Mathematics · Glossary

What is convex set?

Also known as: convex hull

Definition 17.11 University Mathematics — Year 2 · Chapter 17 — Affine Spaces

A subset CC of an affine space is convex when it contains every barycenter with nonnegative weights of its points — equivalently, every segment [A,B]={bar(A,1t;B,t):t[0,1]}\intcc{A}{B} = \{\operatorname{bar}(A, 1-t; B, t) : t \in \intcc{0}{1}\} between its points. The convex hull conv(S)\operatorname{conv}(S) is the set of all nonnegative-weight barycenters of points of SS — the smallest convex set containing SS.

Examples

Example 17.12 (Epigraphs are convex sets)

The region C={(x,y):yx2}C = \{(x, y) : y \geq x^2\} above the parabola is convex: for (x1,y1),(x2,y2)C(x_1, y_1), (x_2, y_2) \in C and t[0,1]t \in \intcc01, the convexity inequality of the square function gives

((1t)x1+tx2)2(1t)x12+tx22(1t)y1+ty2,\bigl((1-t)x_1 + tx_2\bigr)^2 \leq (1-t)x_1^2 + tx_2^2 \leq (1-t)y_1 + ty_2 ,

so the barycenter stays above the parabola. The computation is general: {yf(x)}\{y \geq f(x)\} is convex exactly when ff is a convex function — convex sets and convex functions (Chapter 8) are two faces of one notion, epigraphs being the dictionary. This is the geometric reason support lines exist for convex functions, the fact that will prove Jensen’s inequality in Chapter 22.

Example 17.13 (Redundant generators of a convex hull)

Let S={(0,0),(2,0),(2,2),(0,2),(1,1)}S = \{(0,0), (2,0), (2,2), (0,2), (1,1)\}. The fifth point is the barycenter

(1,1)=bar((0,0),12; (2,2),12),(1,1) = \operatorname{bar}\bigl((0,0), \tfrac12;\ (2,2), \tfrac12\bigr),

so it already lies in the hull of the other four: conv(S)\operatorname{conv}(S) is the square with the four corners as vertices. In general, a point of SS that is a nonnegative-weight barycenter of the other points of SS can be deleted without changing the hull; the points that can never be deleted (here the four corners) are the extreme points of the hull. Determining them is a pure barycenter computation: (2,0)(2,0), say, cannot be written as bar\operatorname{bar} of the remaining points with nonnegative weights, because the first coordinate would force all weight onto points with x=2x = 2, and the second coordinate then fails. Convexity questions reduce, again and again, to solving small weighted systems.

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