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Let $X$ be a topological vector space with topology $T$.

When is the weak topology on $X$ the same as $T$? Of course we always have $T_{weak} \subset T$ by definition but when is $T \subset T_{weak}$?

Assume that $X$ is any topological space, not necessarily normed.

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At least for normed space: if and only if the space is finite dimensional (consider the weak closure of $\{x,\lVert x\rVert=1\}$). –  Davide Giraudo Apr 29 '12 at 19:33
    
@DavideGiraudo Thanks Davide, I know : ) (Hence the "not necessarily normed") –  Matt N. Apr 29 '12 at 19:33
    
Did you make a conjecture for the general case? –  Davide Giraudo Apr 29 '12 at 19:39
    
@DavideGiraudo No. ${}{}{}{}{}$ –  Matt N. Apr 29 '12 at 19:41
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Quick Google search leat me to this: Sidney A. Morris: A topological group characterization of those locally convex spaces having their weak topology; Mathematische Annalen, Volume 195, Number 2 (1971), 330-331, DOI: 10.1007/BF01423619. –  Martin Sleziak Apr 30 '12 at 11:52
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3 Answers

up vote 11 down vote accepted

Quick Google search led me to the paper Sidney A. Morris: A topological group characterization of those locally convex spaces having their weak topology; Mathematische Annalen, Volume 195, Number 2 (1971), 330–331, DOI: 10.1007/BF01423619. The following result is given in this paper:

Theorem. Let $E$ be a locally convex Hausdorff real topological vector space. Then $E$ has its weak topology if and only if every discrete subgroup (of the additive group structure) of $E$ is finitely generated.

The proof relies on a result from the paper Sidney A. Morris: Locally Compact Abelian Groups and the Variety of Topological Groups Generated by the Reals; Proceedings of the American Mathematical Society, Vol. 34, No. 1 (Jul., 1972), pp. 290–292; DOI: 10.1090/S0002-9939-1972-0294560-4.


Added: (t.b.) Here's a sketch of the argument:

  1. Suppose that $E$ has the weak topology, and that $\Gamma \subset E$ is a discrete subgroup of the additive group. Then there exists a neighborhood $U$ of $0\in E$ such that $U \cap \Gamma = \{0\}$. In other words, there are continuous linear functionals $\varphi_1,\dots, \varphi_n$ and $\varepsilon \gt 0$ such that $\{0\} = \Gamma \cap \{x \in E\,:\,\max_{i} |\varphi_i (x)| \lt \varepsilon\}$. But this means that $\gamma \mapsto (\varphi_1(\gamma),\dots,\varphi_n (\gamma))$ is an injective map from $\Gamma$ to a discrete subgroup of $\mathbb{R}^n$, and discrete subgroups of $\mathbb{R}^n$ are finitely generated.

  2. Suppose $E$ does not have the weak topology. Then the topology on $E$ is strictly finer than the weak topology, hence there is a symmetric convex open neighborhood $U$ of $0 \in E$ which is not a neighborhood of $0$ in the weak topology, so $U$ cannot contain any linear subspace of finite co-dimension. The Minkowski functional (gauge) of $U$ thus defines a continuous norm $\|\cdot\|$ on some infinite-dimensional subspace $F$ of $E$. A classical argument of Mazur (see e.g. Lindenstrauss–Tzafriri Theorem 1.a.5, and Albiac–Kalton, Theorems 1.4.4, 1.3.9 and 1.3.10) then provides us with a basic sequence in $F$ which we may assume (after changing the norm to an equivalent one if necessary) to be a monotone basic sequence $(x_n)_{n=1}^\infty$. The monotonicity requirement implies that the subgroup generated by $(x_n)_{n=1}^\infty$ is discrete and linear independence implies that it is infinitely generated.

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I've posted this as a CW, in case anyone feels like going through details and elaborate this answer more; feel free to do so. –  Martin Sleziak Apr 30 '12 at 14:44
    
Very nice, thank you! –  Matt N. Apr 30 '12 at 14:46
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I added a sketch of the argument. I think the proof I give of part 1 is somewhat simpler than the one in the paper. –  t.b. Apr 30 '12 at 23:37
    
@t.b. Nice, thank you! –  Matt N. May 1 '12 at 6:47
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Posted my question at MO so that any possible answer does not get lost in the comments: mathoverflow.net/q/156538/13356 –  AlexE Feb 3 at 9:09
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I'd like to add that if we assume $X$ to be a normed vector space (over $\mathbb R$) then we have $T_{norm} = T_{weak}$ if and only if $X$ is finite dimensional. To see why this is the case:

