Weak Topologies and Continuous Maps

Proposition 1.4.1   Let $S$ be a topological space, $X$ a vector space, $F$ a separating family of linear functionals on $X$, and $\phi:S \rightarrow X$ a map. Then $\phi$ is continuous (where $X$ has the $\sigma(X,F)$-topology) iff $f \circ \phi : S \rightarrow \mathbb{F}$ is continuous for all $f \in F$

Proof.

$\Leftarrow$

Since $\phi$ is continuous and $f$ is continuous (where $X$ has $\sigma(X,F)$ topology), $f \circ \phi$ is continuous.

$\Rightarrow$

Let $V$ be a basic open set; i.e., $\exists x_0 \in X$, $f_1, \ldots, f_n \in F$ and $\epsilon _1, \ldots, \epsilon _n > 0$ such that

$$V := \{ x \in X : \vert f_i(x - x_0)\vert < \epsilon _i \}$$

Then,

$$\phi^{-1}(V) := \{ s \in S : \vert f_i(\phi(s) - x_0)\vert < \epsilon _i \}$$

$$= \{ s \in S : \vert f_i \phi(s) - f_i(x_0)\vert < \epsilon _i \} = \bigcap (f_i \circ \phi)^{-1}( B_{\epsilon _i}(f_i(x_0)) ).$$

This is open, as a finite intersection of inverse images of open sets.

$\qedsymbol$

Corollary 1.4.2   Let $X, Y$ be an LCS and $T:X \rightarrow Y$ a continuous linear map. Then $T:X \rightarrow Y$ is continuous when $X$ and $Y$ are given their weak topologies, i.e. $\sigma(X,X^*)$, $\sigma(Y,Y^*)$.

Proof. Let $f \in Y^*$. Since $T$ is continuous and $f$ is continuous, $f \circ T : X \rightarrow \mathbb{F}$ is continuous when $X$ has its original topology. So, $f \circ T$ is also continuous when $X$ has the $\sigma(X,X^*)$ topology. Thus, for all $f: Y \rightarrow \mathbb{F}$, continuous in the $\sigma(Y,Y^*)$ topology. $f \circ T$ is continuous, where $X$ has the $\sigma(X,X^*)$ topology. By the proposition, $T$ is (wk, wk)-continuous. $\qedsymbol$

Remark:
Let $X$ be a LCS over $\mathbb{C}$. Let $X_\mathbb{R}$ be the same LCS considered as an LCS over $\mathbb{R}$ and then $X_\mathbb{R}^*$ is the continuous $\mathbb{R}$-linear functionals into $\mathbb{R}$.
Fact:
$\sigma(X,X^*)$ and $\sigma(X_\mathbb{R}, X_\mathbb{R}^*)$ are the same topology (Exercise IV 1.4).

Theorem 1.4.3   Let $X$ be a LCS and $A \subset X$ be convex. Then the closure of $wk A$ and $\sigma(X,X^*)$-closure of $A$, $wk-cl A$ are equal.

Proof. Since the $\sigma(X,X^*)$ topology is contained in the original topology, and so the set $\mathcal{S}$ of wk-closed subset of $X$ containing $A$ and $\mathcal{T}$, the set of closed subsets of $X$ containing $A$ satisfy $\mathcal{S}\subset \mathcal{T}$. So,

$$cl A = \bigcap_{T \in \mathcal{T}} T \subset \bigcap_{T \in \mathcal{S}} T = wk-cl A$$

Now, suppose $x \notin cl A$. By the HB-Sep Thm, there is $f \in X^*$, $\alpha \in \mathbb{R}$, $\epsilon > 0$ such that $\forall a \in cl A$,

$$Re f(a) \leq \alpha < \alpha + \epsilon \leq Re f(x)$$

In particular,

$$A \subset \{ y \in X :$$    Re $$f(y) \leq \alpha \},$$

a wk-closed set and $x$ is not in this set. So, $x \notin wk-cl A$. Thus, $cl A \supset wk-cl A$. $\qedsymbol$

Monday, January 21, 2006:

Notation:
Consider $T:X \rightarrow Y$. Since $X$ and $Y$ can have several topologies, we say $T$ is $(\sigma,\tau)$-continuous if it is continuous when $X$ has the topology $\sigma$ and $Y$ has the topology $\tau$.

If $X$ is a Banach space, we use $\Vert\cdot\Vert$ for the topology induced by the norm. In particular, we use $\Vert\cdot\Vert$ for the usual topology on $\mathbb{F}$.

Corollary 1.4.4   For $X$, $Y$, LCS witih topologies $\sigma$ and $\tau$, if $T:X \rightarrow Y$ is $(\sigma,\tau)$ continuous, then it is $(wk, wk)$ continuous.

Proof. Let $f \in Y^*$. Then $f \circ T$ is $(\sigma, \vert\cdot\vert)$ continuous and so $f \circ T$ is $(wk, \vert\cdot\vert)$-continuous by the definition of wk. Thus, for every $f \in Y^*$, $f \circ T$ is $(wk, \vert\cdot\vert)$ continuous. By the proposition, $T$ is $(wk, wk)$-continuous. $\qedsymbol$

Theorem 1.4.5   If $X, Y$ are Banach space and $T:X \rightarrow Y$ is linear, then $T$ is $(\Vert\cdot\Vert,\Vert\cdot\Vert)$-continuous iff $T$ is $(wk, wk)$-continuous

Proof.

$\Leftarrow$

Done.

$\Rightarrow$

Let $f \in Y^*$. Then $f$ is $(wk, \vert\cdot\vert)$-cts and so $f \circ T$ is $(wk, \vert\cdot\vert)$-cts. By the definition of the weak topology, $f \circ T$ is $(\Vert\cdot\Vert,\Vert\cdot\Vert)$-cts. That is, for each $f \in Y^*$, $f(T(B_1(0))$ is bounded (in $\mathbb{F}$). By PUB (2nd corollary), $T(B_1(0))$ is bounded in $Y$; i.e., $\Vert T\Vert < \infty$.

$\qedsymbol$

In fact, we can formulate and prove versions of PUB for LCS. Recall $B \subset X$, an LCS, is bounded if for every $U$, open set containing 0, there is $\epsilon > 0$ so that $\epsilon B \subset U$.

Theorem 1.4.6   Weak Version of PUB
Let $X$ be a Banach space, $Y$ a normed space, and $\mathcal{A}\subset B(X,Y)$. If, for each $x \in X$, $\{ Ax : A \in \mathcal{A}\}$ is weakly bounded (in the weak topology), then $\mathcal{A}$ is norm bounded.

The key idea to this proof is to show $U \subset Y$ is weakly bounded iff $\forall f \in Y^*$, $f(U)$ is bounded. Now, apply the last corollary of PUB.