# Is Reflexivity Necessary for the Weak and Weak* Topologies to Coincide?

Let $X$ be a normed vector space, not necessarily Banach. Suppose that $X$ is not reflexive, implying the existence of such $\varphi\in X^{**}$ ($X^{**}$ being the double dual of $X$) of that for any $x\in X$ there exists some $f_x\in X^*$ ($X^*$ being the dual of $X$) such that $\varphi(f_x)\neq f_x(x)$. My question is: in this case, is the weak* topology on $X^*$ necessarily strictly weaker than the weak topology on $X^*$?

Another thread answers this question in the affirmative for Banach spaces. That is, if $X$ is Banach and the weak and weak* topologies on $X^*$ coincide, then $X$ must be reflexive; or, equivalently, if $X$ is Banach but not reflexive, then the weak* topology on $X^*$ must be strictly weaker than the weak topology on $X^*$. That proof relies on the fact that if $X$ is Banach and $X^*$ reflexive, then $X$ is reflexive. However, this latter result can be shown to fail if $X$ is not Banach.

My strategy was to take the $\varphi\in X^{**}$ specified above and try constructing such an open set based on it in the weak topology on $X^*$ that is not open in the weak* topology, but I'm stuck with this approach.

Any ideas or hints would appreciated.

UPDATE: I got it. The claim is true even without assuming that $X$ is Banach. That is, if the weak and weak* topologies on $X^*$ coincide, then $X$ is necessarily reflexive (and, consequently, also Banach, but the point of my question was not assuming completeness in the proof of reflexivity). The proof is conceptually neat but the rigorous details are surprisingly abstruse, so I won't replicate it here. The idea is based on lemma 4 and proposition 5 of these notes, in case anyone else is interested. (Disclaimer: I personally didn't really like the way the induction step was carried out in lemma 4, so I used the Hahn–Banach theorem instead to generate the desired linear combination).

Thank you guys for viewing this question and upvoting it. :-)

• Interesting fact, maybe relevant here: the weak* topology is the product topology (cf vg Brezis). Jan 26, 2014 at 9:44
• @Martín-BlasPérezPinilla Do you mean the (relative) topology of pointwise convergence on $\mathbb K^X$, restricted to bounded linear functions on $X$? Jan 26, 2014 at 10:04
• @triple_sec The main advantage of the induction step is its use for arbritrary locally convex spaces, in which Hahn-Banach is not applicable. Basically the notes prove the "standard theorem", that the dual of the weak dual is again the original space. Here your approach is successful as well: Assuming your open set produced by $\varphi$ is weak*-open too, would force $\varphi$ to be an evaluation functional.
– Vobo
Jan 26, 2014 at 10:41
• Right. Used in the proof of en.wikipedia.org/wiki/Banach%E2%80%93Alaoglu_theorem, for example. Jan 26, 2014 at 18:18

The only continuous linear functionals on $X^*$ equipped with the weak star topology are the evaluation functionals $$X^*\to\Bbbk,~\phi\mapsto \phi(x)$$ where $x\in X$ and $\Bbbk$ is the ground field ($\Bbb R$ or $\Bbb C$.) In fact, if $Y$ is a linear space, $\mathcal{L}$ is a collection of linear functionals on $Y$, and $Y_{\mathcal L}$ is the coarsest topology on $Y$ that makes all the linear functionals in $\mathcal L$ continuous, then $(Y_{\mathcal L})^*=\mathrm{Vect}(\mathcal{L})$.
So if $X$ is not reflexive, the Banach space $(X^*,|\cdot|)$ has more continuous linear functionals, and therefore has more open subsets.
• That's right, but what I didn't see a priori (even though it did feel intuitively true) is that the evaluation functionals are the only ones in $X^{**}$ that are continuous. They are continuous, but what if requiring that they be continuous necessarily makes all other functionals in $X^{**}$ continuous as well in some non-reflexive spaces? This turned out not to be the case, so that the maximum coarseness of the weak* topology on $X^*$ really has its bite in the sense that it doesn't make more functions continuous than it “should,” but this wasn't obvious to me at first. Jan 26, 2014 at 9:34
• @Martín-BlasPérezPinilla It's still an evaluation functional, as $\sum_{i=1}^n f(x_i)=f(\sum_{i=1}^n x_i)$ for all $f\in X^*$, since $f$ is linear. Also, $X$ is a vector space, so that $\sum_{i=1}^n x_i\in X$. Therefore, a finite sum of evaluation functionals is also an evaluation functional, corresponding to the sum of the underlying “evaluators” in $X$. Jan 26, 2014 at 10:07
• True (silly questin, indeed!), I was thinking in $X$=some function space and $ev:X\longrightarrow{\Bbb K}$. Jan 26, 2014 at 18:21