CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

How statement and proof provenance work

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Every continuous function on $[0, 1]$ is uniformly approximated by everywhere-differentiable functions whose derivative vanishes at a prescribed point

Statement

Let $c \in (0, 1)$ , let $f \in C ([0, 1], R)$ and let $ε > 0$ . Then there is a function $g : R \to R$ , differentiable at every real point (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set), with

$g^{'} (c) = 0 and \sup_{x \in [0, 1]} ∣ f (x) - g (x) ∣ < ε .$

Since a function differentiable at every real point is continuous there, the restrictions to $[0, 1]$ of the everywhere-differentiable functions with vanishing derivative at $c$ are uniformly dense in $C ([0, 1], R)$ .

The source states this for $c = 1 / 2$ and for differentiability on $(0, 1)$ ; the statement above is the altered form obtained by letting the point be arbitrary and by producing an approximant differentiable on all of $R$ , which is what the construction below actually delivers. Nothing in the proof uses $0 < c < 1$ ; the restriction to $(0, 1)$ is kept only so that $c$ is an interior point of the interval on which the approximation is measured.

Facts & Assumptions

Given: A point $c \in (0, 1)$ , a function $f \in C ([0, 1], R)$ and a real $ε > 0$ .

[L1]

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

[L2]

Let $a_{0}, \dots, a_{n - 1}$ be real and let $P : R \to R$ be the polynomial function $P (x) = \sum_{j < n} a_{j} x^{j}$ . Then $P$ is differentiable at every $c \in R$ , and $P^{'} (c) = \sum_{1 \leq j < n} j a_{j} c^{j - 1}$ , the term of index $0$ being $0$ (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term).

[L3]

Let $A \subseteq R$ , let $c \in A$ be a limit point of $A$ , let $u, v : A \to R$ be differentiable at $c$ and let $α \in R$ . Then $u + v$ is differentiable at $c$ with $(u + v)^{'} (c) = u^{'} (c) + v^{'} (c)$ , and $α u$ is differentiable at $c$ with $(α u)^{'} (c) = α u^{'} (c)$ (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L4]

Let $A, B \subseteq R$ , let $v : A \to R$ with $v [A] \subseteq B$ and let $u : B \to R$ . Let $c \in A$ be a limit point of $A$ at which $v$ is differentiable, put $b : = v (c)$ , and suppose $b$ is a limit point of $B$ at which $u$ is differentiable. Then $u \circ v$ is differentiable at $c$ and $(u \circ v)^{'} (c) = u^{'} (v (c)) v^{'} (c)$ (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[L5]

The functions $\sin$ and $\cos$ are differentiable on $R$ , with $(\sin x)^{'} = \cos x$ and $(\cos x)^{'} = - \sin x$ ; also $\sin 0 = 0$ and $\cos 0 = 1$ (The derivatives of sine and cosine are cosine and minus sine).

[L6]

For every real $x$ , $\sin^{2} x + \cos^{2} x = 1$ ; consequently $∣ \sin x ∣ \leq 1$ and $∣ \cos x ∣ \leq 1$ (Parity and the Pythagorean identity for sine and cosine).

[L7]

A function differentiable at a point is continuous at that point (A function differentiable at $c$ is continuous at $c$ ).

[L8]

For every $ε > 0$ in a complete ordered field there is a natural number $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

Proof

technique · direct

1.1givenL1choose

By [L1] choose a polynomial $p$ with $\sup_{x \in [0, 1]} ∣ p (x) - f (x) ∣ < ε / 3$ .

2.1step 1.1L2

$p$ is differentiable at every real point; put $a : = p^{'} (c)$ . Every real point is a limit point of $R$ , so the derivatives below are all defined symbols.

3.1step 1.1step 2.1algebra

Case $a = 0$ . Put $g : = p$ . Then $g$ is differentiable at every real point with $g^{'} (c) = a = 0$ , and $\sup_{x \in [0, 1]} ∣ f (x) - g (x) ∣ < ε / 3 < ε$ , which is the assertion.

3.2step 2.1L8choose

Case $a \neq 0$ . Then $∣ a ∣ > 0$ , so $ε / (3 ∣ a ∣) > 0$ , and by [L8] there is a natural number $λ \geq 1$ with $1 / λ < ε / (3 ∣ a ∣)$ .

4.1step 3.2choose

Define $v : R \to R$ by $v (x) = λ (x - c)$ and $g : R \to R$ by $g (x) = p (x) - (a / λ) \sin (v (x))$ .

5.1step 4.1L2algebra

$v (x) = λ (x - c)$ is the polynomial function with $a_{0} = - λ c$ and $a_{1} = λ$ , so [L2] makes it differentiable at every real point with $v^{'} (x) = 1 \cdot λ \cdot x^{0} = λ$ ; substituting $x = c$ gives $v (c) = 0$ .

5.2step 3.2step 4.1L6algebra

For every $x \in [0, 1]$ , $∣ g (x) - p (x) ∣ = (∣ a ∣ / λ) ∣ \sin (v (x)) ∣ \leq ∣ a ∣ / λ < ε / 3$ , using $∣ \sin ∣ \leq 1$ and step 3.2.

6.1step 5.1L4L5

Since $\sin$ is differentiable at every real point and every real point is a limit point of $R$ , the chain rule applies to $\sin \circ v$ at every real $x$ and gives $(\sin \circ v)^{'} (x) = \cos (v (x)) λ$ .

6.2step 1.1step 5.2algebra

For every $x \in [0, 1]$ , $∣ f (x) - g (x) ∣ \leq ∣ f (x) - p (x) ∣ + ∣ p (x) - g (x) ∣ < ε / 3 + ε / 3 = 2 ε / 3$ , so $2 ε / 3$ is an upper bound for $∣ f - g ∣$ on $[0, 1]$ and therefore $\sup_{x \in [0, 1]} ∣ f (x) - g (x) ∣ \leq 2 ε / 3 < ε$ .

7.1step 2.1step 4.1step 6.1L3algebra

By [L3], $g$ is differentiable at every real point, with $g^{'} (x) = p^{'} (x) - (a / λ) λ \cos (v (x)) = p^{'} (x) - a \cos (v (x))$ .

8.1step 2.1step 5.1step 7.1L5algebra

At $x = c$ we have $v (c) = 0$ and $\cos 0 = 1$ , so $g^{'} (c) = p^{'} (c) - a \cdot 1 = a - a = 0$ .

9.1step 3.1step 8.1step 6.2

In both cases a function $g : R \to R$ differentiable at every real point has been produced with $g^{'} (c) = 0$ and $\sup_{x \in [0, 1]} ∣ f (x) - g (x) ∣ < ε$ , which is the first assertion.

10.1step 9.1L7∎

Such a $g$ is continuous at every real point, so its restriction to $[0, 1]$ lies in $C ([0, 1], R)$ ; since $f \in C ([0, 1], R)$ and $ε > 0$ were arbitrary, these restrictions are uniformly dense in $C ([0, 1], R)$ , which is the second assertion.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

37 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. M. Erdman, A Companion to Real Analysis, Proposition 21.2.10 (standard reference, not scraped)

Every continuous function on [0,1][0,1] is uniformly approximated by everywhere-differentiable functions whose derivative vanishes at a prescribed point