Definition D1
f is said to be convex on an interval I when f(x) < f(a) + [f(b) − f(a)](b − a)⁻¹ (x − a) for every a < b in I and every x with a < x < b.
Proof.
The right side is the value on the line joining the two points (a, f(a)) and (b, f(b)), that is, the value on the chord. Convexity is defined by comparing the graph and the chord.convention
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Definition D2
When f′ is differentiable on an interval I, its derivative is written f″ and called the second derivative.
Proof.
f″ is the derivative not of f but of f′.convention
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Theorem T1
If f′ is differentiable on an interval I and f″ > 0 on I, then f′ increases on I.
Proof.
(1)f″ is the derivative of f′.D2
(2)The derivative of f′ is positive on I, so f′ increases on I.(1), assumption, A2
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Theorem T2
If f is continuous on [a, b], differentiable on (a, b), and f′ increases on (a, b), then f is convex on [a, b].
Proof.
(1)Take a < x < b. f is continuous on [a, x] and [x, b] and differentiable on each open interval.assumption, convention
(2)By the mean value theorem, there is p ∈ (a, x) with f(x) − f(a) = f′(p)(x − a).(1), A1
(3)By the mean value theorem, there is q ∈ (x, b) with f(b) − f(x) = f′(q)(b − x).(1), A1
(4)p < x < q and f′ increases, so f′(p) < f′(q).(2), (3), assumption
(5)0 < x − a and 0 < b − x, so 0 < (x − a)(b − x).(1), A6, A8
(6)Multiplying both sides of (4) by (x − a)(b − x) and substituting (2)(3) gives [f(x) − f(a)](b − x) < [f(b) − f(x)](x − a)(2), (3), (4), (5), A5, A12, A13, E
(7)Adding (x − a)f(x) + (b − x)f(a) to both sides and using (b − x) + (x − a) = b − a gives (b − a)f(x) < (b − x)f(a) + (x − a)f(b)(6), A9, A11, A13, A17, A18, A19, A20
(8)0 < b − a, so 0 < (b − a)⁻¹, and multiplying both sides by it gives f(x) < [(b − x)f(a) + (x − a)f(b)](b − a)⁻¹(7), A6, A7, A5, A12, A13, A14, A15
(9)The right side equals f(a) + [f(b) − f(a)](b − a)⁻¹(x − a).A11, A12, A13, A14, A15, A17, A18, A19, A20
(10)Therefore f lies below the chord and is convex on [a, b].(8), (9), D1
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Theorem T3
If f is continuous on [a, b], twice differentiable on (a, b), and f″ > 0 on (a, b), then f is convex on [a, b].
Proof.
(1)f″ > 0, so f′ increases on (a, b).assumption, T1
(2)f′ increases, so f lies below the chord.(1), T2
(3)The chain has two links: f″ > 0 ⟹ f′ increases ⟹ convex. There is no route that skips the middle link.(1), (2)
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Corollary C
The converse of T3 fails. f(x) = x⁴ is convex, yet f″(0) = 0.
Proof.
(1)f′(x) = 4x³ and f″(x) = 12x², so f″(0) = 12 · (0 · 0) = 0.A4, D2, A16
(2)If x₁ < x₂, then x₁³ < x₂³, and 0 < 4, so 4x₁³ < 4x₂³.A3, A10, A5
(3)Therefore f′ increases on ℝ.(2)
(4)f′ increases, so f is convex.(3), T2
(5)Therefore the second derivative of a convex function need not be positive at every point.(1), (4)
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