Notes on Complex Analysis
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§11.4 Hartogs’ Phenomenon

11.4 Hartogs’ Phenomenon

One of the most prominent results of complex function theory is Hartogs’ phenomenon, which reveals a fundamental difference between holomorphic functions of one variable and those of several variables. First discovered by Friedrich Hartogs in 1906, it essentially states that holomorphic functions in several complex variables exhibit a surprising rigidity.

To understand the sense of Hartogs’ phenomenon, consider the simple example of 𝑧↦1𝑧 in one complex variable. This function is holomorphic on ℂ∗ with a singularity at 0 that cannot be removed. However, in the case of several complex variables, Hartogs’ phenomenon asserts that if a function is holomorphic in a domain that excludes a compact subset, it can be extended to a holomorphic function on the entire domain, including the excluded subset. One immediate conclusion from this is that isolated singularities are always removable in higher dimensions.

Theorem 11.4.1 (Hartogs’ Extension Theorem).

Let Ω⊆ℂ𝑛 (𝑛≥2) be a domain and 𝐾⊆Ω be a compact subset such that Ω∖𝐾 is connected. Then any holomorphic function 𝑓:Ω∖𝐾→ℂ has a unique extension to a holomorphic function 𝑓̃:Ω→ℂ such that 𝑓̃≡𝑓 on Ω∖𝐾.

Hartogs, in 1906, first proved his extension theorem using an integral formula, but was considered to be incomplete with gaps. In 1939, Fueter gave a proof for the case 𝑛=2, and later Bochner and Martinelli developed more general integral kernel methods for higher dimensions. Finally, Ehrenpreis, in 1961, provided a succinct and analytic proof using the 𝜕‾ operator: by multiplying the function by a smooth cutoff or bump function and then solving a 𝜕‾-problem to correct the non-holomorphy introduced by the cutoff, one can construct the desired extension.

The general study of complex function theory in multiple variables is made practical with the definition of differential forms in multiple variables and in particular the 𝜕‾-problem, whose utility in a single variable has already been preluded to many times before.

Only certain cases of Theorem 11.4.1Theorem 11.4.1Theorem 11.4.1 are proved here.

Proposition 11.4.2.

Let Ω⊆ℂ𝑛 be a Reinhardt domain centered at 𝟎. Let 𝑓:Ω→ℂ be holomorphic. Then 𝑓 admits the unique Laurent expansion

𝑓(𝒛)=∑𝒌∈ℤ𝑛𝑎𝒌𝒛𝒌,𝑎𝒌∈ℂ,(11.4.1)
converging absolutely and uniformly on compact subsets of Ω.
Proof.

Let 𝒘=(𝑤1,…,𝑤𝑛)∈Ω with each 𝑤𝑗≠0. Since Ω is a Reinhardt domain, for every 𝜽=(𝜃1,…,𝜃𝑛)∈[−π,π]𝑛 the point

𝑧𝑗=𝑤𝑗ei𝜃𝑗,𝑗∈ℕ≤𝑛

lies in Ω; moreover, the set of all such points, 𝐾, is compact in Ω. Assuming that a Laurent expansion exists and converges uniformly on this compact set, termwise integration on both sides of 𝑓(𝒛)𝒛𝒌+(1,…,1)=∑𝒋∈ℤ𝑛𝑎𝒋𝒛𝒋−𝒌−(1,…,1) gives

𝑎𝒌=1(2πi)𝑛∫𝐾𝑓(𝒛)𝒛𝒌+(1,…,1)d𝒛=𝒘−𝒌(2π)𝑛∫[−π,π]𝑛𝑓(𝑤1ei𝜃1,…,𝑤𝑛ei𝜃𝑛)e−i(𝑘1𝜃1+…+𝑘𝑛𝜃𝑛)d𝜃1⋯d𝜃𝑛,

which shows that the coefficients 𝑎𝒌 are uniquely determined.

To prove the existence of such an expansion, again fix 𝑤∈Ω. Because Ω is Reinhardt, there exists 𝜀>0 such that the polyannulus

Ω(𝒘,𝜀)={𝒛=(𝑧1,…,𝑧𝑛)∈ℂ𝑛:|𝑤𝑗|−𝜀<|𝑧𝑗|<|𝑤𝑗|+𝜀,𝑗∈ℕ≤𝑛}

is contained in Ω. On this set one may perform one-variable Laurent expansions (Theorem 4.1.3Theorem 4.1.3Theorem 4.1.3) successively in each coordinate, holding the others fixed. Since 𝑓 is holomorphic on Ω, this procedure yields

𝑓(𝒛)=∑𝒌∈ℤ𝑛𝑎𝒌(𝒘)𝒛𝒌,

a Laurent series converging uniformly on a neighborhood of 𝑤. If 𝒘′∈Ω(𝒘,𝜀), then the same construction gives

𝑓(𝑧)=∑𝒌∈ℤ𝑛𝑎𝒌(𝒘′)𝑧𝒌.

