ν. Further let f be a meromorphic function in C and denote by D(f) C the domain of holomorphy of f. Then f is called real if for each z D(f) D(f ) the relation f(z) = f (z) holds. If f is a rational function, say f(z) = p(z) q(z) (p, q relatively prime, q monic), then f is real if and only if p and q have real coefficients. In analogy to the Pick matrix (4) we introduce the following notation. Definition 2 Let f be a complex valued function defined on some subset D(f) of C +, let k IN and ζ,...,ζ k D(f). Denote by IP ζ,...,ζ k (f) the matrix IP ζ,...,ζ k (f) = ( f(ζi ) f (ζ j ) ζ i ζ j It follows immediately that f N π ν if and only if for any k IN and points ζ,...,ζ k D f the Pick matrix IP ζ,...,ζ k (f) has signature (π, ν ) with π π and ν ν, and if for some k and some points ζ,...,ζ k D f in both relations equality holds. Theorem Let f be a real rational function. For k IN, k deg f and distinct points ζ,...,ζ k D f the signature of the Pick matrix IP ζ,...,ζ k (f) does not depend on k and ζ,...,ζ k. If signip ζ,...,ζ k (f) = (π f, ν f ), then f N π f ν f. ) k j,i=.

4 4 Proof : Let m = deg f, k m and let K (z, z ) for z, z D f be the kernel K (z, z ) = p(z)q(z ) p(z m )q(z) = a z z rs z r z s. r,s=0 Further let ζ,...,ζ k be distinct points of D f. The number of positive (negative, respectively) eigenvalues of the Pick matrix IP ζ,...,ζ k (f) equals the number of positive (negative, respectively) squares of the hermitian form H ζ,...,ζ k (y) = k r,s= K (ζ r, ζ s )y r y s defined for y = (y,...,y k ) C k. We consider the hermitian form H(x) = m r,s=0 a rs x r x s defined for x = (x 0,...,x m ) C m, and the linear mapping U ζ,...,ζ k : C k C m represented with respect to the canonical bases by the matrix U ζ,...,ζ k = ζ ζ k ζ m ζk m For y C k the following relation can be checked immediately: H ζ,...,ζ k (y) = H(U ζ,...,ζ k y). As U ζ,...,ζ k has rank m and maps a definite subspace (of C k considered with the form H ζ,...,ζ k ) injectively onto a definite subspace (of C m considered with the form H) we find that the signature of IP ζ,...,ζ k (f) equals the signature of H. Therefore it does not depend on k and ζ,...,ζ k. If points ζ,..., ζ k with k < m are given, choose arbitrary points ζ k+,..., ζ m. Then signip ζ,...,ζ m (f) = (π f, ν f ) and therefore π f (ν f, respectively) are upper bounds for the signature numbers of IP ζ,...,ζ k (f). Corollary If k > degf, for any points ζ,...,ζ k IP ζ,...,ζ k (f) is singular. D f the Pick matrix

5 5 Corollary 2 Any real rational function f is contained in some set N π ν. Here the pair (π, ν) is the signature of IP ζ,...,ζ k (f) where k = deg f and ζ,...,ζ k are degf distinct points of D f. Further π + ν degf holds. In Section 5 it will follow that in the last relation the equality sign holds. 3 The unique solution Consider now the interpolation problem (3) with data (z, w ),...,(z n, w n ) C + C and the corresponding Pick matrix (4). If signip = (π 0, ) and ν < or π < π 0 obviously no solution of the problem (3) exists in N π ν. Let us introduce some notation (following e.g. []): If P = (p ij ) i=,...n is a j=...m (n m)-matrix, n n, m m and {i,...,i n } {,...,n}, {j,...,j m } {,...,m} then ) ( i,...,i n P = (p il j j,...,j k ) l=,...n. m k=...m Let P r = P ( ),...,r,...,r for r n, m and define P0 = (). For a quadratic matrix P = (p ij ) n i,j= we consider the matrix P = (p ij) n i,j= of the algebraic complements of P: ( ),..., i, i +,...,n p ij = ( ) i+j IP,...,j, j +...n. In the sequel P will be called the reciprocal matrix of P. Lemma Consider the interpolation problem (3) with given data (z, w ),..., (z n, w n ) C + C and let l = rankip +. The data (z, w ),...,(z n, w n ) can be ordered such that the Pick matrix IP has the following properties:. IP l 0, 2. if IP i = 0 for some i with 0 < i < l, then IP i IP i+ 0. For real symmetric matrices this lemma is proved e.g. in [4], 98 and remains valid in the complex hermitian case. In the following we suppose that the data points are ordered according to Lemma. Further, let l = rankip + and define numbers for i =,...,l as ( ),...,i, i +,..., l = ( ) i+l IP. (5),...,l

