# Exercise with locally convex topological vector spaces

i was trying to solve this exercise from Royden:

Let X be a locally convex topological vector space, let $Y \subset X$ be a closed subspace and $x_0 \in X-Y$.Prove that there exists a continuous linear functional $\varphi: X \rightarrow \mathbb{R}$ such that $\varphi(y) = 0$ if $y \in Y$ and $\varphi(x_0) \ne 0$

I don't understand where to use the fact that $Y$ is closed. My attempt was to construct a positive homogeneous and subadditive functional $p_y(x) = 1 - \chi_{Y}$. It's not hard to see that it has the desired properties. Moreover it is clear that $p_y(x_0) = 1$. Thus one can take $\varphi(\lambda x_0) = \lambda$ and extend it to a linear functional $\varphi \le p_y$ on $X$. Observing that $p_y(x) \le 1$ we have that $\varphi$ is bounded on $X$ and thus, because $X$ is locally convex topological vector spaces, continuous.

Isn't it right?

• What formulation of the hahn banach theorem are you using? I only ask because your function $p_y$ is not sublinear. – Aweygan Dec 28 '16 at 15:12
• Why not? Let $x,y \in X$ if $x+y \in Y$ it's obvious since $p_y(x+y) =0$ otherwise, since $Y$ is a subspace of X, we must have $x \not\in Y$ or $y \not\in X$. Thus subadditivity is true also in this case. – jJjjJ Dec 28 '16 at 18:27
• I forgot to tag @Aweygan – jJjjJ Dec 28 '16 at 20:11
• $p_y$ is just the indicator function for $X\setminus Y$. Indicator functions are not positive homogeneous. – Aweygan Dec 28 '16 at 20:16
• @Aweygan right. Sorry, was stupid – jJjjJ Dec 28 '16 at 20:19

Since both the sets $Y$ and $\{x_0 \}$ are closed and $\{x_0\}$ is compact then from separation theorem there exists a linear functional $f: X\to \mathbb{R}$ and real numbers $a< b$ such that $f(y)<a$ for $y\in Y$ and $f(x_0 ) =b .$ Now, take any $u\in Y$ then $$f(ny )<a$$ for all $n\in \mathbb{N}$ hence $f(y)\leqslant 0$ and analogously $$f(-ny )<a$$ thus $f(y) \geqslant 0$ and finally $$f(y)=0.$$
• Yes that's ok, but I asked if my solution is correct or not, since I don't use the fact that $Y$ is closed – jJjjJ Dec 27 '16 at 17:14