The Adjoint of a Linear Operator

Prove Proposition 1.3: if $\mathscr {X}$ and $\mathscr {Y}$ are Banach spaces, $A, B \in \mathcal{B}(\mathscr {X}, \mathscr {Y})$, and $\alpha , \beta \in \mathbb {F}$, then $(\alpha A + \beta B)^* = \alpha A^* + \beta B^*$. Moreover, $A^*: (\mathscr {Y}^*, \mathrm{wk}^*) \to (\mathscr {X}^*, \mathrm{wk}^*)$ is continuous.

Complete the proof of Proposition 1.4: if $\mathscr {X}$ and $\mathscr {Y}$ are Banach spaces and $A \in \mathcal{B}(\mathscr {X}, \mathscr {Y})$, then:

Verify the statement made in (1.5): let $(X, \Omega , \mu )$ and $M_\phi : L^p(\mu ) \to L^p(\mu )$ be as in Example III.2.2. If $1 \le p < \infty$ and $1/p + 1/q = 1$, then $M_\phi^*: L^q(\mu ) \to L^q(\mu )$ is given by $M_\phi^* f = \phi f$. That is, $M_\phi^* = M_\phi$.

Verify the statement made in (1.6): let $K$ and $k$ be as in Example III.2.3. If $1 \le p < \infty$ and $1/p + 1/q = 1$, then $K^*: L^q(\mu ) \to L^q(\mu )$ is the integral operator with kernel $k^*(x, y) \equiv k(y, x)$.

Verify the statement made in (1.7): let $X$, $Y$, $\tau$, and $A$ be as in Example III.2.4. Then $A^*: M(Y) \to M(X)$ is given by

for every Borel subset $\Delta$ of $X$ and every $\mu$ in $M(Y)$.

Let $1 \le p < \infty$ and define $S: l^p \to l^p$ by $S(\alpha_1, \alpha_2, \ldots ) = (0, \alpha_1, \alpha_2, \ldots )$. Compute $S^*$.

Let $A \in \mathcal{B}(c_0)$ and for $n \ge 1$, define $e_n$ in $c_0$ by $e_n(n) = 1$ and $e_n(m) = 0$ for $m \ne n$. Put $\alpha_{mn} = (Ae_n)(m)$ for $m, n \ge 1$.

Let $A \in \mathcal{B}(l^1)$ and for $n \ge 1$ define $e_n$ in $l^1$ by $e_n(n) = 1$, $e_n(m) = 0$ for $m \ne n$. Put $\alpha_{mn} = (Ae_n)(m)$ for $m, n \ge 1$.

(Bonsall [1986]) Let $\mathscr {X}$ be a Banach space, $Z$ a nonempty set, and $u: Z \to \mathscr {X}$. If there are positive constants $M_1$ and $M_2$ such that (i) $\left\| u(z)\right\| \le M_1$ for all $z$ in $Z$ and (ii) for every $x^*$ in $\mathscr {X}$, $\sup \left\{ \left|\langle u(z), x^* \rangle \right|: z \in Z\right\} \ge M_2 \left\| x^*\right\|$; then for every $x$ in $\mathscr {X}$ there is an $f$ in $l^1(Z)$ such that $(*)\ x = \sum \left\{ f(z)u(z): z \in Z\right\}$ and $M_2 \inf \left\| f\right\|_1 \le \left\| x\right\| \le M_1 \inf \left\| f\right\|_1$, where the infimum is taken over all $f$ in $l^1(Z)$ such that $(*)$ holds.

(Bonsall [1986]) Let $m$ be normalized Lebesgue measure on $\partial \mathbb {D}$ and for $\left|z\right| < 1$ and $\left|w\right| = 1$ let $p_z(w) = (1 - \left|z\right|^2)/\left|1 - \bar{z}w\right|^2$. So $p_z$ is the Poisson kernel. Show that if $f \in L^1(m)$, then there is a sequence $\left\{ z_n\right\} \subseteq \mathbb {D}$ and a sequence $\left\{ \lambda_n\right\}$ in $l^1$ such that $(*)\ f = \sum_{n=1}^{\infty } \lambda_n p_{z_n}$. Moreover, $\left\| f\right\|_1 = \inf \sum_{n=1}^{\infty } \left|\lambda_n\right|$, where the infimum is taken over all $\left\{ \lambda_n\right\}$ in $l^1$ such that $(*)$ holds.

If $\mathscr {X}$ and $\mathscr {Y}$ are Banach spaces and $B \in \mathcal{B}(\mathscr {Y}^*, \mathscr {X}^*)$, then there is an operator $A$ in $\mathcal{B}(\mathscr {X}, \mathscr {Y})$ such that $B = A^*$ if and only if $B$ is $\mathrm{wk}^*$-continuous.

If $\mathscr {X}$ is a Banach space and $\mathscr {M}$ and $\mathscr {N}$ are closed subspaces, show that the following statements are equivalent.