= Convex set
{title2=$(1-t)x+ty\in C\quad(0\leq t\leq1)$}
{wiki}
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 when
$$
x,y\in C,\quad 0\leq t\leq1\quad\Longrightarrow\quad(1-t)x+ty\in C.
$$
The 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.<Convex Optimization by Stephen Boyd and Lieven Vandenberghe>[§2.1.4, pp. 23–25: convex sets and finite convex combinations]{ref}
The <triangle inequality> verifies the disk example directly: if $\|x\|,\|y\|\leq1$, then $\|(1-t)x+ty\|\leq(1-t)\|x\|+t\|y\|\leq1$. 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.
\Image[/convex-disk-and-nonconvex-circle-segments.png]
{title=A disk contains every chord, while a circle omits its chords}
{description=Original comparison of the closed <unit disk> and the <unit circle>. The circle's <midpoint> counterexample is explicit; the shaded disk panel illustrates one <line segment>, with the general statement established by the <triangle inequality>.}
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>.
Back to article page