By uniqueness of Laurent coefficients (in the one-variable expansions), it follows that 𝑎𝒌(𝒘′)=𝑎𝒌(𝒘). Thus the coefficient functions 𝑎𝒌(𝒘) are locally constant on Ω, and since Ω is connected they are constant throughout Ω; we therefore write them simply as 𝑎𝒌.

Consequently,

𝑓(𝒛)=∑𝒌∈ℤ𝑛𝑎𝒌𝒛𝒌

holds in a neighborhood 𝑈𝒘 of every point 𝒘 in Ω. If 𝐾⊂Ω is compact, then {𝑈𝒘:𝒘∈𝐾} is an open cover of 𝐾, and hence it admits a finite subcover. Then the uniform convergence of the series on 𝐾 follows from the uniform convergence on each 𝑈𝒘 in the subcover. The absolute convergence follows from the absolute convergence of the one-variable Laurent expansions used in the construction.

⁠□

Proposition 11.4.3.

Let Ω⊂ℂ𝑛 be a Reinhardt domain centered at 𝟎 such that Ω satisfies the condition that for every 𝑗∈ℕ≤𝑛, there exists a point in Ω of the form 𝑤𝑗𝒆𝑗 (𝒆𝑗 is the 𝑗-th unit vector and 𝑤𝑗∈ℂ∗). Then any holomorphic function 𝑓:Ω→ℂ has the expansion

𝑓(𝒛)=∑𝒌∈ℤ≥0𝑛𝑎𝒌𝒛𝒌,
which converges locally uniformly and absolutely on Ω.
Proof.By Proposition 11.4.2Proposition 11.4.2Proposition 11.4.2, 𝑓 has an expansion matching the form of (11.4.1)(11.4.1)(11.4.1). Now for a fixed 𝑗∈ℕ≤𝑛, if not all 𝑎𝒌’s with 𝑘𝑗<0 are zero, then fixing all variables except 𝑧𝑗 gives a Laurent expansion that does not uniformly converge on all compact neighborhoods of 𝑤𝑗𝒆𝑗∈Ω (either a pole or an essential singularity with respect to 𝑧𝑗). Thus, we are left only with 𝒌 containing nonnegative components, and the conclusion holds. ⁠□

Theorem 11.4.4 (Hartogs’ Extension Theorem for Reinhardt Domains).

Let Ω⊂ℂ𝑛 be a Reinhardt domain centered at 𝟎 such that Ω satisfies the condition that for every 𝑗∈ℕ≤𝑛, there exists a point in Ω of the form 𝑎𝑗𝒆𝑗 (𝒆𝑗 is the 𝑗-th unit vector and 𝑎𝑗∈ℂ∗). Let 𝑓:Ω→ℂ be holomorphic. Then 𝑓(𝒛) can be analytically continued to the complete Reinhardt domain Ω̃ defined by

Ω̃={(𝜌1𝑧1,𝜌2𝑧2,…,𝜌𝑛𝑧𝑛)∈ℂ𝑛:0≤𝜌𝑗≤1,𝑗∈ℕ≤𝑛,(𝑧1,…,𝑧𝑛)∈Ω}
In other words, ∃𝑓̃:Ω̃→ℂ holomorphic such that 𝑓̃≡𝑓 on Ω.
Proof.

By Proposition 11.4.3Proposition 11.4.3Proposition 11.4.3, 𝑓 has the expansion

𝑓(𝒛)=∑𝒌∈ℤ≥0𝑛𝑎𝒌𝒛𝒌,

which absolutely converges on Ω. Substituting (𝜌1𝑧1,…,𝜌𝑛𝑧𝑛) for 𝒛∈Ω with 0≤𝜌𝑗≤1 for each 𝑗∈ℕ≤𝑛 gives

|𝑓̃(𝜌1𝑧1,…,𝜌𝑛𝑧𝑛)|=|∑𝒌∈ℤ≥0𝑛𝑎𝒌(𝜌1𝑧1)𝑘1⋯(𝜌𝑛𝑧𝑛)𝑘𝑛|≤∑𝒌∈ℤ≥0𝑛|𝑎𝒌𝒛𝒌|,
which converges. The function 𝑓̃:Ω̃→ℂ is then holomorphic as it is given by a power series expansion, and it satisfies 𝑓̃≡𝑓 on Ω. ⁠□

Example 11.4.5.

Set 𝑟>0 and let

Ω(𝑟)=𝐵𝑛∖𝑟𝐵𝑛‾={𝒛:𝑟<‖𝒛‖<1}
define a Reinhardt domain. Then any holomorphic function 𝑓:Ω(𝑟)→ℂ can be analytically continued to the entire unit ball 𝐵𝑛 by Theorem 11.4.4Theorem 11.4.4Theorem 11.4.4.