6 6 We associate with the given data the rational function and the set l w i (z z j ) i= j= f min (z) =, (6) l (z z j ) i= Z min = {z i= Lemma 2 The function f min is real. j= l (z z j ) = 0}. j= Proof : We show that for each z {z,...,z l,z,...,z l } Z min the relation f min (z) = f min (z) holds. Indeed ( ) ( ) ( ) ( ) w i z z i= i λ j z z j= j λ j w j z z j= j z z i= i f min (z) f min (z) = ( ) ( ). The numerator equals i= i,j= z z i λ j w j w i z j z i j= λ j w j w i z j z i z z i= i λ j z z j= j ( ) = z z i z z j j= λ j z z j i= w j w i z j z i. Both terms on the right hand side vanish as can be seen by expanding the determinant of the singular matrix IP l : j= λ j w j w i z j z i = 0 and for i =,...,l (j =,...,l, respectively). i= w j w i z j z i = 0 Theorem 2 Let (z, w ),...,(z n, w n ) C + C, signip = (π 0, ) and rankip = l < n for the corresponding Pick matrix (4). Then the problem (3) has a solution f N π 0 if and only if one of the following three (equivalent) conditions is satisfied:

7 7 (i) degf min = l. (ii) λ j 0 for j =,...,l. (iii) z j Z min for j =,...,l or, written more explicitly, i= z j z i 0 for j =,...,l. In this case f = f min which is given by (6). Proof : First we prove the uniqueness statement. Suppose that the function f N π 0 satisfies (3) : f(z i ) = w i for i =,...,n. Arrange the data according to Lemma and consider for z D f and w = f(z) the determinant w w w z z... l w w l w w w z l z z l z z z w IP l,z = w l w z z l... l w l w l w l w w l z l z l z l z l z z l. (7) w w l w z z l... l w l w l w l w w l z l z l z l z l z z l w w w z... l w w l w w w z z l z z l z z z Applying Jacobi s sign rule to the sequence = IP 0, IP,..., IP l 0, IP l = 0, IP l,z it follows that IP l,z has the same sign as IP l. Sylvester s identity implies thus i= w w i z z i ( 0 IP l,z IP l = IP,...,l,l l,z 2 w w = i z z i, i=,...,l,l+ = 0 which yields w = f(z) = f min (z). ) 2 = In the following we show that f min satisfies the interpolation conditions (3) if and only if (i), (ii) or (iii) is valid. To establish the necessity of these conditions we note that (i) is an immediate consequence of Theorem, (ii) is implied by (i) in any case and (i) also implies (iii) because f min takes finite values at points z j for j =,...,l. (8)

8 8 Suppose now that (ii) is satisfied. Then for j =,...,l we have f min (z j ) = w j. If we assume (iii) we find f min (z j ) = i= i= w i w j z j z i + w j = w j for j =,...,l. z j z i Thus (i),(ii) or (iii) imply f min (z j ) = w j for j =,...,l. As λ l 0 we also have f min (z l ) = w l. Now let j {l +,...,n} and consider the determinant (7) for f = f min and z = z j. Then (8) implies that IP l,zj = 0 if and only if w is a solution of the linear equation i= l (z z j ) w = w i i= j= l (z z j ). This is the case for exactly one value of w, which is then equal to f min (z j ), or for all w or for no w. As IP l,zj = 0 for w = w j the last possibility cannot occur. The second possibility is also excluded because Theorem already applies to f min and we find deg f min = l. The only case left is the first one and here we have f min (z j ) = w j. Finally we again apply Theorem to obtain f N π 0. æ 4 Interpolation and inner product spaces Now we associate with a complex function f defined on D(f) C + an inner product space. Definition 3 Let f : D(f) C, D(f) C +. Denote by H f the following linear space of all formal sums H f = { z D(f) j= x z e z x z C, x z = 0 for all but finitely many z D(f)}, equipped with the inner product defined by the relation [e z, e w ] Hf = f(z) f (w) z w (z, w D(f)).

