ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-30

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$1 / z$ is not uniformly approximable by polynomials on the unit circle

Example

Let $γ (t) = e^{i t}$ for $0 \leq t \leq 2 π$ . The function $z \mapsto 1 / z$ on the unit circle is not the uniform limit there of any sequence of polynomials.

Facts & Assumptions

Given: The unit-circle contour $γ (t) = e^{i t}$ 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).

[L2]

The exponential parametrizes the unit circle (Complex sine, cosine, hyperbolic sine, and hyperbolic cosine from the complex exponential).

Verification

technique · contradiction

1.1

Suppose polynomials $p_{n}$ converge uniformly to $1 / z$ on $γ^{*}$ . Then contour integration along the fixed rectifiable contour $γ$ preserves the limit, so $\int_{γ} p_{n} (z) d z \to \int_{γ} d z / z$ .

givenassume-contra

2.1

By [L1], every $\int_{γ} p_{n} (z) d z$ is $0$ . But [L2] gives $\int_{γ} d z / z = \int_{0}^{2 π} i d t = 2 π i$ . This contradiction shows that no such polynomial sequence exists.

L1L2step 1.1discharge-contradiction∎

Depends on

Used by

FALSE: Runge's theorem gives polynomial approximation on every compact set False statement

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

J. Lebl, Guide to Cultivating Complex Analysis, Exercise 9.1.2 (standard reference, not scraped)

1/z1/z is not uniformly approximable by polynomials on the unit circle