Analysis Problem of the Day 92

Today’s problem appeared as Problem 4 on the UCLA Spring 2014 Analysis Qual:

Problem 4. a) For a sequence $\{a_n\} \subseteq [0,1]$ and $f \in C([0,1]),$ define

$\phi(f):=\sum_{n=1}^\infty 2^{-n} f(a_n).$

Prove that there is no $g \in L^1([0,1])$ such that $\phi(f) = \int f(x)g(x)dx$ for all $f \in C([0,1]).$

b) Each $g \in L^1([0,1])$ defines a continuous linear functional on $L^\infty([0,1])$ by

$T_g(f)= \int f(x)g(x)dx.$

Show that there are continuous linear functionals on $L^\infty([0,1])$ that are not of this form.

Solution: a) Pick some $a_m$ and define $f_n(x):=\max(1-n|a_m-x|),0), x \in [0,1].$ Then, $|f_n| \leq 1, f_n \in C([0,1]),$ and as $n \to \infty,$ $f_n$ decreases monotonically to zero a.e. on $[0,1]$ and therefore $f_n \to 0$ in $L^1$ by the Monotone Convergence Theorem. However, $\phi(f_n) \geq 2^{-m} f_n(a_m) = 2^{-m}$ for all $m.$ Thus, if such a $g \in L^1([0,1])$ were to exist, then $|\phi(f_n)| = |\int f_n(x) g(x)dx| \leq \int_{\text{supp }f_n} |g(x)|dx,$ and since for $\epsilon < 2^{-(m+1)}$ there exists $\delta>0$ such that $\mu(A)<\delta$ implies $\int_A |g|<\epsilon,$ since $\mu(\text{supp }f_n) \to 0,$ this yields a contradiction as $n \to \infty.$ Thus, $\phi$ is not given by integration against an $L^1$ function.

b) Note that $\phi$ given in (a) is a bounded linear functional on a subspace $C([0,1]) \subseteq L^\infty([0,1])$ that is not given by integrating against an $L^1$ function, so if one could extend $\phi$ to a bounded linear functional on all of $L^\infty,$ one would be done. This can be accomplished using the Hahn-Banach theorem, and the only thing that needs to be shown is that there exists a seminorm on $L^\infty([0,1])$ bounding $\phi$ on $C([0,1]).$ But clearly, $p(f):=\|f\|_\infty$ is such a seminorm, since $|\phi(f)| \leq \|f\|_\infty$ for $f \in C([0,1]).$ Thus, by the Hahn-Banach theorem, there exists a bounded extension of $\phi$ to all of $L^\infty([0,1])$ that is therefore not given by integration against a function in $L^1,$ and so we are done.

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