Example of a bounded linear functional $T:L^ \infty \to \mathbb{R}$ which cannot be expressed by an integral $\int_{[a,b]}fg$ for an integrable $f$

Question

Let $m$ be the Lebesgue measure on $[a,b]$.

What would be an example of a bounded linear functional $T:L^ \infty \to \mathbb{R}$ which cannot be expressed by an integral $\int_{[a,b]}fg$ for an integrable $f$?

I know that, if $1 \leq p < \infty$, then any bounded linear functional $T:L^p \to \mathbb{R}$ can be expressed by an integral $\int_{[a,b]}fg$ for an $f \in L^q$. (Where $q$ is the conjugate of $p$.) But the proof I am familiar with relies on the fact that $p$ is finite.

Answer 1

My answer is very similar to the solution described in Folland's real analysis book.

Consider the map $L : C([a, b]) \to \mathbb{R}$ given by $$L(f) = f(a).$$ Note that, here $C([a,b])$ is the space of continuous functions on $[a,b]$. We note that $C([a,b])$ can be viewed as a vector subspace of $L^\infty([a,b])$.

Remark: Technically, $C([a,b])$ is not a subspace of $L^\infty([a,b])$ since $C([a,b])$ is a collection of functions whereas $L^\infty([a,b])$ is a collection of equivalence classes. On the other hand, one can map each function $f\in C([a,b])$ to it's equivalence class in $L^\infty([a,b])$. Thus, we see that $C([a,b])$ corresponds to a vector subspace $Y$ of $L^\infty([a,b])$. We are choosing to view $C([a,b])$ as this subspace $Y$. Then $L$ is defined on $Y$ by the map $L([f]) = f(a)$ where $f$ is the continuous representative of the equivalence class $[f]$.

Returning to our problem, we note that $$ L(f) \leq \lVert f\rVert_{\infty}. $$ By the Hahn-Banach Theorem, there exists a linear functional $$T:L^\infty([a,b]) \to \mathbb{R}$$ extending $L$ such that $T(f) \leq \lVert f\rVert_\infty$ for every $f\in L^\infty([a,b])$. That is, $T : L^\infty([a,b]) \to \mathbb{R}$ is a continuous linear functional such that $$ T\mid_{C([a,b])} = L. $$ Suppose for a contradiction that there exists a function $g\in L^1([a,b])$ such that $$ T(f) = \int_{[a,b]} fg $$ for all $f\in L^\infty([a,b])$. Consider the sequence of functions $$ f_n(x) = \max\left(0, 1 - n(x-a)\right). $$ By definition, we have $$T(f_n) = f_n(a) = 1$$ for each $n\in\mathbb{N}$. On the other hand, it follows from the dominated convergence theorem that $$ \lim_{n\to\infty} T(f_n) = \int_{[a,b]} f_n g = 0, $$ which is absurd.

Answer 2

You may take a sequence of linear functionals, $T_n\in(L^\infty)'$, which for $n>\frac{1}{b-a}$ and $f\in L^\infty$ is defined by: $$ T_n f = n\int_a^{a+1/n} f(x) \, dx.$$ Then $|T_n f| \leq \|f\|_\infty$ so by weak-* compactness (Banach-Alaoglu), the sequence has a weak-* accumulation point $S\in (L^\infty)'$.

Being in a non-separable case there need not be a convergent subsequence and in view of the abstract nature of the construction (Axiom of Choice being involved), it is often quite difficult to do any calculations with such an S. In the present case we may, however, extract sufficient information.

First, the interpretation of being a weak-* accumulation point is that to any given $f\in L^\infty$ there is a subsequence $(n_k)$, which in general will depend upon $f$, so that $\lim_k T_{n_k} f = S f$.

1. If we take $f=1$ we see that $T_n 1 = 1$ so that $S1 = \lim T_{n_k} 1 = 1$. In particular, $S$ is non-zero in $(L^\infty)'$.

2. If $f$ has support in $[a+\varepsilon,b]$ for $\varepsilon>0$ we see that $T_nf=0$ for $n$ large enough, implying that $Sf=0$.

Suppose now, that $Sf=\int_a^b gf \, dx$ for some $g\in L^1$. The last part implies that $g\equiv 0$ (Lebsegue a.e.) so $S=0$, thus contradicting the first part. So $S$ may not be described in this way.