9 9 If no confusion can occur the index H f at the inner product will be dropped. Note that by definition the elements e z for z D(f) are linearly independent. If the function f is an extension of some function g (that is D(g) D(f) and f D(g) = g) then H g can be considered in a canonical way as subspace of H f. Recall that for an inner product space L the isotropic subspace is the set L = L L. If L {0} the space L is called degenerated. Consider now the interpolation problem (3) with data (z, w ),...,(z n, w n ) C + C. Let IP be the corresponding Pick matrix, signip = (π 0, ) and δ = n π 0. With the given data we associate the inner product space H ˆf for the function ˆf defined as follows: D( ˆf) = {z,...,z n }, ˆf(zi ) = w i for i =,...,n. (9) In the following the space H ˆf is denoted by H. The dimension of each maximal positive (negative, respectively) subspace of H is π 0 (, respectively) and the dimension of H is δ (> 0). If f is a solution of the interpolation problem (3) we evidently have H H f. With each nonzero neutral vector h of H we shall associate a real rational function f h which will coincide with f min from (6). Definition 4 If h H \ {0}, h = n h i e z i, denote with f h the rational function i= n h n i w i (z z j ) i= j= f h (z) = n n, (0) (z z j ) hi i= j= defined on its domain of holomorphy D fh in C +. Lemma 3 Suppose that the interpolation problem (3) has a solution f in N π 0. Then there are δ linearly independent vectors h,...,h δ H, such that f = f h i for i =,...,δ. Proof : The solution in N π 0 is given by formula (6), which can be written as f min (z) = n i= µ () n i w i (z z j ) j= n n i= µ () i j= (z z j )

10 0 with µ () i = for i =,...,l, µ () l = λ l 0 and µ () i = 0 for i = l +,...,n. From the definition (5) of the numbers we see that (µ (),...,µ () l, µ() l ) T is a column of the reciprocal matrix of IP ( ),...,l,l,...,l,l. Together with IPl 0 and rankip = l this implies that IP (µ (),...,µ () l, µ() l, 0,...,0) T = 0 holds. In other words the vector h = n i= µ () i e zi is isotropic: h H. In the proof of Theorem 2 we have ordered the points z,...,z n according to Lemma. The conditions required in Lemma concern only the properties of the first l points. Thus the order of the remaining points is arbitrary, which means we could make the same calculations with the point z (l )+r (r = 2,...,δ) instead of z l. The function f min constructed in this way can be written as f min (z) = n i= n i= µ (r) i w i n µ (r) i j= n j= (z z j ) (z z j ) where (µ (r),...,µ (r) IP (,...,l,l +r,...,l,l +r µ (r) n l, µ(r) l +r )T ), µ (r) l +r = 0. Again we get and therefore h r = n is again a column of the reciprocal matrix of =... = µ (r) l +r = µ(r) l +r+ =... = 0 and µ(r) l IP (µ (r),...,µ (r) l, 0,...,0, µ(r) l +r, 0,...,0)T = 0, i= µ (r) i e zi H. Obviously the vectors h,...,h δ are linearly independent. Theorem 3 Suppose that the interpolation problem (3) with data (z, w ),..., (z n, w n ) C + C has a solution f N π 0 where (π 0, ) is the signature of the Pick matrix IP, n > π 0 +. Then for arbitrary h H, h 0 we have f = f h. Proof : Lemma 3 implies that the set L ( H ) of all h with f(z) = f h (z) contains δ = n π 0 = dimh linearly independent vectors. The theorem will be proved if we show that L (together with 0) is a subspace.

11 If α C, α 0 and h L then f αh (z) = f h (z). If h, h 2 L, h h 2 consider the representation (0) of f h and f h 2 : f h (z) = p (z) q (z) and f h 2 (z) = p 2(z) q 2 (z). Observe that the polynomials p, q and p 2, q 2 need not be relatively prime. As h h 2 the polynomial q +q 2 is not identically zero. Outside a circle containing all zeros of q, q 2 and q + q 2 we find from h, h 2 L that p (z) = f(z)q (z) and p 2 (z) = f(z)q 2 (z). Thus p (z) + p 2 (z) q (z) + q 2 (z) = f(z) holds which shows that h + h 2 L. In the following we use the geometry of the inner product spaces H and H f to show that there are no solutions of the problem (3) in the sets N π for π 0 < π < π + δ and N ν for < ν < ν + δ where again δ is the defect of IP. Lemma 4 Let f be a complex valued continuous function defined on D(f) C + such that H f is degenerated and let h H f \ {0}, say h = m h i e ζ i H f. Finally suppose that no point z D(f) with m D(f). Then for z D(f) hi i= m j= i= (z ζ i ) = 0 lies isolated in m h i f (z i ) m (z z i ) i= j= f(z) = m h m. () i (z z i ) i= That is, f is the restriction of the rational function on the right hand side of () to D(f). Proof : For z D(f) we find j= This gives m 0 = [e z, h ] = hi i= f(z) f (ζ i ) z ζ i. m m hi (f(z) f (ζ i )) (z ζ i ) = 0 i= j=

