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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.

Pascal's Wager
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    This could be relevant. – Physical Mathematics Apr 25 '20 at 00:00
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    Previous answers seem to characterize the entire dual space, which is nice, but I think a good answer for this question would explicitly exhibit one element that cannot be expressed as $f \mapsto \int f g$ (or at least as explicitly as possible). – JonathanZ Apr 27 '20 at 19:37
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    Whats an example of a non $\ell^1$ element of $\ell^\infty(\Bbb N)^*$? You will need to reference points in $\beta\Bbb N- \Bbb N$ / non-principal ultra-filters of $\Bbb N$. Is that an explicit "example"? I wouldn't be surprised if you need some similar non-constructive things for the $L^\infty$ dual elements. – s.harp Apr 27 '20 at 19:50
  • Maybe this helps. – Calculix Apr 27 '20 at 20:08
  • This is a duplicate I am pretty sure: https://math.stackexchange.com/questions/596720/l1-and-l-infty-are-not-reflexive/596726 – Sorfosh Apr 27 '20 at 22:16
  • There is no example: constructing any function requires use of Hahn-Banach Theorem on $L^\infty([0,1])$ which is nonseparable. Hahan-Banach on non-separable spaces is inherently non-constructive. – rubikscube09 Apr 27 '20 at 23:13

2 Answers2

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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.

Quoka
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    i usually dont give a rat's behind about functions versus equivalence classes, but I might actually care now. for ur solution, u want $L^\infty$ to be a set of functions. but then how do u know the extended $T$ satisfies $Tf_1 = Tf_2$ if $f_1 = f_2$ a.e.? – mathworker21 Apr 27 '20 at 22:32
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    @mathworker21 We assume that $L^\infty$ is a collection of equivalence classes rather than functions, otherwise it's not a Banach space and we can't apply the Hahn-Banach theorem. Technically, $C([a,b])$ isn't a subspace of $L^\infty([a,b])$ but corresponds to a subspace. I'll add some clarifications. – Quoka Apr 27 '20 at 23:00
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    This is not an explicit example (can you evaluate the functional on $\sin(\frac1{x-a})$?), it is a proof that functional exists that has a certain form when restricted to a certain subset. But this is as "explicit" as will be possible. – s.harp Apr 27 '20 at 23:10
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    @mathworker21 No, we want $L^\infty$ to be a set of equivalence classes of functions. Otherwise how is it a normed linear space? Now $C[a,b]$ is a set of functions (usually). But the map $I:C[a,b] \to L^\infty,$ defined by $f\to [f],$ where $[f]$ is the usual equivalence class, is a linear isometry from $C[a,b]$ onto $I(C[a,b]) \subset L^\infty.$ So really it's not $C[a,b],$ but $I(C[a,b]),$ where $T$ starts from. – zhw. Apr 27 '20 at 23:13
  • @s.harp i dont see word "explicit" in the OP – mathworker21 Apr 28 '20 at 03:44
  • @Quoka okay, $+1$ and nicely written. However, I think you should just write your solution directly by saying "define $T$ on the vector subspace of all $L^\infty$ functions that are a.e. continuous by $L(f) = f(a)$, where $f$ is the continuous representative. This is a continuous linear functional on this subspace". – mathworker21 Apr 28 '20 at 03:48
  • @mathworker21 Thank you! I think I'll leave the solution as is for now, simply because (in most analysis books I've seen) these types of identifications are very common. Though I would agree that, ideally, we shouldn't rely on identifications. It has the potential to confuse people and (in some cases) leaves room for mistakes. – Quoka Apr 28 '20 at 06:49
  • @Quoka no,no,no. You're generalizing too much. Like I said, I nearly never care about the distinction and am very very fine with the identification. But for this exact situation, I think making the distinction is somewhat important. Like, usually the justification for being able to ignore the distinction is very basic (e.g. integrals don't detect the difference), but here's it's a bit more subtle. With that said, you of course may still feel free to disagree; I just wanted to make my point clear. – mathworker21 Apr 28 '20 at 06:52
  • I think the remark makes it clear enough that one is indeed working with a "equivalence class" version of $C([a,b])$, even zhw's comment makes that even clearer. – Olorun Apr 28 '20 at 15:04
  • @Quoka I am working on a similar problem, but I am trying to show that the linear functional defined by $L(f) = f(2/3)$ cannot be expressed as such an integral. Is it possible to modify the example you used to work with the linear functional I am dealing with? – user193319 Oct 07 '21 at 19:31
  • @user193319 It should work fine. I haven't checked this, but have you tried redefining $f_n(x)$ as the maximum of $0$ and $1 - n\lvert x-2/3\rvert$? The idea is to have a sequence of continuous functions which tend to $0$ pointwise almost everywhere but are always equal to $1$ at $2/3$. – Quoka Oct 14 '21 at 00:50
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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.

H. H. Rugh
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