# Is a Hahn-Banach extension always continuous?

We proved the following version of the Hahn-Banach extension theorem in a course I'm taking:

Theorem (Hahn-Banach): Let $$X$$ be a real vector space and $$q : X \to \mathbb{R}$$ be sublinear. Let $$U \subseteq X$$ be a linear subspace and $$l : U \to \mathbb{R}$$ be a linear functional that satisﬁes $$l(u) \leq q(u)$$ for all $$u \in U$$. Then there exists a linear extension $$L : X \to \mathbb{R}$$ of $$l$$ that satisﬁes $$L(x) \leq q(x)$$ for all $$x \in X$$.

Is it true that this Hahn-Banach extension $$L$$ is also continuous, hence a $$\mathbb{R}$$-linear functional?

The way we use the theorem later in this course seems to rely on getting a functional out of the extension. Additionally, the Wikipedia entry states that Hahn-Banach yields a linear functional.

Some of my thoughts/work so far:

I see that, if $$q$$ is continuous, then $$L$$ is also continuous: Let $$(x_n) \subseteq X$$ with $$x_n \to 0$$. Then $$L(x_n) \leq q(x_n) \to 0$$ $$-L(x_n) = L(-x_n) \leq q(-x_n) \to 0$$ So $$|L(x_n)| \leq \max\{q(x_n), q(-x_n)\} \to 0$$. Hence, $$L$$ is continuous at $$0$$, and therefore continuous.

But a sublinear function isn't necessarily continuous... so this doesn't help, right?

From what I showed above, we see that $$-q(-x) \leq L(x) \leq q(x) \qquad \forall x \in X.$$ Not sure if this inequality might help.

• Since there is no norm given, what do you mean by continuous? – SmileyCraft Jan 10 '19 at 21:33
• The version of H-B you quote does not assume any topology on the space $X$. Of course, a standard example is when $X$ is normed and $q$ is the norm function. Then the extension is continuous for the norm topology. – Angina Seng Jan 10 '19 at 21:34
• The term "functional" does not necessarily imply continuity, though sometimes this is implicit in context. – Eric Wofsey Jan 10 '19 at 21:35
• Oh, I think I was getting confused because we defined linear functionals in this course as elements of $\mathcal{L}( X, \mathbb{K})$, the bounded linear operators from normed space $X$ to $\mathbb{K} \in \{\mathbb{R}, \mathbb{C}\}$. Looking around online, though (mathworld.wolfram.com/LinearFunctional.html), I see that's a more restrictive definition that normal. – zxmkn Jan 10 '19 at 21:50

The formulation of Hahn-Banach that you have does not require any topology on $$X$$. One obvious application is the case where $$X$$ does have a norm and you use the theorem to extend functionals while preserving the norm.
But the form with the seminorm $$q$$ is the one that allows one to obtain the geometric form of Hahn-Banach (i.e., the separation theorems) which is often the most useful.
Here is a counterexample: Let $$X$$ be an infinite-dimensional normed vector space. Then, there is a discontinuous linear functional $$q$$. Evidently, $$q$$ is sublinear. Now, take $$U = \{0\}$$ and $$l(0) = 0$$. However, the only linear function $$L : X \to \mathbb R$$ satisfying $$q$$ is $$L = q$$ which is discontinuous.
• Thanks, but I don't quite see how you conclude $L = q$. Could you elaborate a bit on how that follows? – zxmkn Jan 11 '19 at 9:24
• Oh, never mind. Now I see why $L=q$. [Suppose not, since if $L(x) < q(x)$, then $L(-x) > q(-x)$, which violates the requirement that $q$ dominate $L$.] – zxmkn Jan 11 '19 at 9:33