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.

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    $\begingroup$ This could be relevant. $\endgroup$ Commented Apr 25, 2020 at 0:00
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    $\begingroup$ 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). $\endgroup$
    – JonathanZ
    Commented Apr 27, 2020 at 19:37
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    $\begingroup$ 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. $\endgroup$
    – s.harp
    Commented Apr 27, 2020 at 19:50
  • $\begingroup$ Maybe this helps. $\endgroup$
    – Calculix
    Commented Apr 27, 2020 at 20:08
  • $\begingroup$ This is a duplicate I am pretty sure: math.stackexchange.com/questions/596720/… $\endgroup$
    – Sorfosh
    Commented Apr 27, 2020 at 22:16

2 Answers 2


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.

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    $\begingroup$ 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.? $\endgroup$ Commented Apr 27, 2020 at 22:32
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    $\begingroup$ @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. $\endgroup$
    – Quoka
    Commented Apr 27, 2020 at 23:00
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    $\begingroup$ 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. $\endgroup$
    – s.harp
    Commented Apr 27, 2020 at 23:10
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    $\begingroup$ @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. $\endgroup$
    – zhw.
    Commented Apr 27, 2020 at 23:13
  • $\begingroup$ @s.harp i dont see word "explicit" in the OP $\endgroup$ Commented Apr 28, 2020 at 3:44

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.


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