If $V$ is a locally convex topological space, the Hahn-Banach theorem shows that a continuous linear functional on a closed subspace can be extended to a continuous linear functional on all of $V$, that is, an element of the dual space $V^*$.

If it is known that $V=U^*$, it has a locally convex pre-dual $U \subseteq V^*$, is there a simple additional hypothesis that allows the extending linear functional to be chosen from $U$? One possible motivation is to try to apply Hahn-Banach on the dual of a non-reflexive space like that of continuously $k$-differentiable functions on the interval.

I'm not at the moment well versed in the distinction between weak, strong, or other kinds of duals, so I'll settle for one that gives the simplest answer.

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    $\begingroup$ I'm not sure this is the kind of answer you want: it is necessary and sufficient for the functional to be continuous with respect to the weak*-topology on $V$ induced by $U$. (If it is continuous with respect to the topology induced by the weak*-topology on the subspace, it extends uniquely to the weak*-closure, then Hahn-Banach for the weak*-topology applies). $\endgroup$
    – Martin
    Commented May 25, 2013 at 22:10
  • $\begingroup$ It might be. Is this criterion effectively checkable? For example, what is the weak*-topology on Radon measures on $\mathbb{R}$ (which is dual to $C(\mathbb{R})$)? Take compact support where necessary. $\endgroup$ Commented May 25, 2013 at 23:43
  • $\begingroup$ As mentioned above, you can allways take the weak-star topology on the dual and use the Hahn-Banach theroem in this context. However, you seem to want a more intrinsic topology in concrete situations. Normally, one would like a complete LC structure. This is certainly possible for most concrete examples, in particular, for the two mentioned here. In fact, the dual of any Banach space has such a topology---the so-called bounded weak-star topology (this takes care of your first example). $\endgroup$
    – jbc
    Commented May 26, 2013 at 5:06
  • $\begingroup$ In the second case you have the dual of a Frechet space and the topology of uniform convergence on compacta fits the bill---(consult the theorem of Banach-Dieudonne). I am assuming that the phrase "compact support" refers to the measures. $\endgroup$
    – jbc
    Commented May 26, 2013 at 5:11

1 Answer 1


Nobody answered here, so I hope, the following will not be perceived as just a self-advertisement.

I believe, the theory of stereotype spaces is the easiest way to memorize these things. If $V$ is a stereotype space, it is automatically dual to some locally convex space $U$, namely to the so-called stereotype dual space $V^\star$, so that $$ V=U^\star,\qquad U=V^\star $$ (here every star $\star$ means the space of linear continuous functionals with the topology of uniform convergence on totally bounded sets, and every equality = means an isomorphism of locally convex spaces). So every continuous functional on $V$ belongs to $U$, and what you need will be fulfilled: if you take a closed subspace $X$ in $V$ and endow $X$ with the topology induced from $V$ (induced in the sense of the theory of topological vector spaces), then each linear continuous functional on $X$ can be extended to a linear continuous functional on $V$, i.e. to an element of $U$.

The class $\sf Ste$ of stereotype spaces is very wide, in particularly, it contains all Banach spaces, all Fréchet spaces, all quasicomplete barreled spaces. And it is closed under natural operations like taking limits and colimits (with some nuances however, since limits and colimits in $\sf Ste$ are not quite the same as in the category of locally convex spaces). Because of this for a locally convex space $V$ the property of being stereotype is usually easily checkable, since usually spaces are constructed from comparably simple components like Banach spaces (with the help of operations like taking limits, co-limits, subspaces, or quotient spaces). Besides this there is a nice criterion of stereotypy.

  • $\begingroup$ Thanks for your answer. But for my purposes, the question was satisfactorily answered in the comments by Martin and jbc. $\endgroup$ Commented Jul 21, 2013 at 21:07

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