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Spring 2017 - Problem 4

construction

For $n \geq 1$ , let $\func{a_n}{\pco{0,1}}{\set{0,1}}$ denote the $n$ -th digit in the binary expansion of $x$ , so that

x = \sum_{n \geq 1} a_n\p{x} 2^{-n} \quad\text{for all } x \in \pco{0, 1}.

(We remove any ambiguity from this definition by requiring that $\liminf_{n\to\infty} a_n\p{x} = 0$ for all $x \in \pco{0, 1}$ .) Let $M\p{\pco{0,1}}$ denote the Banach space of finite complex Borel measures on $\pco{0, 1}$ and define linear functions $L_n$ on $M\p{\pco{0,1}}$ via

L_n\p{\mu} = \int_0^1 a_n\p{x} \,\diff\mu\p{x}.

Show that no subsequence of the sequence $L_n$ converges in the weak-* topology on $\p{M\p{\pco{0,1}}}^*$ .

Solution.

Let $\set{L_{n_k}}_k$ be a subsequence. If $x$ is as given in the problem, consider the Dirac measure $\delta_x$ , which gives $L_{n_k}\p{\delta_x} = a_{n_k}\p{x}$ . Thus, if we have $a_{n_k}\p{x}$ alternate in $k$ , then this sequence does not converge. More concretely, let

x = \sum_{k=1}^\infty \p{1 + \p{-1}^k}2^{-n_k}

and so $L_{n_k}\p{\delta_x} = 1 + \p{-1}^k$ , which does not converge.