Definition D1
f is differentiable at x when the limit lim (t→0) [f(x+t) − f(x)]/t exists, and that value is written f′(x).
Proof.
(1)The difference quotient is defined only for t ≠ 0. F/t denotes F·t⁻¹.convention
(2)When the limit exists, its value is unique. Therefore the notation f′(x) is legitimate.A1, A2
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Definition D2
Fix a real number c and, for t ≠ 0, set r(t) = [f(x+t) − f(x)]/t − c.
Proof.
(1)r is the difference between the difference quotient and c. This expression is defined whatever c is, and r depends on c.convention
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Lemma L1
For every real c and every t ≠ 0, f(x+t) − f(x) = t·c + t·r(t)
Proof.
(1)r(t) + c = [f(x+t) − f(x)]·t⁻¹D2, A8, E
(2)Multiplying both sides by t gives t·(r(t) + c) = ([f(x+t) − f(x)]·t⁻¹)·t(1), E
(3)Since t ≠ 0, the right side is f(x+t) − f(x)(2), A4, A5, A7, A6
(4)The left side is t·r(t) + t·c(2), A3
(5)Therefore f(x+t) − f(x) = t·c + t·r(t)(3), (4), A21
(6)This identity uses only the four arithmetic operations. It holds for any real c, as long as t is nonzero.(5)
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Lemma L2
If f is differentiable at x and c = f′(x), then lim (t→0) r(t) = 0
Proof.
(1)r(t) = [f(x+t) − f(x)]/t − f′(x)assumption, D2
(2)By hypothesis, the limit of the difference quotient is f′(x).assumption, D1
(3)Therefore for every ε > 0 there is a δ > 0 such that 0 < |t| < δ implies |[f(x+t) − f(x)]/t − f′(x)| < ε.(2), A1
(4)That left side is |r(t) − 0|.(1), (3), A19
(5)Therefore lim (t→0) r(t) = 0.(3), (4), A1
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Theorem T1
If f is differentiable at x and f′(x) ≠ 0, then there is an η₀ > 0 such that for all η with 0 < η < η₀, f(x − ηf′(x)) < f(x).
Proof.
(1)Set g = f′(x). By hypothesis g ≠ 0.assumption, D1
(2)Since g ≠ 0, |g| > 0, and g² = |g|² = |g|·|g| > 0.(1), A10, A11, A19
(3)Setting ε = |g|/2 gives ε > 0.(2), A18
(4)Applying L2 with c = g, there is a δ > 0 such that 0 < |t| < δ implies |r(t)| < |g|/2.(3), L2, A1
(5)Set η₀ = δ·|g|⁻¹. Since δ > 0 and |g| > 0, we have η₀ > 0.(2), (4), A17
(6)Suppose 0 < η < η₀ and set t = −ηg.assumption, (5)
(7)|t| = |−ηg| = η|g|, and η|g| < η₀|g| = δ.(6), A14, A15, A11, A4, A5, A7, A6
(8)Since η > 0 and |g| > 0, we have |t| > 0, that is, t ≠ 0.(6), (2), (7), A10, A11, A19
(9)Applying L1 with c = g and this t gives f(x − ηg) − f(x) = (−ηg)·g + (−ηg)·r(−ηg)L1, (6), (8)
(10)(−ηg)·g = −(ηg·g) = −ηg²A9, A4
(11)|(−ηg)·r(−ηg)| = η|g|·|r(−ηg)| < η|g|·(|g|/2) = ηg²/2(4), (7), (8), A14, A15, A11, A10, A4
(12)Since a ≤ |a|, (−ηg)·r(−ηg) < ηg²/2(11), A16, A13
(13)Therefore f(x − ηg) − f(x) < −ηg² + ηg²/2(9), (10), (12), A12
(14)Since ηg² > 0, we have ηg²/2 < ηg², and adding −ηg² to both sides gives −ηg² + ηg²/2 < 0(2), (6), A11, A18, A12, A20, A21
(15)Therefore f(x − ηf′(x)) − f(x) < 0, that is, f(x − ηf′(x)) < f(x)(1), (13), (14), A13, A12, A20
(16)η₀ comes from δ, and δ depends on f and x. All this theorem says is that such an η₀ exists; it is not a claim about arbitrary η.(4), (5), (15)
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Corollary C
If f′(x) = 0, then for every η, x − ηf′(x) = x.
Proof.
(1)η·f′(x) = η × 0 = 0assumption, A19, A5
(2)x − 0 = xA19
(3)Therefore the update stays in place. The hypothesis f′(x) ≠ 0 in T1 cannot be dropped.(1), (2)
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