False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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FALSE: an arbitrary pointwise supremum of subharmonic functions is subharmonic

Statement refuted

The pointwise supremum of an arbitrary family of subharmonic functions is always subharmonic.

Facts & Assumptions

Given: For each $n \geq 1$ , the function $u_{n} (z) = \max {n Re z, - 1}$ on the unit disc.

[L1]

Finite maxima of subharmonic functions are subharmonic; in particular, the maximum of a harmonic function and a constant is subharmonic (Positive linear combinations and finite maxima preserve subharmonicity).

[L2]

The upper-envelope theorem requires a locally bounded-above family before taking a supremum and then regularizing it (The upper-semicontinuous regularization of a locally bounded-above subharmonic supremum is subharmonic).

Refutation

technique · direct

1.1

The function $Re z$ is harmonic on the unit disc, so $n Re z$ is harmonic for each $n$ . By [L1], each [L1, given, algebra] $u_{n} (z) = \max {n Re z, - 1}$ is subharmonic.

L1givenalgebra

2.1

Their pointwise supremum is [step 1.1, algebra] $u (z) = \sup_{n} u_{n} (z) = {\begin{cases} Re z, & Re z < 0, \\ 0, & Re z = 0, \\ + \infty, & Re z > 0. \end{cases}$ This function is not even finite-valued on the right half-disc, so it cannot be subharmonic in the page's convention.

step 1.1algebra

3.1

Therefore the arbitrary-supremum claim is false. Step 2.1 is also exactly why [L2] insists on local boundedness above and upper-semicontinuous regularization.

L2step 2.1∎

Depends on

Used by

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Sources

Sheldon Axler, Paul Bourdon, and Wade Ramey, Harmonic Function Theory, 2nd ed. (standard reference, not scraped)