Continuous functional s.t. $f(x_0)\ne 0$

1. Aug 31, 2014

DavideGenoa

I read that in any locally convex topological space $X$, not necessarily a Hausdorff space but with linear operations continuous, for any $x_0\ne 0$ we can define a continuous linear functional $f:X\to K$ such that $f(x_0)\ne 0$.

I cannot find a proof of that anywhere and cannot prove it myself. Please correct me if I am wrong, but I think that if $A$ is closed in $X$ and $x_0\in A$ there exist a continuous linear functional rigorously separating $x_0$ and $A$, but I am not sure whether we can use that...

I have also tried with Minkowski functional, but with no result...
$\infty$ thanks for any help!

Last edited by a moderator: Aug 31, 2014
2. Aug 31, 2014

micromass

This follows from the Hahn-Banach theorems. One of the versions of this theorems state that

A proof can be found in Kadison & Ringrose "Fundamentals of the Theory of Operator Algebras, Vol I", Theorem 1.2.14 page 22.

Now you only need to define a suitable operator $\rho_0$ on $\textrm{span}\{x_0\}$ and apply the previous theorem.

3. Aug 31, 2014

DavideGenoa

...and we can choose $\rho_0:\alpha x_0\mapsto \alpha$.
However, I think it may be necessary that $V$ be Hausdorff, since theorem 1.2.6, on which 1.2.14 is based, needs the hypothesis. $\infty$ thanks!

4. Aug 31, 2014

micromass

Yes, you need Hausdorff. Indeed, assume we have for each $x_0\neq 0$ a continuous functional $f:V\rightarrow K$ such that $f(x_0)\neq 0$.

Then take $x,y\in V$ distinct points. We can find a continuous linear functional $f:V\rightarrow K$ such that $f(x-y)\neq 0$. Thus we have $f(x)\neq f(y)$, and thus we can find open sets $U$ and $V$ in $K$ such that $f(x)\in U$ and $f(x)\in V$ and $U\cap V = \emptyset$. Then $x\in f^{-1}(U)$ and $y\in f^{-1}(V)$ and $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint open sets in $V$. As such $V$ is Hausdorff.

5. Aug 31, 2014

DavideGenoa

Functional analysis is so fascinating...
Thank you so much!!!