From Extension to Separation
The previous tutorial stated the Hahn-Banach Theorem as an extension result: a linear functional dominated by a sublinear function can be extended to the whole real vector space without losing domination. A central geometric consequence is that an open convex set can be separated from any point outside it by a continuous linear functional. This formulation turns the existence of a functional into a way to distinguish convex regions.
The separation statement has a strict inequality at every point of the open set. It does not, in general, promise a single positive gap that works uniformly for all points in that set. The distinction matters for unbounded sets and even for simple intervals. We will state the result precisely, obtain it from the dominated extension theorem, and examine what its inequalities do and do not imply.
The orientation of the inequality is a choice: replacing \(L\) by \(-L\) reverses it. The point \(x_0\) need not be on the boundary of \(U\), and \(U\) need not be bounded. The continuity conclusion is essential to this normed-space statement; the algebraic dominated extension theorem by itself does not mention a topology.
The Gauge of an Open Convex Neighborhood
To derive separation, we use a function that records how much a convex neighborhood must be dilated to reach a given vector. This is the Minkowski functional, also called the gauge. Its sublinearity makes the dominated extension form of Hahn-Banach applicable.
The gauge is finite because \(V\) contains a ball about zero. We record the properties needed for separation, including the relation between the gauge and the set \(V\).
Proof. Since \(V\) is open and contains \(0\), there is \(r>0\) such that \(B(0,r)\subseteq V\). For any \(x\in X\), if \(t>\|x\|/r\), then \(\|x/t\|<r\), so \(x/t\in V\) and \(x\in tV\). Thus the set in the infimum is nonempty, \(p_V(x)\) is finite, and taking the infimum over such \(t\) gives \(p_V(x)\leq\|x\|/r\). Also \(p_V(x)\geq0\), and \(p_V(0)=0\).
For \(s>0\), the condition \(sx\in tV\) is equivalent to \(x\in (t/s)V\). Taking infima gives \(p_V(sx)=s p_V(x)\); the equality also holds for \(s=0\). To prove subadditivity, let \(\varepsilon>0\). By the definition of infimum, choose \(s,t>0\) with \(x\in sV\), \(y\in tV\), \(s<p_V(x)+\varepsilon\), and \(t<p_V(y)+\varepsilon\). Write \(x=sv\) and \(y=tw\) for \(v,w\in V\). Convexity gives \[ \frac{x+y}{s+t}=\frac{s}{s+t}v+\frac{t}{s+t}w\in V, \] so \(x+y\in(s+t)V\). Therefore \[ p_V(x+y)\leq s+t<p_V(x)+p_V(y)+2\varepsilon. \] Since this holds for every \(\varepsilon>0\), \(p_V(x+y)\leq p_V(x)+p_V(y)\). Thus \(p_V\) is sublinear.
If \(x\in V\), then \(p_V(x)<1\). For \(x=0\) this follows from \(p_V(0)=0\). If \(x\ne0\), openness gives \(\delta>0\) small enough that \((1+\delta)x\in V\). Hence \(x\in\frac{1}{1+\delta}V\), and \(p_V(x)\leq\frac{1}{1+\delta}<1\). Conversely, if \(p_V(x)<1\), the definition of infimum gives some \(t<1\) with \(x\in tV\). Since \(0\in V\), convexity implies \(tV\subseteq V\) for \(0<t<1\). Therefore \(x\in V\). This proves \(V=\{p_V<1\}\). \(\square\)
The bound \(p_V(x)\leq\|x\|/r\), together with sublinearity, also shows continuity. Indeed, \(p_V(x)-p_V(y)\leq p_V(x-y)\leq\|x-y\|/r\), and interchanging \(x\) and \(y\) gives \(|p_V(x)-p_V(y)|\leq\|x-y\|/r\).
Proof of the Separation Statement
We translate the open set so that the point to be separated becomes the origin. The translated set misses zero. A gauge then supplies a sublinear bound, and the dominated extension theorem produces a functional whose value on one particular vector is large enough to force the desired strict inequality.
Proof of the Hahn-Banach Separation Theorem. Let \(C=U-x_0=\{u-x_0:u\in U\}\). This set is open, convex, nonempty, and does not contain \(0\). Choose \(a\in C\), and set \(V=C-a\). Then \(V\) is open and convex and contains \(0\). Moreover, \(-a\notin V\): if \(-a=c-a\) for some \(c\in C\), then \(c=0\), contradicting \(0\notin C\).
Let \(p=p_V\) be the Minkowski functional of \(V\). By the gauge lemma, \(p\) is finite and sublinear and \(V=\{z:p(z)<1\}\). Since \(-a\notin V\), it follows that \(p(-a)\geq1\). In particular, \(-a\ne0\). On the one-dimensional subspace \(M=\mathbb{R}(-a)\), define \[ f(t(-a))=t\,p(-a)\qquad(t\in\mathbb{R}). \] This is well-defined and linear because \(-a\ne0\). It is dominated by \(p\): for \(t\geq0\), positive homogeneity gives \(f(t(-a))=p(t(-a))\). For \(t<0\), put \(s=-t>0\). Subadditivity at zero gives \(0=p(0)\leq p(-a)+p(a)\), so \(-p(-a)\leq p(a)\). Consequently, \[ f(t(-a))=-s p(-a)\leq s p(a)=p(sa)=p(t(-a)). \]
Apply the Hahn-Banach Theorem, Real Dominated Extension Form, to extend \(f\) to a linear functional \(F:X\to\mathbb{R}\) satisfying \(F(z)\leq p(z)\) for every \(z\in X\). This functional is continuous. In fact, if \(B(0,r)\subseteq V\), the gauge bound gives \(p(z)\leq\|z\|/r\). Applying domination to \(z\) and \(-z\) yields \[ F(z)\leq\frac{\|z\|}{r} \quad\text{and}\quad -F(z)=F(-z)\leq\frac{\|z\|}{r}, \] so \(|F(z)|\leq\|z\|/r\). Also \(F(-a)=p(-a)\geq1\), so \(F\) is nonzero.
