For a fixed function $h\in H(\mathbb{D})$ and a fixed complex number $λ$ let $f$ (It is analytic) be a solution of $$(λI − C)f = h.$$ Using Taylor expansions, it is easy to see that the operator $(λI − C)$ is bijective unless $1/λ$ is a positive integer and that the range of the operator $((1/n)I − C), n ∈ Z^{+}$, does not contain the function $z→z^n−1$. My question how do you conclude the fact using Taylor expansion??? Sorry $C$ is the Cesaro operator $$Cf (z) = \frac{1}{z}\int_{0}^z\frac{f (ζ)}{1−ζ}dζ.$$
2026-03-28 11:13:37.1774696417
surjectivity of an operator which is defined on some set of analytic functions
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