Measurability of Banach space valued functions

There is only an indirect proof of the existence of a function {f\colon [0, 1]\to \mathbb R} that is not Lebesgue measurable. But it’s easy to give an explicit example when the codomain of {f} is a Banach space: just let {b(t)} be the sequence of the binary digits of {t}, considered as an element of the sequence space {\ell_\infty}.

Why is {b} not measurable? Recall that a Banach space-valued function {f} is (Bochner) measurable iff there is a sequence of simple functions {\sum v_k \chi_{A_k}} (finite sum, measurable {A_k}, arbitrary vectors {v_k}) that converges to {f} almost everywhere. This property implies that, with an exception of a null set, the range of {f} lies in the separable subspace spanned by all the vectors {v_k} used in the sequence of simple functions. But {b} has the property {\|b(t)-b(s)\|=1} whenever {t\ne s}, so the image of any uncountable set under {b} is nonseparable.

Another way to look at this: on the interval [0, 1) the function {b} is injective and its range has discrete topology, which implies that every subset of [0, 1) is the preimage of some open subset of {\ell_\infty} under {b}.

The binary-digits functions can also be used to illustrate an issue with the duality of Lebesgue-Bochner spaces {L_p(0, 1; X)} where {X} is a Banach space and {1\le p<\infty}. (So, {f} belongs to this space iff it is Bochner measurable and the {L^p} norm of {\|f\|\colon [0, 1]\to [0, \infty)} is finite.) In general we do not have the expected relation {L_p(0, 1; X)^* = L_q(0, 1; X^*)} with {1/p+1/q=1}. The natural isometric embedding of {L_q(0, 1; X^*)} into {L_p(0, 1; X)^*} is still there: any {g\in L_q(0, 1; X^*)} acts on {L_p(0, 1; X)} by {f\mapsto \int \langle f(t), g(t) \rangle\, dt}. But unless {X^*} has the Radon–Nikodym property, these are more bounded linear functionals on {L_p(0, 1; X)}.

To construct such a functional, let {b_n(t)} be the {n}-th binary digit of {t}. Given {f\in L_1(0, 1; \ell_1)}, write it in coordinates as {(f_1, f_2, \dots)} and define {\varphi(f) = \sum_n \int_0^1 f_n b_n}. This is a bounded linear functional, since {|\varphi(f)|\le \sum_n \int_0^1 |f_n| = \|f\|}. But there is no function {g\in L_\infty(0, 1; \ell_\infty)} that represents it, i.e., {\varphi(f) = \int_0^1 \langle f(t), g(t)\rangle \,dt = \sum_n \int_0^1 f_n g_n }. Indeed, if such {g} existed then by considering {f} with only one nonzero coordinate, we find that {g_n} must be {b_n}, using the duality {L_1^* =  L_\infty} in the scalar case. But the function {[0, 1]\to \ell_\infty} with the components {(b_1, b_2, \dots)} is not measurable, as shown above.

This example, which applies to all {1\le p<\infty}, also serves as a reminder that the duality relation {L_p(0, 1; X)^* = L_q(0, 1; X^*)} depends on the dual space {X^*} having the Radon-Nikodym property (RNP), not {X} itself. Indeed, {X=\ell_1} has the RNP; its dual does not.

The importance of {X^*} having the RNP becomes clear once one tries to follow the usual proof of {L_p^*=L_q}. Given {\varphi\in L_p(0,1;X)^*}, we can define an {X^*}-valued measure {\tau} on {[0, 1]} by {\tau(A)(v) = \varphi( v\chi_A)} where {A\subset [0, 1]} is Lebesgue measurable and {v\in X}. This measure has reasonable finiteness and continuity properties coming from {\varphi} being a bounded functional. Still, the existence of a density {g\colon [0, 1]\to X^*} of the measure {\tau} depends on the structure of the Banach space {X^*}.

Almost norming functionals, Part 2

Let E be a real Banach space with the dual E^*. Fix \delta\in (0,1) and call a linear functional e^*\in E^* almost norming for e if |e|=|e^*|=1 and e^*(e)\ge \delta. In Part 1 I showed that in any Banach space there exists a continuous selection of almost norming functionals. Here I will prove that there is no uniformly continuous selection in \ell_1.

Claim. Let S be the unit sphere in \ell_1^n, the n-dimensional \ell_1-space.  Suppose that f\colon S\to \ell_{\infty}^n is a map such that f(e) is almost norming e in the above sense. Then the modulus of continuity \omega_f satisfies \omega_f(2/n)\ge 2\delta.

(If an uniformly continuous selection was available in \ell_1, it would yield selections in \ell_1^n with a modulus of continuity independent of n.)

Proof. Write f=(f_1,\dots,f_n). For any \epsilon\in \{-1,1\}^n we have n^{-1}\epsilon \in S, hence

\sum\limits_{i=1}^n \epsilon_i f_i(n^{-1}\epsilon)\ge n\delta for all \epsilon\in \{-1,1\}^n. Sum over all \epsilon and change the order of summation:

\sum\limits_{i=1}^n \sum\limits_{\epsilon}\epsilon_i f_i(n^{-1}\epsilon)\ge n2^n\delta

There exists i\in\{1,2,\dots,n\} such that

\sum\limits_{\epsilon}\epsilon_i f_i(n^{-1}\epsilon) \ge 2^n \delta

Fix this i from now on. Define \tilde \epsilon to be the same \pm vector as \epsilon, but with the ith component flipped. Rewrite the previous sum as

\sum\limits_{\epsilon} -\epsilon_i f_i(n^{-1}\tilde \epsilon)\ge 2^n\delta

and add them together:

\sum\limits_{\epsilon}\epsilon_i [f_i(n^{-1}\epsilon)-f_i(n^{-1}\tilde \epsilon)]\ge 2^{n+1}\delta

Since \|n^{-1}\epsilon-n^{-1}\tilde \epsilon\|=2/n, it follows that 2^n \omega_f(2/n) \ge 2^{n+1}\delta, as claimed.

Almost norming functionals, Part 1

Let E be a real Banach space with the dual E^*. By the Hahn-Banach theorem, for every unit vector e\in E there exists a functional e^*\in E^* of unit norm such that e^*(e)=1. One says that e^* is a norming functional for e. In general, one cannot choose e^* so that it depends continuously on e. For example, the 2-dimensional space with \ell_1 norm does not allow such a continuous selection.

Fix \delta\in (0,1) and call a linear functional e^* almost norming for e if |e|=|e^*|=1 and e^*(e)\ge \delta. In any Banach space there exists a continuous selection of almost norming functionals.

Continue reading “Almost norming functionals, Part 1”