Linear Functionals

Complete the proof of Proposition 5.4: if $\mathcal{X}$ is a normed space, $\mathcal{X}^*$ is a Banach space. (It has been shown in the text that $\mathcal{X}^*$ is complete; only the fact that $\mathcal{X}^*$ is a normed space remains.)

Show that $\mathcal{X}^*$ is a normed space.

Give an example of a measure space $(X, \Omega , \mu )$ that is not $\sigma$-finite for which the conclusion of Theorem 5.6 is false, where Theorem 5.6 states: if $(X, \Omega , \mu )$ is a $\sigma$-finite measure space and $g \in L^\infty (X, \Omega , \mu )$, define $F_g : L^1(\mu ) \to \mathbb {F}$ by $F_g(f) = \int fg \, d\mu$. Then $F_g \in L^1(\mu )^*$ and the map $g \mapsto F_g$ defines an isometric isomorphism of $L^\infty (\mu )$ onto $L^1(\mu )^*$.

Let $\left\{ \mathcal{X}_i : i \in I\right\}$ be a collection of normed spaces. If $1 \leq p < \infty$, show that the dual space of $\oplus_p \mathcal{X}_i$ is isometrically isomorphic to $\oplus_q \mathcal{X}_i^*$, where $1/p + 1/q = 1$.

If $\mathcal{X}_1, \mathcal{X}_2, \ldots$ are normed spaces, show that $(\oplus_0 \mathcal{X}_n)^*$ is isometrically isomorphic to $\oplus_1 \mathcal{X}_n^*$.

Let $n \geq 1$ and let $C^{(n)}[0,1]$ be defined as in Example 1.10 (the collection of functions $f : [0,1] \to \mathbb {F}$ having $n$ continuous derivatives, with norm $\left\| f\right\| = \sup_{0 \leq k \leq n}\left\{ \sup \left\{ \left|f^{(k)}(x)\right| : 0 \leq x \leq 1\right\} \right\}$). Show that $\left\| f\right\| = \sum_{k=0}^{n-1} \left|f^{(k)}(0)\right| + \sup \left\{ \left|f^{(n)}(x)\right| : 0 \leq x \leq 1\right\}$ is an equivalent norm on $C^{(n)}[0,1]$. Show that $L \in (C^{(n)}[0,1])^*$ if and only if there are scalars $\alpha_0, \alpha_1, \ldots , \alpha_{n-1}$ and a measure $\mu$ on $[0,1]$ such that $L(f) = \sum_{k=0}^{n-1} \alpha_k f^{(k)}(0) + \int f^{(n)} \, d\mu$. If $C^{(n)}[0,1]$ is given this new norm, find a formula for $\left\| L\right\|$ in terms of $\left|\alpha_0\right|, \left|\alpha_1\right|, \ldots , \left|\alpha_{n-1}\right|$, and $\left\| \mu \right\|$?

Give $\mathcal{X} = C([0,1])$ the norm $\left\| f\right\| = \int \left|f(t)\right| \, dt$ and define $L : \mathcal{X} \to \mathbb {F}$ by $L(f) = f(\tfrac {1}{2})$. Show directly (without using Theorem 5.6) that $L$ is not bounded. Now prove this as a consequence of (5.6), where Theorem 5.6 states: if $(X, \Omega , \mu )$ is a $\sigma$-finite measure space and $g \in L^\infty (X, \Omega , \mu )$, define $F_g : L^1(\mu ) \to \mathbb {F}$ by $F_g(f) = \int fg \, d\mu$. Then $F_g \in L^1(\mu )^*$ and the map $g \mapsto F_g$ defines an isometric isomorphism of $L^\infty (\mu )$ onto $L^1(\mu )^*$.