Real and complex analysis

346 REAL AND COMPLEX ANAL YSlSand21t 1 f" _..tog I g(re i9 ) I dO = log I g(O) I, (2)we see that log I 9 I satisfies the hypotheses of Theorem 11.30 and is therefore thePoisson integral of a real measure p.. Thusf(z) {I eit + = cB(z) exp -it- z dp.(t) } ,re - zwhere c is a constant, I c I = 1, and B is a Blaschke product.Observe how the assumption that the integrals of log+ I 9 I are bounded(which is a quantitative formulation of the statement that I 9 I does not get tooclose to (0) implies the boundedness of the integrals of log- I 9 I (which says thatI 9 I does not get too close to 0 at too many places).If p. is a negative measure, the exponential factor in (3) is in H OO • Apply theJordan decomposition to p.. This shows:To everyfe N there correspond twofunctions b l and b 2 e H OO such that b 2 hasno zero in U andf= b l /b 2 •Since b! =F 0 a.e., it follows thatfhas finite radial limits a.e. Also, f* =F 0 a.e.Is log If* I e I!(T)? Yes, and the proof is identical to the one given inTheorem 17.17.However, the inequality (3) of Theorem 17.17 need no longer hold. Forexample, ifgf(z) = exp ~ ~}, (4)then IIfllo = e, I f* I = 1 a.e., andlog I f(O) I = 1 > 0 = 21t 1 f" _..tog I f*(ei~ I dt. (5)(3)The Shift Operator17.20 Invariant Subspaces Consider a bounded linear operator S on a Banachspace X; that is to say, S is a bounded linear transformation of X into X. If aclosed subspace Y of X has the property that S(Y) c Y, we call Y an S-invariantsubspace. Thus the S-invariant subspaces of X are exactly those which aremapped into themselves by S.The knowledge of the invariant subspaces of an operator S helps us to visualizeits action. (This is a very general-and hence rather vague-principle: Instudying any transformation of any kind, it helps to know what the transformationleaves fixed.) For instance, if S is a linear operator on an n-dimensionalvector space X and if S has n linearly independent characteristic vectors Xl' .•. ,