## Separation theorem: Euclidean space vs. infinite dimensional space

Separating hyperplane theorem (separation theorem in short) is an elementary result in functional analysis and convex analysis and is important to optimization. Here we only discuss the “weak version”. In Euclidean space $\mathbb{R}^n$ we have the following result.

THEOREM 1. Let $A$ and $B$ be two nonempty convex disjoint subsets in $\mathbb{R}^n$. Then there exist $x\in\mathbb{R}^n$ and $\gamma\in\mathbb{R}$ such that $\langle x,a \rangle\leq\gamma\leq\langle x,b \rangle$ for any $a\in A$, $b\in B$. Note $\langle\cdot , \cdot\rangle$ is the standard inner product on $\mathbb{R}^n$.

The proof of THEOREM 1 depends on the fact that any nonempty convex set in $\mathbb{R}^n$ has nonempty relative interior. This is not the case if the underlying space is infinite dimensional.

It seems rare to see this general result in standard texts of functional analysis e.g.Rudin, what we can see is a weaker one:

THEOREM 2. Given a real topological vector space $X$. Let $A$ and $B$ be two nonempty convex disjoint subsets in $X$. If $A$ is open, then there exists $\Lambda\in X^{*}\setminus \{0\}$ and $\gamma\in\mathbb{R}$ such that $\Lambda x<\gamma\leq \Lambda y$ for any $x\in A$ and $y\in B$.

The condition that one of $A$, $B$ has nonempty interior cannot be dropped by considering the following example:

EXAMPLE 3 (The example and verification are not originated by me).

Consider $X=l^1$. Let $B=\mathbb{R}\cdot (1,0,0,\cdots)$. Take $A$ to be $\displaystyle\bigcap_{n=0}^{\infty}{E_n}$, where $E_n=\{x\in l^1:x_0\geq|n^3 x_n - n|\}$.

It can be verified directly that $A$ is a closed convex set. Also $A$ and $B$ are disjoint since if $x=(x_0,0,0,\cdots)\in A$, then $x_0\geq |n|$ for all $n$ which contradicts $x\in l^1$.

Now suppose $A$ and $B$ can be separated by some nontrivial continuous linear functional. Then $A-B$ and $\{0\}$ can also be separated by the same functional. If we can show $A-B$ is a dense subset of $l^1$, then by limit arguments one can show the functional must be trivial, which is a contradiction.

Therefore, it suffices to show $A-B$ is a dense subset of $l^1$. Indeed, let $x=(x_0,x_1,\cdots)$ and $\varepsilon>0$ be given. Choose a large positive integer $N$ such that $\displaystyle\sum_{n=N+1}^{\infty}\frac{1}{n^2}+\sum_{n=N+1}^{\infty}|x_n|<\varepsilon$. Then take a large number $a$ such that $\displaystyle \frac{a}{N^3}\geq\max\{|x_1-1|,|x_2-\frac{1}{2^2}|,\cdots,|x_N-\frac{1}{N^2}|\}$. Thus take $\displaystyle y=(a,x_1,\cdots,x_N,\frac{1}{(N+1)^2},\frac{1}{(N+2)^2},\cdots)-(a-x_0,0,0,\cdots)$. It follows from the choice of $a$ above that $\displaystyle (a,x_1,\cdots,x_N,\frac{1}{(N+1)^2},\frac{1}{(N+2)^2},\cdots)\in A$. Hence $y\in A-B$. The choice of $N$ above guarantees $\|x-y\|<\varepsilon$. We are done.

Clearly $B$ has empty interior. In fact $A$ also has empty interior. Indeed, pick $x\in A$ and consider $B(x,\varepsilon)$. Choose $N$ large such that $\displaystyle \frac{1}{N^2}+\frac{x_0}{N^3}<\frac{\varepsilon}{4}$ and $\displaystyle |x_N|<\frac{\varepsilon}{4}$. Then one can check $\displaystyle z=(x_0,x_1,\cdots,x_N+\frac{\varepsilon}{2},x_{N+1},\cdots)\in B(x,\varepsilon)\setminus A$ since under the choice of $N$ we have $z\notin E_N$, where $E_N$ is defined above.

Remark: We can replace $l^1$ by $l^p$, $1\leq p<\infty$ to construct a similar example.

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### 2 Responses to Separation theorem: Euclidean space vs. infinite dimensional space

1. KKK says:

Nice. Two minor questions.
Is your $y$ in your counterexample equal to $(a, x_1, \cdots, x_N, \frac{1}{(N+1)^2}, \frac{1}{(N+2)^2}, \cdots)$?
Is it $z\notin E_N$ instead of $x$ in your second last paragraph?

• Hon Leung says:

Thanks.

They are typos.