Analysis Problem of the Day 60

Today’s problem appeared as Problem 3 on the UCLA Spring 2023 Analysis Qual:

Problem 3. Let $(X,\mathcal{M})$ be a measure space and let $V$ be a separable real Banach space.

a) Show that there exists a sequence $L_1,L_2,... \in V^*$ such that $\|L_n\|=1$ for all $n$ and

$\|v\|=\sup_n |L_n(v)|, \quad \text{ for all }v \in V.$

b) Show that $\phi: (X,\mathcal{M}) \to (V,\mathcal{B}(V))$ is measurable iff it is weakly measurable, i.e. $\phi$ is measurable if and only if $\phi \circ L: X \to \mathbb{R}$ is $(\mathcal{M},\mathcal{B}(\mathbb{R}))$ measurable for all $L \in V^*.$

Solution: a) Note that it suffices to show $\sup_n |L_n(\widehat{v})|=1$ for all $\|\widehat{v}\|=1.$ Since $V$ is separable, there exists a countable dense subset of the unit sphere in $V,$ i.e. a sequence $v_1,...,v_n \in V$ of unit vectors such that for all $\|v\|=1,$ for any $\epsilon >0$ there exists $v_n$ such that $\|v-v_n\|<\epsilon.$ Moreover, by Hahn-Banach, there exist linear functionals $L_n$ such that $L_n(v_n)=1$ and $\|L_n\|=1.$ More explicitly, if one defines $L_n$ as the extension of the linear functional on $\mathbb{R} v_n$ given by $L_n(v_n)=1$ and bounded by the seminorm $p(v)=\|v\|$ on $V,$ it follows that $\|L_n\|=1$ and $L_n(v_n)=1,$ as desired.

Then, for any $\|v\|=1,$

$|L_n(v)| = |L_n(v-v_n)+L_n(v_n)| \geq 1-\epsilon$

for any $\epsilon >0$ for an appropriate choice of $n,$ and $|L_n(v)| \leq \|L_n\|\|v\|=1$ for all $n.$ It follows that $\sup_n |L_n(v)|=1$ for all $\|v\|=1,$ and so the claim follows.

b) Clearly, if $\phi$ is measurable, since the composition of measurable functions is measurable and continuous functions are measurable, one has $L \circ \phi$ is measurable for all $L \in V^*.$ Conversely, since $V$ is separable, it is second countable and has a topological base of open balls. It thus suffices to show that $\phi^{-1}(B(x,r)) \in \mathcal{M}$ for any ball. I claim that

$B(x,r) = \bigcap_n L_n^{-1}((L_n(x)-r,L_n(x)+r)),$

which would imply that

$\phi^{-1}(B(x,r)) = \bigcap_n (L_n \circ \phi)^{-1} (L_n(x)-r,L_n(x)+r),$

which is a countable intersection of $\mathcal{M}$ -measurable sets and therefore also $\mathcal{M}$ -measurable. Indeed, if $y \in B(x,r),$ then

$\|x-y\|<r \implies |L_n(x)-L_n(y)| < r \implies L_n(y) \in (L_n(x)-r,L_n(x)+r).$

Conversely, if $y \in \bigcap_n L_n^{-1}((L_n(x)-r,L_n(x)+r)),$ one has $|L_n(\frac{x-y}{r})|<1$ for all $n.$ But if $\|\frac{x-y}{r}\| \geq 1,$ for sufficiently small $\epsilon >0,$ there exists a $c \geq 1$ and $v_n$ such that $\|\frac{x-y}{r} -cv_n\|<\epsilon,$ so that $|L_n(\frac{x-y}{r}-cv_n)| \geq c-|L_n(\frac{x-y}{r})| > \epsilon,$ which is a contradiction since $\|L_n\| =1.$ Thus, $\|\frac{x-y}{r}\|<1,$ i.e. $y \in B(x,r).$ We thus conclude that $\phi$ is measurable if and only if it is weakly measurable.

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