The M untzGSz asz Theorem

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Theorem . For each z0 D, the function (z) z0 z 1 z0z is an inner function and Mz0 {f H : f (z0) 0} H2. Proof. The function is clearly in H. Moreover, it is continuous on the closure of D. Therefore, to show that is inner, it suffices to show that (z) 1 when z 1. For this, note that z 1 implies zz 1, so that z0 z
z0 z 1 z0 z

           1.

1 z0z
z(z z0)
z z z0
To show that Mz0 H 2, rst note that (z0)f (z0) 0 for all f H2, so H2 Mz . For the other inclusion, note that f (z0) 0 implies that f (z) (z)g(z) for some function g analytic in D. Let

inf (z) : z D, z 1 z0 . 2 Clearly 0. Thus

for r...

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Published On:
04/19/2021
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