Linear Functionals

Let $X$ be a vector space over $\mathbb{F}$. Call $M$, a linear subspace, a hyperplane if $\dim(X/M) = 1$.

Example:
If $f: X \rightarrow \mathbb{F}$, then $\ker F$ is a hyperplane (if $f \neq 0$). In fact, $f$ induces an isomorphism of $\mathbb{F}$ and $X/M$.

Properties

1. Every hyperplane arises in this way. If $\dim(X/M) = 1$, then there is an isomorphism
   $$T : X/M \rightarrow \mathbb{F}$$
   If $Q:X \rightarrow X/M$ is the quotient map, then $T \circ Q$ is a bounded linear functional with kernel equal to $M$.
2. Two linear functionals have the same kernel if and only if they are (nonzero) multiples of each other. In particular, every functional with kernel $M$ is a scalar multiple of $T \circ Q$.
   Proof. Suppose $f,g : X \rightarrow \mathbb{F}$, $\ker f = \ker g$. Pick $x_0 \in X$ such that $f(x_0) = 1$, So $g(x_0) \neq 0$. For $x \in X$, let $\alpha = f(x)$ and then $y = x - \alpha x_0 \in \ker F$. So $0 = g(y) = g(x) - f(x)g(x)$. Thus,

   $$g = (g(x_0)) f$$

   $\qedsymbol$

3. Every hyperplane is either dense or closed.
   Proof. If $M \neq \bar M$, then pick $y \in \bar M \setminus M$. Since $\dim(X/M) = 1$, $y+M \neq M$. So, given $x + M \in X/M$, there is $\alpha \in \mathbb{F}$ such that $x+M = \alpha(y+M)$. Thus, $x = \alpha y + M \in \bar M$. So, $\bar M = X$. $\qedsymbol$
   Example: of a not closed hyperplane. Let $X = c_0$, the space of sequences converging to 0 with the infinity norm. Let
   $$x = (1, \frac12, \frac13, \ldots)$$
   and let $e_1, e_2, \ldots$ be the usual basis. Since $\{ x_1, e_1, e_2, \ldots \}$ is a linearly independent set. There is a a $B \subset c_0$ so that $\{ x, e_1, e_2, \ldots \} \cup B$ is a linear basis for $c_0$. Thus, for $y \in c_0$, we can write $y$ as
   $$\alpha_0 x + \alpha_1 e_1 + \alpha_2 e_2 + \cdots + \sum_{b \in B} \alpha_b b$$
   with at most finitely many indices nonzero.
   Define a map $f : c_0 \rightarrow \mathbb{F}$ by $f(y) = \alpha_0$. Then this is a linear map, so $\ker f$ is a hyperplane. Notice $x \notin \ker f$ as $f(x) = 1$. But $e_i \in \ker f$ for $i = 1,2, \ldots$. Further, $x$ is in the closed span of the $e_i$, so $x \in \overline{ker f} \setminus \ker f$.

Theorem 10.0.1   For $x$ a normed space, $f: X \rightarrow \mathbb{F}$ is continuous if and only if $\ker f$ is closed.

Proof. For the forward direction, $\ker f = f^{-1}(\{0\})$ is closed if $f$ is continuous.
If $\ker f$ is closed, then $Q:X \rightarrow X/\ker f$ is continuous and let $T:X / \ker f \rightarrow \mathbb{F}$ be an isomorphism. Then $T \circ Q$ is continuous and, by property $2$, $f$ is a scalar multiple of $T \circ Q$. $\qedsymbol$

Definition 10.0.2   For $X$ a normed space, the dual space of $X$, denoted $X^*$ (or $X^\char93$) is the vector space of all bounded linear functionals, equipped with the norm $\Vert f\Vert = \sup_{\Vert x\Vert \leq 1} \vert f(x)\vert$. That is, $X^* = B(X,\mathbb{F})$.

Proposition 10.0.3   For $X$ a normed vector space, $X^*$ is a Banach space.

Examples of Dual Spaces:

1. Riesz Representation Theorem for $L^p$, $1 < p < \infty$. As usual, $L(X,\Omega, \mu)$ is a measure space. For every $F \in L^p(X)^*$, there is a unique $g \in L^q(X)$ (where $\frac1p + \frac1q = 1$) so that
   $$F(f) = \int_X fg d\mu, \forall f \in L^p(X)$$
   moreover $\Vert F\Vert _{(L^p)^*} = \Vert g\Vert _{L^q}$.
2. If $(X,\Omega,\mu)$ is a $\sigma$-finite measure space, then for every $F \in L^1(X)^*$, there is a unique $g \in L^\infty(X)$ so that
   $$F(f) = \int_X fg d\mu, \, \forall f \in L^1(X)$$
   Moreover, $\Vert F\Vert _{(L^1)^*} = \Vert g\Vert _{L^\infty}$
3. $(L^\infty)^*$ is not $L^1$.
4. Let $X$ be a locally compact space. For every $F \in (C_0(X))^*$, there is a regular Borel measure $\mu$ on $X$ so that
   $$F(f) = \int_X f d\mu, \, \forall f \in C_0(X)$$
   Define for such a $\mu$
   $$\Vert\mu\Vert = \sup_ \{ \sum_{i=1}^n \vert\mu(S_1)\vert : S_i \in \Omega, \bigcup S_i = X, \mu(S_i \cup S_j) = 0, i \neq j \}$$
   Then $\Vert F\Vert _{C_0(X)^*} = \Vert\mu\Vert$.
   

Important Special Cases:

1. $(c_0)^* = \ell^1$
2. $(\ell^1)^* = \ell^\infty$
3. $(\ell^p)^* = \ell^q$, $\frac1p + \frac1q = 1$.

Q: Find a functional on $\ell^\infty$ that does not come from an element of $\ell^1$.
If $x = (x_1, x_2, \ldots) \in \ell^1$, then $F_x \in (\ell^\infty)^*$ by

$$F_x(y_1, y_2, \ldots) = \sum_{i=1}^\infty x_i y_i$$

Define $G(y_1,y_2, \ldots) = \limsup_i y_i$. Then, $G \in (\ell^\infty)^\char93$, but

$$G((0,0,\ldots, 0,1,0, \ldots) = 0$$

for all $i \in \mathbb{N}$. But if $F_x( 0, 0, \ldots, 0, 1, 0, \ldots ) = 0$, then $x_i 0$. So if $G = F_x$ then $x = (0,0,0, \ldots )$ and $G = F_x$ is the zero functional. But $G(1,1,1,\ldots) = 1 \neq 0$.