# Extension of continuous linear functionals from a closed subspace to the whole locally-convex space

Consider a Hausdorff, real (but I believe this not to be relevant), locally-convex topological linear space $X$ and a closed linear subspace $Y$. Let $f \in Y^*$ (the topological dual). I want to show that there exist $F \in X^*$ (not necessarily unique) such that $F \big| _Y = f$.

The only thing that this makes me think of is the Hahn-Banach theorem. However, I do not know how to use it, since that theorem makes use of an auxiliary seminorm, the role of which is to dominate the functional to be extended. In my problem, there is no such dominating seminorm, so how should I fabricate it? Or is there an alternative approach not involving the Hahn-Banach theorem?

• – Fei Li Aug 30 '16 at 9:36

As $f$ is continous (for the induced topology) there is an open neighborhood of the origin, $U$ so that $|f (y)|\leq 1$ for $y\in U\cap Y$. Pick a balanced convex neighborhood $A$ so that $0\in A\subset U$ and construct the Minkowski functional $$\mu_A(x)=\inf \{ t>0 : t^{-1} x \in A\}.$$ Then $\mu_A$ is a semi-norm on $X$ for which you may continue with Hahn-Banach in the usual way [see e.g. Rudin, Functional Analysis, Chap 1 for the Minkowski part]
• Could you please clarify one last point: why is it true that $| f(y) | \le \mu _A (y) \ \forall y \in Y$? – Alex M. Aug 30 '16 at 9:52
• The set of such $t$'s is non-empty ($A$ is absorbing). If $t^{-1} y\in A\cap Y$ then $t^{-1}y\in U \cap Y$ so $|f(t^{-1} y)|\leq 1$ or $|f(y)|\leq t$. Now take inf over such $t$'s. – H. H. Rugh Aug 30 '16 at 10:07
• I was ignoring the fact that $A$ can be chosen to be absorbing. – Alex M. Aug 30 '16 at 10:13
• No problem, in fact any open nghd of $0$ in a topological vector space is absorbing (due to continuity of multiplication with scalars). – H. H. Rugh Aug 30 '16 at 10:22