| R. Narasimhan, Complex analysis of one variable, Basel, 1985. |
...., for all 2 [0; 0 ] Remark 6.2. Suppose that H 2 H 1 (C m Thetam ) Then, as is well known, lim #0 H( i ) H(i ) exists for almost all 2 R. It is easy to show that r(H(s) is a subharmonic function on C 0 . Using standard results on subharmonic functions (see for example Narasimhan [25], p. 227) it is not difficult to prove that supfr(H(s) j s 2 C 0 g = ess supfr(H(i ) j 2 Rg : As a consequence, 6.1) will be satisfied if ess supfr(H(i ) j 2 Rg 1. The proof of Theorem 6.1 requires some preparation. If H 2 M 0 (C m Thetam ) and s 0 is a pole of H, then trivial ....
....ff and H = D Gamma1 N is a leftcoprime factorization over H ff , then the zeros of det D and det D in C ff coincide (counting multiplicities) Proof. It is well known that the ring H ff is a Bezout domain, i.e. every finitely generated ideal is principal (see for example Narasimhan [25], p. 136) Now M ff is the quotient field of H ff and statements ( and ( follow from Vidyasagar [32] p. 330. Statement ( is proved in [32] p. 76 for rational matrices. An inspection of the proof in [32] shows that it only utilizes the fact that the elementary divisor theorem holds ....
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R. Narasimhan, Complex Analysis in One Variable, Birkhauser, Boston, 1985.
....fl sufficiently small we get: x = 2 R(H) We note that if a = b and the topological circle L : R(ff) a bounds the open disk R(H) ae Omega 0 , then the function R(z) Gamma a tends to zero on fl; this contradicts Theorem 2. Remark 7. Alternatively one can use F. and M. Rees theorem (see e.g. [Nar]) for this part of the proof. Therefore we can assume that either a 6= b (Case 1) or a = b and the topological circle L = R(ff) a bounds an open disk which contains R(H) and a nonempty part E of the limit set of G 0 (Case 2) In the former case the arc ff separates a part E of Omega 0 from ....
R. Narasimhan, "Complex analysis in one variable", Birkhauser Boston, Boston, Mass., 1985. 6
....2 U . In fact, while it is trivial that the Hautus conditions are necessary for the solvability of the Bezout equations, sufficiency follows from the fact that the elementary divisor theorem holds for holomorphic matrices, i.e. any holomorphic matrix is equivalent to its Smith form (see Narasimhan [16], p. 139) Moreover, the same argument applies globally, and thus (N; D; N) is bi coprime if and only if it satisfies the generalized Hautus conditions in s for all s 2 C cl 0 . 3 Proposition 2.3 Let N 2 H m Thetan Gamma , D 2 H n Thetan Gamma and N 2 H n Thetal Gamma , suppose ....
R. Narasimhan. Complex Analysis in One Variable, Birkhauser, Boston, 1985.
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R. Narasimhan, Complex analysis of one variable, Basel, 1985.
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R. Narasimhan, Complex analysis in one variable, Birkhauser, 1985.
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