Convex Set in $\mathbb{R}^n$ ¶

Definition¶

A set $C\subseteq\mathbb{R}^n$ is convex if for every $\boldsymbol{x},\boldsymbol{y}\in C$ and every $\lambda\in[0,1]$ $[0, 1]$ ,

\lambda \boldsymbol{x}+(1-\lambda)\boldsymbol{y}\in C.

Equivalently, the whole line segment between any two points of $C$ lies inside $C$ .

Interpretation¶

A convex set contains, with any two of its points, the whole straight segment between them. In optimization terms: moving in a straight line from one feasible point to another never leaves the feasible region.

Two shaded regions. Left: a convex set in which the segment between two points x and y stays inside. Right: a set with a dent, where part of the segment between x and y leaves the set.

In the left set the segment from $\boldsymbol{x}$ to $\boldsymbol{y}$ stays inside, and $\boldsymbol{z}$ is one of its points $\lambda\boldsymbol{x}+(1-\lambda)\boldsymbol{y}$ $λ) y$ . The right set has a dent, and part of the drawn segment runs outside it. One such segment proves that a set is not convex; convexity needs every pair of points.

Drag $\boldsymbol{x}$ and $\boldsymbol{y}$ inside the set and move $\lambda$ : a set is convex when every such segment stays inside. Find a segment that leaves the L-shape.

As $\lambda$ runs from $1$ to $0$ , the point $\lambda\boldsymbol{x}+(1-\lambda)\boldsymbol{y}$ moves along the segment from $\boldsymbol{x}$ to $\boldsymbol{y}$ (parametric representation of a line); the two weights are nonnegative and add up to $1$ . If $\boldsymbol{a}^{\top}\boldsymbol{x}\le b$ and $\boldsymbol{a}^{\top}\boldsymbol{y}\le b$ $b$ , then $\boldsymbol{a}^{\top}(\lambda\boldsymbol{x}+(1-\lambda)\boldsymbol{y})\le\lambda b+(1-\lambda)b=b$ $b$ , so a half-space is convex, and so is every intersection of half-spaces, such as the feasible region of a linear program. Unions behave differently: two disjoint discs together are not convex.

In $\mathbb{R}$ a segment is an interval $[x,y]$ , which is why convexity of a function was defined on an interval. A convex function on $\mathbb{R}^n$ needs a convex domain for the same reason: $f(\lambda\boldsymbol{x}+(1-\lambda)\boldsymbol{y})$ must be defined.

Convex sets need not be closed or bounded: an open ball, a half-space and $\mathbb{R}^n$ itself are convex.

Key Results¶

Intersections of convex sets are convex.
Euclidean open balls, closed balls, half-spaces, and affine subspaces are convex. The same ball conclusion holds for norm-induced metrics, but not for arbitrary metrics.
A feasible region described by linear equality and weak linear inequality constraints is convex.

Examples¶

The set $\{(x,y)\colon x+y\le 1\}$ is convex because it is a half-space.
The unit disk $\{(x,y)\colon x^2+y^2\le 1\}$ is convex, but the unit circle $\{(x,y)\colon x^2+y^2=1\}$ is not.

Prerequisites¶

Sources¶

Math IV: Multivariate Calculus · Week 4, Lecture 1

Aliases: convex set in rn, convex set Keywords: convex subset, line segment property Symbols: element of, lambda, real numbers, subset

Last updated: 20 October 2026 03:02

Convex Set in Rn\mathbb{R}^nRn¶