Dual Spaces for LCS

As with Banach spaces, the dual space of a LCS $X$ is $B(X,\mathbb{F})$, i.e., the vector space of continuous linear functionals. But there is no norm. It is denoted $X^*$.

Many results from the theory of duals of Banach spaces carry over. For example, hyperplanes are either closed or dense, a functional is continuous iff it is continuous at one point iff the kernel is closed.

Proposition 1.2.1   A functional $f$ on a LCS $X$ is continuous iff there are $p_1, \ldots, p_n \in \mathcal{P}$ and $\alpha_1, \ldots, \alpha_n \geq 0$ such that $\forall x \in X$,

$$\vert f(x)\vert \leq \sum_{i=1}^n \alpha_i p_i(x)$$

where $\mathcal{P}$ is the family of seminorms for $X$.

Proof. Let

$$U := \{ x \in X : \vert f(x)\vert < 1 \}$$

Note that $U$ is open. Since $0 \in U$, there is a basic open set $B$ with $B \subset U$. As $B$ is a basic open set, there are $p_1, \ldots, p_n \in \mathcal{P}$ and $\epsilon _1, \ldots, \epsilon _n$ such that

$$B = \{ x \in X : p(x) < \epsilon _i, i = 1, \ldots, n \}$$

Choose $\alpha_1, \ldots, \alpha_n \geq 0$ such that

$$\frac{1}{n \alpha_i} > \epsilon _i$$

Thus, for $x \in B$,

$$\alpha_i p_i (x) < \frac1n.$$

Let

$$P(Y) = \sum_{i=1}^n \alpha_i p_i(y)$$

If $P(y) = 0$, then $P(\delta y) = 0$ for all $\delta \geq 0$. I.e., $\delta y \in B \subset U$ for all $\delta \geq 0$. Thus,

$$\delta f(y) = f(\delta y) \leq 1, \forall \delta \geq 0$$

Thus, $f(y) = 0$.

If $P(y) = \alpha \neq 0$, then $P(\frac1\alpha y ) = 1$. and so $P(\delta y) \leq 1$ for all $\delta$, $0 \leq \delta < \frac1\alpha$. So, $\delta y \in B \subset U$ so

$$\delta f(y) = f(\delta y) \leq 1, \forall \delta, 0 \leq \delta < \frac1\alpha$$

By continuity of scalar multiplication,

$$f(\frac1\alpha y) \leq 1$$

Therefore,

$$f(y) \leq \alpha = P(y)$$

$\qedsymbol$

Example: A new way to think of a dual space.
Recall the Riesz Representation Theorem for $\mathcal{C}(x)$. If $X$ is a locally compact Hausdorff space and $M(X)$ is the normed vector space of regular Borel measures on $X$ with total variation norm, then define $F:M(X) \rightarrow C(X)^*$ by

$$F_\mu(f) = \int f d\mu$$

$f \in \mathcal{C}(X)$, $\mu \in M(X)$.
Then, $F$ is isometric and onto.

Let $X$ be completely regular. (i.e., $\forall x \in A \subset X$, $A$ open, there is $f \in C(X)$ such that $f(x) = 1$, $f\vert _{A^c} \equiv 0$. We make $C(X)$ into a LCS by using the family of seminorms

$$p_K(f) = \sup_{x \in K} \vert f(x)\vert$$

for $K \subset X$, compact.

Theorem 1.2.2   Every element of $C(X)^*$ has the form

$$L(f) = \int_K f d\mu$$

where $K \subset X$ is compact and $\mu \in M(K)$.
Conversely, every such $K$ and $\mu$ give rise to an element of $C(X)^*$.

Proof. Given $K$ and $\mu$ as above, observe

$$L(f) \leq \int \vert f\vert d\vert\mu\vert \leq \sup_{x \in K} \vert f(x)\vert \Vert\mu\Vert$$

where $\Vert\mu\Vert$ is the total variation norm; $\vert\mu\vert(X)$.

$$\leq \Vert\mu\Vert p_K(f)$$

So, by the proposition above, $L \in C(X)^*$.
Conversely, suppose $L \in C(X)^*$ By the proposition, there are compact sets $K_1, \ldots, K_n$ and $\alpha_1, \ldots, \alpha_n > 0$ such that

$$\vert L(f)\vert \leq \sum_{i=1}^n \alpha_i p_{K_i}(f)$$

Let $K = \bigcup_i K_i$; $\alpha = \sum_{i=1}^n \alpha_i$. Then,

$$\sum_{i=1}^n \alpha p_{K_i}(f) \leq \alpha p_K(f)$$

In particular, if $f\vert _K \equiv 0$, $L(f) = 0$.

Define $F:C(K) \rightarrow \mathbb{F}$ by, for $g \in \mathcal{C}(K)$, letting $\tilde g$ be any continuous extension to $X$ and setting

$$F(g) := L(\tilde g )$$

Notice that $\hat g$ is a second extension, then

$$\tilde g - \hat g\vert _K \equiv 0$$

so $L(\tilde g) = L(\hat g)$. It's routine to show that $F$ is linear.
For $g \in \mathcal{C}(K)$,

$$\vert F(g)\vert = \vert L(\tilde g)\vert \leq \alpha p_K(\tilde g) = \alpha \Vert g\Vert _\infty$$

So $F$ is continuous when $\mathcal{C}(K)$ is given the $\Vert\cdot\Vert _\infty$ norm.
By the Riesz Representation Theorem, there exists a $\mu \in M(K)$ such that

$$F(g) = \int g d\mu$$

For $f \in \mathcal{C}(X)$, $f\vert _K \in \mathcal{C}(K)$ so

$$L(f) = F(f\vert _K) = \int_K f d\mu$$

$\qedsymbol$