Exercise 16, § 1

Let $F$ be a Banach space, $F'$ its dual. For every $p$ such that $1 \leq p \leq +\infty$, denote by $\Lambda_{F'}^p(T, \mu)$ (or simply by $\Lambda_{F'}^p$) the set of mappings $f$ of $T$ into $F'$ such that, for every $z \in F$, the numerical function $\langle z, f \rangle$ belongs to $L^p(T, \mu)$. One has $\Lambda_{F'}^p \supset L_{F'}^p$, and $\Lambda_{F'}^1$ is the space of scalarly essentially $\mu$-integrable functions with values in $F'$ ($F'$ being equipped with $\sigma(F', F)$). One knows that, denoting by $\theta(z)$ the class of $\langle z, f \rangle$ in $L^p(\mu)$, the mapping $z \mapsto \theta(z)$ is continuous (No. 4, Lemma 2); let $M_p(f)$ be its norm; this is a semi-norm on the space $\Lambda_{F'}^p$. In order that $M_p(f) = 0$, it is necessary and sufficient that $f$ be scalarly locally negligible.

a) In order that $f \in \Lambda_{F'}^p$, it is necessary and sufficient that for every numerical function $g \in L^q$, the function $gf$ be scalarly essentially $\mu$-integrable; one then has $M_1(gf) \leq M_p(f) N_q(g)$.

b) In the space $\Lambda_{F'}^1$, show that the semi-norm $M_1$ is equivalent to the semi-norm $M'_1(f) = \sup |\int_A f d\mu|$, where $A$ runs over the set of measurable subsets of $T$; more precisely, $M'_1 \leq M_1 \leq 2M'_1$. Deduce from this that $M_p$ is a semi-norm equivalent to $M'_p(f) = \sup_{N_q(g) \leq 1} M'_1(gf)$.

c) Take for $F$ a Hilbert space having a countable orthonormal basis, and for $\mu$ the Lebesgue measure on $T = [0, 1]$. Show that there exists a mapping of $T$ into $F$, not integrable but scalarly integrable, that is the limit for the semi-norm $M_1$ of a sequence of measurable step functions (cf. Exer. 11); from this, deduce that on the space $L_{F'}^1$ of $\mu$-integrable functions, the topology defined by the semi-norm $M_1$ is strictly coarser than that defined by $N_1$. Analogous result for the semi-norms $M_p$ and $N_p$, for $1 \leq p < +\infty$.

d) Show that on the space $L_{F'}^\infty$, $M_\infty = N_\infty$.

e) Take for $F$ a Hilbert space having a countable orthonormal basis, arranged in a double sequence $(e_{mn})$. Let $\mu$ be Lebesgue measure on $T = [0, 1]$. Let $u_m$ be the mapping of $T$ into $F$ such that $u_m(t) = e_{mn}$ for $(n-1)2^{-m} \leq t < n2^{-m}$ and for $1 \leq n \leq 2^m$, and $u_m(1) = 0$; set $f_n = \sum_{i=0}^n u_i$. Show that, for $1 \leq p < +\infty$, $(f_n)$ is a Cauchy sequence in $\Lambda_{F'}^p$, but does not converge to any function in $\Lambda_{F'}^p$.

f) Let $(f_n)$ be a Cauchy sequence in the space $\Lambda_{F'}^p$ ($1 \leq p \leq +\infty$); assume that there exists a mapping $f$ of $T$ into $F'$ such that, for every $z \in F$, the sequence of functions $\langle z, f_n \rangle$ converges in measure to $\langle z, f \rangle$ (Ch. IV, §5, No. 11). Show that $f \in \Lambda_{F'}^p$ and that the sequence $(f_n)$ has limit $f$ in $\Lambda_{F'}^p$.

g) Take for $F$ the Banach space $L^1(\mathbf{N})$ of absolutely convergent series; its dual is $F' = L^\infty(\mathbf{N})$ (TVS, IV, §1, Exer. 1). Let $\mu$ be Lebesgue measure on $T = [0, 1]$. Let $f$ be the mapping of $T$ into $F'$ such that $f(0) = 0$ and, for $2^{-n-1} < t \leq 2^{-n}$, $f(t)$ is the sequence all of whose terms are zero, except for the term of index $n$, equal to $2^n/p$ for $1 \leq p < +\infty$. Show that $f$ is measurable for the strong topology on $F'$, and belongs to $\Lambda_{F'}^p$, but that there does not exist any sequence $(f_n)$ of step functions that tends to $f$ in the space $\Lambda_{F'}^p$.