$\Longrightarrow$ Assume $X$ is infinite dimensional. To show that then $T_{norm} \neq T_{weak}$ it's enough to find a set that is closed in one of the two topologies but not in the other. Note that $S := \{x \in X \mid \|x\| = 1 \}$ is closed in $T_{norm}$. But it's not closed in $T_{weak}$ since $0$ is in the weak closure of $S$: Let $U$ be any neighbourhood of $0$ in $T_{weak}$. Then there exists an open set $O$ such that $0 \in O \subset U$. We know that $\bigcup_{\varepsilon>0, r_0 \in \mathbb R} \{ \bigcap_{i=1}^n f_i^{-1} (B(r_0, \varepsilon)) \mid n \in \mathbb N \}$ forms a neighbourhood basis of $T_{weak}$. Hence there exist $f_1, \dots f_n \in X^\ast, \varepsilon > 0, r \in \mathbb R$ such that $0 \in O = \bigcap_{i=1}^n f_i^{-1} (B(r, \varepsilon))$.

Now we define a map $\varphi : X \to \mathbb R^n$, $x \mapsto (f_1(x), \dots, f_n(x))$. This map is linear. Hence $\operatorname{dim}{X} = \operatorname{dim}{\ker \varphi} + \operatorname{dim}{\operatorname{im}{\varphi}}$. We know that its image has dimension at most $n$ so since $X$ has infinite dimension we know that its kernel has to have infinite dimension. In particular, we can find an $x$ in $X$ such that $x \neq 0$ and $f_i(x) = 0$ for all $i \in \{1, \dots n\}$, i.e., $\varphi (x) = 0$. Since $\varphi$ is linear we also have $\varphi (\lambda x) = 0$ for all $\lambda \in \mathbb R$, in particular, for $\lambda = \frac{1}{\|x\|}$. Hence we have found a point $\frac{x}{\|x\|}$ that is in $S$ and also in the neighbourhood $U$ of $0$. Since the neighbourhood $U$ was arbitrary we get that $0$ is in the weak closure of $S$.

$\Longleftarrow$ Let $X$ be finite dimensional. Since by definition we always have $T_{weak} \subset T_{norm}$ it's enough to show that every open ball $B_{\|\cdot\|}(x_0, \varepsilon)$ is weakly open. Since $X$ is finite dimensional we can write every $x$ in $X$ as $x = \sum_{i=1}^n x_i e_i$ where $e_i$ is the basis of $X$. Define $f_i : X \to \mathbb R$ as $f_i : x \mapsto x_i$. Then $f_i$ are in $X^\ast$. Also since $X$ is finite dimensional, all norms on $X$ are equivalent and hence it's enough to show that $B_{\|\cdot\|_\infty}(x_0, \varepsilon)$ is weakly open. But that is clear since

$$ B_{\|\cdot\|_\infty}(x_0 , \varepsilon ) = \{x \in X \mid \max_i |x_i-x_{0i}|< \varepsilon \} = \{x \in X \mid \max_i |f_i(x-x_0)| < \varepsilon \} = \bigcap_{i=1}^n f_i^{-1} (B(x_{0i}, \varepsilon))$$

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On the implication $\Rightarrow$: what you prove in particular is that every weak neighborhood of $0$ of an infinite-dimensional normed space contains a non-trivial linear subspace. Since the open unit ball is norm-open but contains no subspace, it can't be weakly open. On the implication $\Leftarrow$: One can show that the usual topology on $\mathbb{R}^n$ is the only Hausdorff topology that makes $\mathbb{R}^n$ into a topological vector space. –  t.b. May 10 '12 at 17:21
    
@t.b. Thank you very much for this enlightening comment. –  Matt N. May 10 '12 at 18:46
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For a finite dimensional vector space there is only one topology that turns it into a topological vector space, so that weak topology and initial topology are necessarily the same.

For an infinite dimensional locally convex topological vector space T with dual T' this is never the case (edit: actually I'm not sure about this, see below), if the initial topology is the Mackey topology.

The Mackey-Arens theorem says that all topologies consistent with a given duality T, T' are comparable:

  • The weak topology is the weakest, it is the topology of pointwise convergence.

  • The Mackey topology is defined to be the strongest, it is the topology of uniform convergence on every absolutely convex, weakly compact set.

Since T is infinite dimensional, we can pick a sequence $(x_n), n \in \mathbb{N}$, of pairwise different vectors that are part of an algebraic basis of T. This sequence converges to 0 in the weak topology (edit: coming to think of it, I don't know it this is true or how to prove or disprove it. But I will not delete the answer, maybe someone else can improve it :-), but it does not in the Mackey topology. We can check this using the polar $A^°$ of the set $\{x_n, n \in \mathbb{N} \}$, (which is absolutely convex and weakly compact, so that $(x_n)$ would need to converge to 0 on this set, which it obviously does not).

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Since the weak topology is always locally convex by definition, a non-locally convex topological vector space cannot have the weak topology. –  t.b. Apr 30 '12 at 13:06
    
Ugh, right. I deleted my disclaimer about non locally convex spaces. –  Tim van Beek May 1 '12 at 10:34
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