When do weak and original topology coincide?
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.
Solution 1:
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
$$ B_{\|\cdot\|_\infty}(x_0 , \varepsilon ) = \{x \in X \mid \max_i |x_i-x_{0i}|< \varepsilon \} = \{x \in X \mid \max_i |f_i(x-x_0)| < \varepsilon \} = \bigcap_{i=1}^n f_i^{-1} (B(x_{0i}, \varepsilon))$$
Solution 2:
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:
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.
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.
Solution 3:
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).