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A convex combination is a finite weighted average of points in a real vector space. For $n\ge 1$, it has the formThe nonnegative coefficients forbid subtracting a component, and their unit sum keeps the result on the same affine scale as the inputs. With two points it parametrizes their line segment; equal coefficients give their centroid. A convex set contains every finite convex combination of its points. Conversely, the two-point case already gives the defining property of a convex set.^[ref]

The distinction from a linear combination is concrete. Combining $(0,0)$ and $(2,0)$ with coefficients $-1$ and $2$ gives $(4,0)$, outside their line segment, despite the unit coefficient sum. Using two coefficients both equal to one gives $(2,0)$ here, but their sum is two, so it is not a convex combination with those coefficients. A numerical output may happen to coincide with a point admitting a different convex representation; the definition applies to the specified coefficients.

Consider a stipulated continuous-allocation problem. Amounts $x,y$ of two activities use $x+2y$ units of a common resource, with budget six. The feasible regionis a triangle with extreme points $(0,0)$, $(6,0)$ and $(0,3)$. It is the intersection of three closed half-spaces, hence a closed convex set. The general intersection argument is short: a line segment between two points in every constraint set belongs to every constraint set, so it belongs to their intersection.^[ref]

For an original numerical check, take $A=(1,2)$ and $B=(5,1/4)$. Their resource totals are $5$ and $11/2$. Every convex combination $Z(t)=(1-t)A+tB$ is nonnegative and has resource totalIn particular the midpoint is $(3,9/8)$, using $21/4$ units. This verifies the full line segment, rather than just a plotted sample of it.

This is useful whenever divisible quantities share linear limits: the feasible choices remain compatible under averaging. A new linear equation also preserves this property, because averaging two identical constraint values gives the same value. The affine subspace and closed half-space viewpoints explain this without requiring a separate argument for every budget.

A convex set permits every intermediate mixture of two of its points. If two choices meet its constraints, continuously averaging them also meets those constraints. This geometric property makes many feasible regions easier to explore: there is no forbidden gap along a straight move between feasible choices.

Formally, a subset $C$ of a real vector space is convex whenThe parameter $t$ traverses the entire line segment, including both endpoints. The definition does not require a closed subset, a bounded set, or a nonempty interior. The empty set, the whole vector space, and every affine subspace are convex; an open unit disk and a closed unit disk are both convex.^[ref]

The triangle inequality verifies the disk example directly: if $\Vert x\Vert ,\Vert y\Vert \le 1$, then $\Vert (1-t)x+ty\Vert \le (1-t)\Vert x\Vert +t\Vert y\Vert \le 1$. For the open disk, the last inequality is strict. The unit circle fails the same test: $(1,0)$ and $(-1,0)$ belong to it, but their midpoint $(0,0)$ does not. A curved route along the unit circle connects the endpoints, so being path-connected is weaker than being a convex set.

The distinction between a convex set and a convex function matters. The former concerns where one may choose a point; the latter concerns how a value changes along line segments. Convex optimization combines both: averaging preserves feasibility, and a convex function obeys an upper bound by the corresponding average of endpoint values. A convex set alone does not make an arbitrary objective a convex function or eliminate its inferior local minima.

For a proper cone $K$, suppose a point satisfies $Ax-b\in \mathrm{int}K$ and $p_{\ast }$ is finite. This generalized Slater condition gives strong duality and attainment of the conic dual problem's optimum. It does not by itself assert primal attainment.^[ref]

The two worked examples have strict primal points: $(u,v)=(2,2)$ has slack $(2,2,1)$ in the interior of the nonnegative orthant, and $(t,z)=(2,0)$ has slack $(2,1,0)$ in the interior of the second-order cone. Both optima were already certified by explicit feasible pairs. Slater's condition explains why a matching optimal dual variable can be expected under these hypotheses; checking a supplied zero-gap pair needs only weak duality.

Changing cone geometry preserves the bound mechanism. For a positive semidefinite cone, the dual cone pairing is the Frobenius inner product of two matrices. This connects the same certificate idea to semidefinite programming, while the central path uses strictly interior primal and dual points to approach vanishing gap.

In 1995, Lieven Vandenberghe and Stephen Boyd presented a primal-dual interior-point method for convex optimization with matrix inequalities. Their algorithmic study included exploiting problem structure and allowing incomplete Conjugate gradient method solves within an iteration. This illustrates how the primal-dual viewpoint guides computation as well as checking a final answer.^[ref]