The Hahn-Banach Separation Theorem

The Hahn-Banach Separation Theorem

Theorem 1 (The Hahn-Banach Separation Theorem): Let $E$ be a locally convex topological vector space. If $A$ and $B$ are disjoint convex sets and $A$ is open, then there exists a continuous linear form $f$ on $E$ such that $f(A)$ and $f(B)$ are disjoint.
  • Proof: Observe that $A - B$ is an open set since $A$ is open, and $A - b$ is open for each $b \in B$ (as $E$ is a topological vector space), so:
(1)
\begin{align} \quad A - B = \bigcup_{b \in B} (A - b) \end{align}
  • shows that $A - B$ is a union of open sets and is thus indeed open.
  • Also observe that $A - B := A + (-1)B$ is convex. Indeed, let $x, y \in A - B$ and let $\lambda, \mu \in \mathbf{F}$ be such that $\lambda + \mu = 1$. Write $x := a_x - b_x$ and $y := a_y - b_y$ where $a_x, a_y \in A$ and $b_x, b_y \in B$. Then, by the convexity of $A$ and $B$ we have that $\lambda a_x + \mu a_y \in A$ and $\lambda b_x + \mu b_y \in B$, thus:
(2)
\begin{align} \quad \lambda x + \mu y &= \lambda (a_x - b_x) + \mu(a_y - b_y) \\ &= (\lambda a_x + \mu a_y) - (\lambda b_x + \mu b_y) \in A - B \end{align}
  • Lastly observe that $A - B$ does not contain the origin. Indeed, if $o$ were contained in $A - B$, then there would exist $a \in A$ and $b \in B$ such that $o = a - b$, so that $a = b \in A \cap B$, which contradicts the assumption that $A$ and $B$ are disjoint.
  • Since $E$ is a locally convex topological vector space, $A - B$ is an open and convex set in $E$, and $\{ o \}$ is a subspace of $E$ for which $\{ o \} \cap A - B = \emptyset$, we have by the theorem on the The Hahn-Banach Theorem for Vector Spaces Part 3 page that there exists a closed hyperplane $H$ of $E$ containing $o \}$ and such that $H \cap A - B = \emptyset$.
  • Let $f$ be a nonzero linear form such that $f^{-1}(o) = H$. Furthermore, since a linear form is closed if and only its null space is closed, and since $f^{-1}(o) = H$ is closed, we have that $f$ is continuous.
  • Lastly observe that if $k \in f(A) \cap f(B)$ then there exists $a \in A$ and $b \in B$ such that $f(a) = f(b) = k$. Then $f(a - b) = 0$, and so $a - b \in f^{-1}(0) = H$. But $a - b \in A - B$, and $A - B$ and $H$ are disjoint, which is a contradiction.
  • Thus $f(A) \cap f(B) = \emptyset$. $\blacksquare$
Corollary 2: Let $E$ be a locally convex topological vector space. If $B$ is a convex set and $a \not \in \overline{B}$ then there exists a continuous linear form $f$ on $E$ such that $f(a) \not \in \overline{f(B)}$.
  • Proof: Let $B$ be a convex set and let $a \not \in \overline{B}$.
  • Since $E$ is a locally convex topological vector space, every base of neighbourhoods of the origin must consist of sets that are absolutely convex and absorbent.
  • Since $a \not \in \overline{B}$, there exists an open neighbourhood of the origin $U$, such that $(a + U) \cap B = \emptyset$, for which is absolutely convex (and absorbent) from above.
  • Then by The Hahn-Banach Separation Theorem, since $a + U$ and $B$ are disjoint convex sets, there exists a continuous linear form $f$ on $E$ such that $f(a + U) \cap f(B) = \emptyset$. Furthermore, $f$ is a nonzero linear form.
  • Since $a + U$ is open and since $f$ is a nonzero linear form on $E$, by a proposition on the Linear Forms on a Vector Space and its Algebraic Dual page, we have that $f$ is an open map and that $f(a + U)$ is open. Furthermore $f(a + U) = f(a) + f(U)$ is an open neighbourhood of $f(a)$ such that $f(a + U) \cap f(B) = \emptyset$, and thus $f(a) \not \in \overline{f(B)}$. $\blacksquare$
Corollary 3: Let $E$ be a locally convex topological vector space. If $B$ is an absolutely convex set and $a \not \in \overline{B}$ then there exists a continuous linear form $f$ on $E$ such that $|f(x)| \leq 1$ for all $x \in B$, and $f(a) > 1$.
