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.

  • 2
    $\begingroup$ Since there is no norm given, what do you mean by continuous? $\endgroup$ – SmileyCraft Jan 10 at 21:33
  • 2
    $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.