Convexity is the geometry of not losing points while averaging. If two feasible designs are allowed but the straight path between them leaves the feasible region, then interpolation can create a forbidden design even before any optimisation or analysis begins.

[example: Two Feasible Points with a Forbidden Midpoint] Let $S=D_- \cup D_+$, where \begin{align*} D_-=\{p\in\mathbb{R}^2:\|p-(-2,0)\|_2\leq 1\} \end{align*} and \begin{align*} D_+=\{p\in\mathbb{R}^2:\|p-(2,0)\|_2\leq 1\}. \end{align*} The point $x=(-2,0)$ lies in $D_-$ because \begin{align*} \|x-(-2,0)\|_2=\|(0,0)\|_2=0\leq 1, \end{align*} and the point $y=(2,0)$ lies in $D_+$ because \begin{align*} \|y-(2,0)\|_2=\|(0,0)\|_2=0\leq 1. \end{align*} Thus $x,y\in S$. Their midpoint is \begin{align*} \frac{x+y}{2}=\frac{(-2,0)+(2,0)}{2}=\frac{(0,0)}{2}=(0,0). \end{align*} This midpoint is not in $D_-$, since \begin{align*} \|(0,0)-(-2,0)\|_2=\|(2,0)\|_2=2>1, \end{align*} and it is not in $D_+$, since \begin{align*} \|(0,0)-(2,0)\|_2=\|(-2,0)\|_2=2>1. \end{align*} Therefore $(0,0)\notin S$, even though it is the average of two points of $S$. This shows that a set can contain two locally solid pieces while still failing to contain the interpolation forced by averaging. [/example]

The failure in the example is not about smoothness, dimension, or coordinates. It is about whether the set contains each line segment joining two of its points, and that single requirement becomes the organizing principle for [convex geometry](/page/Convex%20Geometry).

## Definition

The basic question is how to formalize the phrase "no gaps along straight interpolation." A definition should mention only the segment joining two selected points, because longer finite averages can then be built from repeated two-point averages. This local segment test is the smallest condition that still controls the global averaging behavior of the set.

[definition: Convex Set] Let $V$ be a real [vector space](/page/Vector%20Space). A convex set in $V$ is a subset $C \subset V$ such that for every $x,y \in C$ and every $t \in [0,1]$, the point \begin{align*} (1-t)x + ty \end{align*} belongs to $C$. [/definition]

The definition says that every chord whose endpoints lie in $C$ lies entirely inside $C$. The parameter $t$ records position along the chord, so convexity is invariant under affine changes of coordinates rather than depending on a chosen origin.

The two-point condition is not enough notation for arguments that average several points at once. We need a way to record finite averaging without introducing an order of pairwise operations, while still preventing cancellation by negative coefficients and preserving affine invariance.

[definition: Convex Combination] Let $V$ be a real vector space. A convex combination of points $x_1,\ldots,x_m \in V$ is a point of the form \begin{align*} \sum_{i=1}^m \lambda_i x_i, \end{align*} where $\lambda_i \geq 0$ for each $i$ and \begin{align*} \sum_{i=1}^m \lambda_i = 1. \end{align*} [/definition]

The weights behave like masses attached to the points. Negative weights would allow cancellation and directions outside the geometric span of the selected points, while the condition that the weights sum to $1$ keeps the construction affine.

There is a small gap between the definition of convexity and the way convex combinations are used: convexity only mentions two points, while a convex combination may involve many. The obstruction is that a finite weighted average is not visibly produced in one step from the definition; it must be assembled by repeatedly joining intermediate points.

To use convexity in calculations, we need to know that this repeated interpolation never leaves the set. Once that is established, checking membership in a convex set can be reduced to expressing a point as a finite convex combination of known points.

[quotetheorem:10092]

This result turns convexity into a finite [algebraic closure](/page/Algebraic%20Closure) property. In practice, it lets us prove that a point remains feasible by writing it as a weighted average of known feasible points.

[example: The Disk Is Convex] Let $C=\{x\in\mathbb{R}^2:\|x\|_2\leq 1\}$. To show that $C$ is convex, take $x,y\in C$ and $t\in[0,1]$; we must prove that $(1-t)x+ty\in C$. By the triangle inequality for the Euclidean norm, \begin{align*} \|(1-t)x+ty\|_2\leq \|(1-t)x\|_2+\|ty\|_2. \end{align*} Since $t\in[0,1]$, both $1-t$ and $t$ are nonnegative, so absolute homogeneity gives \begin{align*} \|(1-t)x\|_2+\|ty\|_2=(1-t)\|x\|_2+t\|y\|_2. \end{align*} Because $x,y\in C$, we have $\|x\|_2\leq 1$ and $\|y\|_2\leq 1$, hence \begin{align*} (1-t)\|x\|_2+t\|y\|_2\leq (1-t)\cdot 1+t\cdot 1. \end{align*} Finally, \begin{align*} (1-t)\cdot 1+t\cdot 1=1-t+t=1. \end{align*} Therefore $\|(1-t)x+ty\|_2\leq 1$, so $(1-t)x+ty\in C$. Every segment between two points of the disk remains inside the disk, and hence the disk is convex. [/example]

The same computation works for balls in any [normed vector space](/page/Normed%20Vector%20Space) because it used only homogeneity and the triangle inequality. Convexity is therefore not a peculiarity of Euclidean circles; it is built into the geometry of norms.

The opposite question is just as important: given an arbitrary set, what is the smallest convex world that contains it? This construction is needed whenever a finite sample, a collection of constraints, or a set of extreme possibilities must be enlarged until averaging becomes legal.

[definition: Convex Hull] Let $V$ be a real vector space and let $A \subset V$. The convex hull of $A$, denoted $\operatorname{conv}(A)$, is the intersection of all convex subsets of $V$ that contain $A$. [/definition]

The intersection definition makes existence automatic because $V$ itself is convex and contains $A$. The more useful description is constructive: take all possible finite convex combinations of points of $A$.

The intersection definition is clean but not directly computable: it tells us what must contain the hull, not how to list its points. To use convex hulls in examples and calculations, we need to know that no points are missing from the finite convex combinations and no extra points have been added.