Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-30
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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1/z is not uniformly approximable by polynomials on the unit circle

Example

Let γ(t)=eit for 0≤t≤2π. The function z↦1/z on the unit circle is not the uniform limit there of any sequence of polynomials.

Facts & Assumptions

Given: The unit-circle contour γ(t)=eit and the function 1/z on γ∗.

[L1]

Every polynomial has a global primitive, so its integral around a closed contour is 0 (The line integral of a continuous function admitting a primitive is that primitive's endpoint increment along every rectifiable path).

Verification

technique · contradiction
1.1givenassume-contra

Suppose polynomials pn converge uniformly to 1/z on γ∗. Then contour integration along the fixed rectifiable contour γ preserves the limit, so ∫γpn(z) dz→∫γdz/z.

2.1L1L2step 1.1discharge-contradiction∎

By [L1], every ∫γpn(z) dz is 0. But [L2] gives ∫γdz/z=∫02πi dt=2πi. This contradiction shows that no such polynomial sequence exists.

Depends on

Used by

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources