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Fall 2010 - Problem 8

operator theory

Let $A\p{\D}$ be the $\C$ -vector space of all holomorphic functions on $\D$ and suppose that $\func{L}{A\p{\D}}{\C}$ is a multiplicative linear functional, that is,

L\p{af + bg} = aL\p{f} + bL\p{g} \quad\text{and}\quad L\p{fg} = L\p{f}L\p{g}

for all $a, b \in \C$ and all $f, g \in A\p{\D}$ . If $L$ is not identically zero, show that there is $z_0 \in \D$ so that $L\p{f} = f\p{z_0}$ for all $f \in A\p{\D}$ .

Solution.

Set $z_0 = L\p{z}$ . Since $L$ is not identically zero, there exists $f \in A\p{\D}$ such that $L\p{f} \neq 0$ . Thus, by multiplicity of $L$ , we have $L\p{f \cdot 1} = L\p{f}L\p{1} \implies L\p{1} = 1$ . Similarly, $L\p{z^n} = \p{L\p{z}}^n = z_0^n$ , and so by linearity of $L$ , we have shown that the claim is true whenever $f$ is a polynomial.

Now let $f \in A\p{\D}$ . Since $f$ is holomorphic, we may apply the factor theorem to get $f\p{z} - f\p{z_0} = \p{z - z_0}g\p{z}$ . By multiplicity of $L$ , $L\p{f\p{z} - f\p{z_0}} = L\p{z - z_0}L\p{g\p{z}} = 0$ . By linearity, we conclude that

L\p{f\p{z}} = L\p{f\p{z_0}} = f\p{z_0}L\p{1} = f\p{z_0}.

To complete the proof, we need to show that $z_0 \in \D$ . If this were not the case, then $f\p{z} = \frac{1}{z - z_0} \in A\p{\D}$ , which implies that $L\p{f\p{z}} = f\p{z_0} = \infty \in \C$ , but this is absurd.