For any \(c\in C\), we have \(c-a\in V\), and hence \(p(c-a)<1\). Domination implies \(F(c-a)\leq p(c-a)<1\). Since \(F(a)=-F(-a)\leq-1\), linearity gives \[ F(c)=F(a)+F(c-a)<F(a)+1\leq0. \] Finally, each \(u\in U\) has \(c=u-x_0\in C\), so \(F(u)-F(x_0)=F(u-x_0)<0\). Taking \(L=F\) proves the stated separation, with \(L\) continuous and nonzero. \(\square\)
Separation of Two Convex Sets
Proof. Define \(U=A-B=\{a-b:a\in A,\ b\in B\}\). It is nonempty and convex. It is open because it is the union, over \(b\in B\), of the open sets \(A-b\). It does not contain \(0\), since \(a-b=0\) would imply \(a=b\), contrary to disjointness. Apply the separation theorem to \(U\) and \(x_0=0\). It gives a nonzero continuous linear functional \(L\) with \(L(a-b)<L(0)=0\). By linearity this is \(L(a)<L(b)\), as required. \(\square\)
The conclusion compares every point of \(A\) with every point of \(B\), but it does not claim that the difference \(L(b)-L(a)\) has a positive lower bound independent of \(a\) and \(b\). That stronger conclusion requires additional hypotheses in many settings; it is not part of the statement proved here.
Worked Examples
Worked Example: Separating a Point from an Open Disk
Let \(X=\mathbb{R}^2\) with its Euclidean norm, let \(U=\{(x,y):x^2+y^2<1\}\), and take \(x_0=(2,0)\). The set \(U\) is nonempty, open, and convex, and \(x_0\notin U\) because \(2^2+0^2=4\not<1\). Define \(L(x,y)=x\). This is linear and continuous; for example, \(|L(x,y)|=|x|\leq\sqrt{x^2+y^2}\). For every \((x,y)\in U\), \(x^2<1\), so \(-1<x<1\). Thus \(L(x,y)<1<2=L(x_0)\). The functional separates the whole disk from the chosen point.
Worked Example: Separating Two Convex Subsets of the Line
Take \(A=(-3,-2)\) and \(B=[0,2]\) in \(\mathbb{R}\). Both are nonempty and convex, they are disjoint, and \(A\) is open. Let \(L(x)=x\), which is nonzero and continuous. For every \(a\in A\), \(a<-2\), while every \(b\in B\) satisfies \(b\geq0\). Therefore \(L(a)=a<-2<0\leq b=L(b)\). This verifies the strict comparison for every pair \(a\in A\), \(b\in B\), not just for selected endpoints.
Worked Example: Strict Separation Without a Uniform Gap
Let \(U=(-\infty,0)\subset\mathbb{R}\), \(x_0=0\), and \(L(x)=x\). The set \(U\) is nonempty, open, and convex, and \(x_0\notin U\). The functional is nonzero, linear, and continuous, and \(L(u)=u<0=L(x_0)\) for every \(u\in U\). Yet there is no \(\delta>0\) such that \(L(x_0)-L(u)\geq\delta\) for all \(u\in U\). Indeed, choosing \(u=-1/n\) gives \[ L(x_0)-L(-1/n)=0-(-1/n)=1/n, \] which tends to \(0\) as \(n\to\infty\). The inequality is strict at every point, but the positive differences can be arbitrarily small.
How to Read the Separation Inequality
A common pitfall is to treat strict separation as if it automatically supplied a uniform margin. The theorem says \(L(u)<L(x_0)\) individually for every \(u\in U\); it does not say that there is a fixed \(\delta>0\) with \(L(u)\leq L(x_0)-\delta\) throughout \(U\). The half-line example shows that this stronger interpretation would be false even in one dimension.
It is also important to keep track of the hypotheses. Openness is what permits the gauge to satisfy \(V=\{p_V<1\}\), which drives the strict inequality in the proof. Convexity gives sublinearity of the gauge. The normed-space structure makes the extended functional continuous, through the bound by \(\|z\|/r\). In the geometric statement, nonemptiness ensures that a point \(a\) can be selected in the translated set. These assumptions play distinct roles and should not be silently omitted.
Check Your Understanding
Use the separation statement, its proof, and the examples to answer the following questions.
- What hypotheses on \(U\) and \(x_0\) are required for the point-separation form of Hahn-Banach?
- Why does \(0\notin C\), with \(V=C-a\), imply that \(-a\notin V\)?
- How does the gauge bound imply that the extended functional in the proof is continuous?
- How does the point-separation theorem give a functional separating an open convex set \(A\) from a disjoint convex set \(B\)?
- In the half-line example, why does strict inequality at every point fail to give a uniform positive gap?