  • Proof: Since $B$ is absolutely convex, it is convex. So by Corollary 2, there exists a continuous linear form $g$ on $E$ such that $g(a) \not \in \overline{g(B)}$.
  • Since $B$ is absolutely convex, we have that $g(B)$ is absolutely convex. Indeed, let $s, t \in g(B)$ and let $\lambda, \mu \in \mathbf{F}$ be such that $|\lambda| + |\mu| \leq 1$. Then take $x, y \in B$ such that $g(x) = s$ and $g(y) = t$. Then since $B$ is absolutely convex we have that $\lambda x + \mu y \in B$. So $g(\lambda x + \mu y) \in g(B)$. But by the linearity of $g$, we have that:
(3)
\begin{align} \quad \lambda s + \mu t = \lambda g(x) + \mu g(y) = g(\lambda x + \mu y) \in g(B) \end{align}
  • So for each $b \in B$, we have that $g(b) \in g(B)$, and so, for each $\lambda \in \mathbf{F}$ with $|\lambda| \leq 1$, we have that
(4)
\begin{align} \quad \lambda g(b) \in g(B) \end{align}
  • And since $g(a) \not \in \overline{g(B)}$, we have that $|g(b)| < |g(a)|$ for all $b \in B$. Thus:
(5)
\begin{align} \quad \sup_{b \in B} |g(b)| < |g(a)| \end{align}
  • Let $\displaystyle{\alpha := \sup_{b \in B} |g(b)|}$.
  • If $\alpha \neq 0$, then let $f$ be the linear form defined for all $x \in E$ by:
(6)
\begin{align} \quad f(x) := \left ( \frac{|g(a)|}{\alpha g(a)} \right ) g(x) \end{align}
  • Then if $x \in B$, we have that:
(7)
\begin{align} \quad |f(x)| = \left ( \frac{|g(a)|}{\alpha |g(a)|} \right ) |g(x)| = \frac{1}{\alpha} |g(x)| \leq 1 \end{align}
  • and:
(8)
\begin{align} \quad f(a) = \left ( \frac{|g(a)|}{\alpha g(a)} \right ) g(a) = \frac{|g(a)|}{\alpha} > 1 \end{align}
  • If $\alpha = 0$, then let $f$ be the linear form defined for all $x \in E$ by:
(9)
\begin{align} \quad f(x) := \left ( \frac{2}{g(a)} \right ) g(x) \end{align}
  • Then if $x \in B$, we have that:
(10)
\begin{align} \quad |f(x)| = \left ( \frac{2}{|g(a)|} \right ) |g(x)| = 0 \leq 1 \end{align}
  • and:
(11)
\begin{align} \quad f(a) = \left ( \frac{2}{g(a)} \right ) g(a) = 2 > 1 \quad \blacksquare \end{align}
Corollary 4: Let $E$ be a real locally convex topological vector space. If $A$ and $B$ are disjoint conve sets and $A$ is open then there exists a continuous linear form $f$ on $E$, and a constant $\alpha$, such that $f(x) > \alpha$ for all $x \in A$, and $f(x) \leq \alpha$ for all $x \in B$.
  • Proof: By the Hahn-Banach separation theorem, there exists a continuous real linear form $f$ on $E$ such that $f(A) \cap f(B) = \emptyset$. Since $A$ and $B$ are convex, so are $f(A)$ and $f(B)$. But $f(A)$ and $f(B)$ are convex sets of real numbers and are thus intervals.
  • Suppose that:
(12)
\begin{align} \quad \sup \{ f(x) : x \in B \} \leq \inf \{ f(x) : x \in A \} \end{align}
  • By setting $\alpha := \inf \{ f(x) : x \in A \}$ we have that $f(x) \leq \alpha$ for all $x \in B$. Furthermore, since $A$ is open and since $f$ is a nonzero linear form, we have that $f(A)$ is open (by a proposition on the Linear Forms on a Vector Space and its Algebraic Dual page). So $f(A)$ is an open interval of real numbers and thus $\alpha \not \in f(A)$. Thus $f(x) > \alpha$ for all $x \in A$.
  • If instead we have that:
(13)
\begin{align} \quad \sup \{ f(x) : x \in B \} \geq \inf \{ f(x) : x \in A \} \end{align}
  • Then instead take $f' := -f$. Then $f'$ is still a continuous linear form for which $f'(A) \cap f'(B) = \emptyset$, and such that:
(14)
\begin{align} \quad \sup \{ f'(x) : x \in B \} \leq \inf \{ f'(x) : x \in A \} \end{align}
  • Then repeat the same argument as before. $\blacksquare$
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