Hyperplane separation theorem

The proof is based on the following lemma:

Proof of lemma: Let ${\displaystyle \delta =\inf\{|x|:x\in K\}.}$ Let ${\displaystyle x_{j}}$ be a sequence in ${\displaystyle K}$ such that ${\displaystyle |x_{j}|\to \delta }$. Note that ${\displaystyle (x_{i}+x_{j})/2}$ is in ${\displaystyle K}$ since ${\displaystyle K}$ is convex and so ${\displaystyle |x_{i}+x_{j}|^{2}\geq 4\delta ^{2}}$. Since

${\displaystyle |x_{i}-x_{j}|^{2}=2|x_{i}|^{2}+2|x_{j}|^{2}-|x_{i}+x_{j}|^{2}\leq 2|x_{i}|^{2}+2|x_{j}|^{2}-4\delta ^{2}\to 0}$

as ${\displaystyle i,j\to \infty }$, ${\displaystyle x_{i}}$ is a Cauchy sequence and so has limit x in ${\displaystyle K}$. It is unique since if y is in ${\displaystyle K}$ and has norm δ, then${\displaystyle |x-y|^{2}\leq 2|x|^{2}+2|y|^{2}-4\delta ^{2}=0}$ and x = y. ${\displaystyle \square }$

Proof of theorem: Given disjoint nonempty convex sets A, B, let

${\displaystyle K=A+(-B)=\{x-y\mid x\in A,y\in B\}.}$

Since ${\displaystyle -B}$ is convex and the sum of convex sets is convex, ${\displaystyle K}$ is convex. By the lemma, the closure ${\displaystyle {\overline {K}}}$ of ${\displaystyle K}$, which is convex, contains a vector ${\displaystyle v}$ of minimum norm. Since ${\displaystyle {\overline {K}}}$ is convex, for any ${\displaystyle n}$ in ${\displaystyle K}$, the line segment

${\displaystyle v+t(n-v),\,0\leq t\leq 1}$

lies in ${\displaystyle {\overline {K}}}$ and so

${\displaystyle |v|^{2}\leq |v+t(n-v)|^{2}=|v|^{2}+2t\langle v,n-v\rangle +t^{2}|n-v|^{2}}$.

For ${\displaystyle 0<t\leq 1}$, we thus have:

${\displaystyle 0\leq 2\langle v,n\rangle -2|v|^{2}+t|n-v|^{2}}$

and letting ${\displaystyle t\to 0}$ gives: ${\displaystyle \langle n,v\rangle \geq |v|^{2}}$. Hence, for any x in A and y in B, we have: ${\displaystyle \langle x-y,v\rangle \geq |v|^{2}}$. Thus, if v is nonzero, the proof is complete since

${\displaystyle \inf _{x\in A}\langle x,v\rangle \geq |v|^{2}+\sup _{y\in B}\langle y,v\rangle .}$

More generally (covering the case v = 0), let us first take the case when the interior of ${\displaystyle K}$ is nonempty. The interior can be exhausted by a nested sequence of nonempty compact convex subsets ${\displaystyle K_{1}\subset K_{2}\subset K_{3}\subset \cdots }$. Since 0 is not in ${\displaystyle K}$, each ${\displaystyle K_{n}}$ contains a nonzero vector ${\displaystyle v_{n}}$ of minimum length and by the argument in the early part, we have: ${\displaystyle \langle x,v_{n}\rangle \geq 0}$ for any ${\displaystyle x\in K_{n}}$. We can normalize the ${\displaystyle v_{n}}$'s to have length one. Then the sequence ${\displaystyle v_{n}}$ contains a convergent subsequence (because the n-sphere is compact) with limit v, which is nonzero. We have ${\displaystyle \langle x,v\rangle \geq 0}$ for any x in the interior of ${\displaystyle K}$ and by continuity the same holds for all x in ${\displaystyle K}$. We now finish the proof as before. Finally, if ${\displaystyle K}$ has empty interior, the affine set that it spans has dimension less than that of the whole space. Consequently ${\displaystyle K}$ is contained in some hyperplane ${\displaystyle \langle \cdot ,v\rangle =c}$; thus, ${\displaystyle \langle x,v\rangle \geq c}$ for all x in ${\displaystyle K}$ and we finish the proof as before. ${\displaystyle \square }$

The number of dimensions must be finite. In infinite-dimensional spaces there are examples of two closed, convex, disjoint sets which cannot be separated by a closed hyperplane (a hyperplane where a continuous linear functional equals some constant) even in the weak sense where the inequalities are not strict.[5]

The above proof also proves the first version of the theorem mentioned in the lede (to see it, note that ${\displaystyle K}$ in the proof is closed under the hypothesis of the theorem below.)

Here, the compactness in the hypothesis cannot be relaxed; see an example in the next section. This version of the separation theorem does generalize to infinite-dimension; the generalization is more commonly known as the Hahn–Banach separation theorem.

We also have:

This follows from the standard version since the separating hyperplane cannot intersect the interiors of the convex sets.

Note that the existence of a hyperplane that only "separates" two convex sets in the weak sense of both inequalities being non-strict obviously does not imply that the two sets are disjoint. Both sets could have points located on the hyperplane.

The theorem does not apply if one of the bodies is not convex.

If one of A or B is not convex, then there are many possible counterexamples. For example, A and B could be concentric circles. A more subtle counterexample is one in which A and B are both closed but neither one is compact. For example, if A is a closed half plane and B is bounded by one arm of a hyperbola, then there is no strictly separating hyperplane:

${\displaystyle A=\{(x,y):x\leq 0\}}$

${\displaystyle B=\{(x,y):x>0,y\geq 1/x\}.\ }$

(Although, by an instance of the second theorem, there is a hyperplane that separates their interiors.) Another type of counterexample has A compact and B open. For example, A can be a closed square and B can be an open square that touches A.

In the first version of the theorem, evidently the separating hyperplane is never unique. In the second version, it may or may not be unique. Technically a separating axis is never unique because it can be translated; in the second version of the theorem, a separating axis can be unique up to translation.

The separating axis theorem (SAT) says that:

Two convex objects do not overlap if there exists a line (called axis) onto which the two objects' projections do not overlap.

SAT suggests an algorithm for testing whether two convex solids intersect or not.

Regardless of dimensionality, the separating axis is always a line. For example, in 3D, the space is separated by planes, but the separating axis is perpendicular to the separating plane.

The separating axis theorem can be applied for fast collision detection between polygon meshes. Each face's normal or other feature direction is used as a separating axis, as well as the cross products. Note that this yields possible separating axes, not separating lines/planes.

If the cross products were not used, certain edge-on-edge non-colliding cases would be treated as colliding. For increased efficiency, parallel axes may be calculated as a single axis.

• Dual cone
• Farkas's lemma
• Kirchberger's theorem
• Optimal control

• Boyd, Stephen P.; Vandenberghe, Lieven (2004). Convex Optimization (pdf). Cambridge University Press. ISBN 978-0-521-83378-3.
• Golshtein, E. G.; Tretyakov, N.V. (1996). Modified Lagrangians and monotone maps in optimization. New York: Wiley. p. 6. ISBN 0-471-54821-9.
• Shimizu, Kiyotaka; Ishizuka, Yo; Bard, Jonathan F. (1997). Nondifferentiable and two-level mathematical programming. Boston: Kluwer Academic Publishers. p. 19. ISBN 0-7923-9821-1.

• Collision detection and response