MthT 430 Notes Derivatives of Convex Functions

A function f is convex on an interval I if every secant line is above the graph on I.

Algebraically, convexity is expressed by: If X₁ < X₂, then for X₁ < X < X₂,

f(X) ≤ f(X1) + f(X2) − f(X1) X2− X1 (X − X1).

1.

Show that if X₁ < X₂ < X₃, then

f(X2) − f(X1) X2 − X1 ≤ f(X3) −f(X1) X3 − X1 , f(X3) −f(X1) X3 − X1 ≤ f(X3) − f(X2) X3 − X2 .

This is an algebraic verification of the geometric observation that slope―P₁ P₂ ≤ slope―P₁ P₃ and slope―P₁ P₃ ≤ slope―P₂ P₃.

In particular, as X₃ decreases to X₂, the difference quotient

f(X3) − f(X2) X3 − X2

decreases and is bounded below by

f(X2) − f(X1) X2 − X1 .

Thus

lim X → X2+  f(X) − f(X2) X − X2 = D+f(X2)

exists. The right hand limit of the difference quotient,

D+f(x) = lim ∆x→ 0+  f(x + ∆x) − f(x) ∆x

is called the right derivative or right derivate of f at x.

Similarly the left derivative

lim X → X2−  f(X) − f(X2) X − X2 = D−f(X2)

exists (and is ≤ D₊f(X₂)).

Thus we have shown: If f is convex on an open interval I = (a,b), then for each x ∈ I, the left derivative D₋f(x) and the right derivative D₊f(x) both exist. If they are the same, then f^′(x₀) exists. Note that if a < x₁ < x₂ < b, then

D−f(x1) ≤ D+f(x1) ≤ D−f(x2) ≤ D+f(x2).

It follows that D₋f(x) and D₋f(x) are nondecreasing functions on I.

2.

Is the following result true? If not, give a counterexample.

Theorem. If f is convex on an open interval I = (a,b), then f is continuous on I.

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