12 2 which yields (), at least if m hi i= m j= (z ζ j ) 0. The following lemma is proved by a straightforward calculation. Lemma 5 Let h H \ {0}, h = n h i e z i and define u(z) as u(z) = n h i i= i= n (z z j ) j= Then for z, w D fh the relation n h i i= n (z z j )e zi. j= [e z, e w ] = f h (z) f h (w) z w = [u(w), u(z)] H (2) holds. Using this lemma we prove Lemma 6 Let h H \ {0}, h = n h i e z i. Then f h N π ν and ν with π π 0 and ν. i= for some numbers π Proof : Let m IN, ζ,...,ζ m D fh with n h i i= n (z z j ) 0 for z = ζ,...,ζ m. (3) j= Consider the space C m provided with the inner product [e i, e j ] C m = f(ζ j ) f (ζ i ) ζ j ζ i for i, j =,...,m, where e i denotes the i-th canonical basis vector of C m. Then Lemma 5 implies that the operator U defined by the relations U e i = u(ζ i ) is an isometry from C m, [.,.] C m into H. Thus π0 (, respectively) is an upper bound for the dimension of positive (negative, respectively) subspaces of C m, [.,.] C m. Because the set of points z D fh satisfying (3) lies dense in D fh the assertion follows.

13 3 Remark Note that Lemma 4 cannot always be applied to the function ˆf defined as in (9) because D( ˆf) is discrete. So we need not necessarily have f h (z i ) = w i for i =,...,n. But if no point z i is a zero of z n n (z z j ), hi i= we find that f h extends ˆf which means that f h (z i ) = w i for all i =,...,n. Then the above lemma shows that f h is a solution in N π 0 of the interpolation problem (3). Thus we can add to the conditions (i), (ii) and (iii) of Theorem 2 the following also equivalent condition. (iv) There exists a vector h H \ {0}, h = n h i e z i, such that no point z i i= (i =,...,n) is a zero of z n n (z z i ). hi i= j= j= Using some results about the geometry of inner product spaces (which can be found e.g. in [8] or [3]) we obtain the following theorem. Theorem 4 Consider the interpolation problem (3) with data (z, w ),..., (z n, w n ) C + C. If δ denotes the defect of the Pick matrix IP and signip = (π 0, ) there is no solution f in any set N π or N ν with π 0 < π < π 0 + δ or < ν < + δ. Proof : Let f be a solution of the problem (3) with f N ν and < ν < +δ. Then we have H H f. Consider the nondegenerated inner product space L = H f /H f. The largest dimension of a negative subspace of L equals ν. In case H H f = {0} we can consider H in a natural way as a subspace of L. If H denotes some -dimensional negative subspace of H, we can decompose L as L = H [ +](H ). Obviously H (H ) and the largest dimension of a negative subspace of (H ) equals ν. Then the dimension of neutral subspaces of (H ) is also bounded by ν, which is impossible as dimh = δ > ν. Thus we have H H f {0}. For any vector h H H f, h 0 Lemma 4 implies f = f h. Lemma 6 yields f h N ν with ν. This is a contradiction to ν > which proves the assertion. The remaining part of the theorem is proved analogously. Remark 2 If a solution of (3) in N π 0 exists it is not necessary to use Lemma 6 in the proof of Theorem 4. Because in this case Theorem 3 already implies

14 4 f h N π 0. 5 Real rational functions Using some ideas of the previous sections we will now complete the results of Section 2. To start with, let f be any real rational function of degree k. Then for n = k+ and any points z,...,z n D f the Pick matrix IP z,...,z n has signature (π f, ν f ) and is singular. Furthermore, f N π f ν f. Thus we can consider f as the unique solution in the set N π f ν f of the interpolation problem with data (z, f(z )),...,(z n, f(z n )). Consider the inner product space H assigned to this interpolation problem. By reversing the argument of Lemma 4 we get: Lemma 7 Let f be a real rational function. Then we have H H f. Proof : Let h H, h 0. By Theorem 3 we have f(z) = f h (z). An easy computation yields [e z, h ] = 0 for each z D f, that is h H f. Theorem 5 Let f be a real meromorphic function in C and consider f restricted to the domain D f. Then H f degenerates if and only if f is a real rational function. In this case dimh f /H f = degf (< ). Proof : The first assertion follows from Lemma 4 together with Lemma 7. To prove the second part of the theorem let l = dimh f /H f and k = deg f. If we consider f as the unique solution of the above constructed interpolation problem we find k = rank IP. Thus we find a k-dimensional nondegenerated subspace of H f which implies l k. Assume that l > k. Linearly independent vectors of H f /H f yield linearly independent vectors in H f. The subspace they generate in H f is nondegenerated. So we can find points ζ,...,ζ m D f, such that rankip m l > k. But this is impossible as deg f = k. Because rankip = π f +ν f the above theorem shows that the inequality π f +ν f deg f given in Corollary 2 is in fact an equality. This immediately implies the following corollary.

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