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

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Polynomials are uniformly dense in $C ([a, b], R)$ for every closed interval

Statement

For $a \leq b$ , every continuous real function on $[a, b]$ is a uniform limit of polynomials.

Facts & Assumptions

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

[L1]

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

[L2]

Closed intervals have the endpoint convention of Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length.

Proof

technique · direct

1.1givenL2algebra

If $a = b$ , the constant polynomial $q (x) = f (a)$ agrees with $f$ on the singleton interval.

1.2givenL2construct

Now suppose $a < b$ and define $g (t) = f (a + (b - a) t)$ on $[0, 1]$ .

1.3L1choose

Choose a polynomial $p$ with $∣ p (t) - g (t) ∣ < ε$ on $[0, 1]$ by [L1].

2.1step 1.1step 1.2step 1.3algebra∎

Then $q (x) = p ((x - a) / (b - a))$ is a polynomial and satisfies $∣ q (x) - f (x) ∣ < ε$ on $[a, b]$ . Together with step 1.1 this proves both cases.

Depends on

Dependency tree · 2 levels

Sources

Bernstein polynomial (Encyclopedia of Mathematics)