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.

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    $\begingroup$ Since there is no norm given, what do you mean by continuous? $\endgroup$ – SmileyCraft Jan 10 at 21:33
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    $\begingroup$ 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. $\endgroup$ – Lord Shark the Unknown Jan 10 at 21:34
  • $\begingroup$ The term "functional" does not necessarily imply continuity, though sometimes this is implicit in context. $\endgroup$ – Eric Wofsey Jan 10 at 21:35
  • $\begingroup$ 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. $\endgroup$ – zxmkn Jan 10 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.

  • $\begingroup$ Thanks, but I don't quite see how you conclude $L = q$. Could you elaborate a bit on how that follows? $\endgroup$ – zxmkn Jan 11 at 9:24
  • $\begingroup$ 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 $L$.] $\endgroup$ – zxmkn Jan 11 at 9:33

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