I want some proof for the following statement :

$L^1$ and $L^{\infty}$ are not reflexive.

Can anyone help me, please? or reference me?

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    $\begingroup$ It's strange to start a bounty on a question whose answer you've already accepted. Acceptance is supposed to indicate that you are satisfied with the answer, but your bounty says you want more. $\endgroup$ – Rahul Dec 10 '13 at 9:04
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    $\begingroup$ For some reason, Erich wants a complete solution written here, instead of having to look in a book, or at another web site. $\endgroup$ – GEdgar Dec 11 '13 at 15:40

Brezis "Fuctional analysis Sobolev spaces and Partial Differential Equations" pag 99-102 There is everything you need


If $V$ is a Banach space we call $V'$ the dual space (see continuous dual space on wikipedia), i.e. the space of linear continuous functionals $\xi \colon V\to \mathbb R$. Then it is well known that there exists a natural injection $$ J \colon V \to V'' $$ defined by $$ J(v)(\xi) = \xi(v) $$ for all $\xi \in V'$. We know that $J$ is an isometry, in particular: linear, continuous and injective (see double dual). We say that $V$ is reflexive if $J$ is also surjective i.e. if every $T\in V''$ can be written as $J(v)$ for some $v\in V$ (see reflexivity on wikipedia). It is not difficult to prove that if $V$ is reflexive, so is $V'$ (in fact being $V''$ isomorphic to $V$ one finds that $V'''$ is isomorphic to $V'$ see properties).

We know that $(L^1)'$ can be represented by $L^\infty$. This means that given any $\xi \in (L^1)'$ there exists $f\in L^\infty$ such that $$ \xi(g) = \int f \cdot g $$ for all $g\in L^1$ (see dual spaces of $L^p$ on wikipedia). What we want to prove is that there exists a linear continuous functional $T\colon L^\infty\to \mathbb R$ which cannot be written as $T = J(f)$ for any $f\in L^1$.

Let $W$ be the subspace of $L^\infty$ composed by continuous functions. Define the linear operator $T:W\to \mathbb R$ by $$ T(f) = f(0) $$ and notice that $T$ is continuous with respect to the norm of $L^\infty$: in fact $|T(f)| = |f(0)| \le \sup |f| = \Vert f\Vert_{L^\infty}$. Hence by Hahn-Banach theorem you can extend it to a continuous linear function on the whole space $T:L^\infty \to \mathbb R$.

Suppose now that $T=J(f)$ for some $f\in L^1$. This means that $T$ can be represented as: $$ T(g) = \int f \cdot g $$ for some $f\in L^1$. Then we know that $\int f\cdot g = g(0)$ for all continuous and bounded $g$. Now notice that given any $x_0\neq 0$ and $\epsilon <|x_0|$, it is possible to find continuous functions $g_k$ with support in $[x_0-\epsilon,x_0+\epsilon]$ which converge in $L^\infty$ to the characteristic function of the interval $[x_0-\epsilon,x_0+\epsilon]$. We hence find that $$ 0 = \frac{1}{2\epsilon}\int f \cdot g_k \to \frac{1}{2\epsilon}\int_{x_0-\epsilon}^{x_0+\epsilon} f(x)\,dx $$ for all $\epsilon>0$. But if $x_0$ is a Lebesgue point (wikipedia) for $f$ such an integral must converge to $f(x_0)$ when $\epsilon\to 0$. Hence $f(x_0)=0$ for all Lebesgue points $x_0\neq 0$ which means (since almost every point is a Lebesgue point) that $f=0$ almost everywhere. But then we should conclude that $T=0$ which is a contradiction.

Hence we have found $T\in (L^\infty)'$ which is not represented by any $f \in L^1$. This means that $(L^\infty)' \supsetneq L^1$ and hence $L^1$ and $L^\infty$ are not reflexive.

  • $\begingroup$ Nice answer! A minor point: when $L^\infty$ is finite dimensional is reflexive if and only if it's finite dimensional. It has been shown somewhere on the site. In this case, the extension of $T$ is necessarily continuous. $\endgroup$ – Davide Giraudo Dec 10 '13 at 16:59
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    $\begingroup$ Note that Hahn-Banach (or some other proposition not provable in ZF) is required for this result. You cannot write down an explicit functional on $L^\infty[0,1]$ that is not in $L^1$. $\endgroup$ – GEdgar Dec 11 '13 at 15:43
  • $\begingroup$ How can you got the equality $\int f g_k=0$? $\endgroup$ – mnmn1993 Nov 11 '17 at 7:55

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