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Casorati-Weierstrass theorem

Statement

Let f have an essential singularity at a. Then for every r>0 with 0<za<r in the domain of f, the image f({0<za<r}) is dense in C.

Equivalently, for every wC and every ε>0, some point z with 0<za<r satisfies f(z)w<ε.

Facts & Assumptions

Given: An essential singularity of f at a and a radius r>0 with f holomorphic on 0<za<r.

[L1]

Essential means neither removable nor a pole (Every isolated singularity is removable, a pole, or essential).

[L2]

A bounded holomorphic function on a punctured disc has a removable singularity (Characterizations of removable singularities).

[L3]

A function on a punctured disc has a pole exactly when its modulus tends to infinity there (Characterizations of poles).

[L4]

Reciprocal and sum rules preserve holomorphy wherever the denominators stay nonzero (Linearity, product, reciprocal, and quotient rules for complex derivatives).

Proof

technique · contradiction
1.1

Suppose, for contradiction, that f({0<za<r}) is not dense in C. Then some wC and some ε>0 satisfy f(z)wε for every z with 0<za<r.

assume-contra
2.1

The function g(z):=1/(f(z)w) is therefore holomorphic on 0<za<r by [L4] and bounded there by 1/ε.

step 1.1L4algebra
3.1

By [L2], the bounded function g extends holomorphically across a. If the extension satisfies g(a)0, then 1/g is holomorphic near a and f=w+1/g is removable there by [L4]. If instead g(a)=0, then 1/g has a pole at a by [L3], so f=w+1/g has a pole there as well.

step 2.1L2L3L4
4.1

Either outcome in step 3.1 contradicts [L1], because an essential singularity is neither removable nor a pole. Therefore the assumption of step 1.1 is false, and every punctured neighbourhood image is dense in C.

step 1.1step 3.1L1discharge-contradiction

Depends on

Used by

Dependency tree · two levels

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Sources