CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02

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Polynomials are uniformly dense in $C ([0, 1], R)$

Statement

For every $f \in C ([0, 1], R)$ and $ε > 0$ , there is a polynomial $p$ with $\sup_{x \in [0, 1]} ∣ p (x) - f (x) ∣ < ε$ .

Facts & Assumptions

Given: $f \in C ([0, 1], R)$ and $ε > 0$ .

[L1]

The Bernstein polynomials of $f$ converge uniformly to $f$ (Bernstein polynomials converge uniformly to every continuous function on $[0, 1]$ ).

Proof

technique · direct

1.1

Choose $n$ with $\sup_{x} ∣ B_{n} (f) (x) - f (x) ∣ < ε$ .

L1choose

2.1

The finite defining sum for $B_{n} (f)$ is a polynomial in $x$ , so $p : = B_{n} (f)$ has the required property.

step 1.1algebra∎

Depends on

Bernstein polynomials converge uniformly to every continuous function on $[0,1]$

Used by

Dependency tree · two levels

4 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

Bernstein polynomial (Encyclopedia of Mathematics) (standard reference, not scraped)

Polynomials are uniformly dense in C([0,1],R)C([0,1],R)