# Duality between $L^p$ spaces and weak or weak* convergence

If the measure $μ$ on $S$ is sigma-finite, then the dual of $L^1(μ)$ is isometrically isomorphic to $L^∞(μ)$ (more precisely, the map $κ_1$ corresponding to $p = 1$ is an isometry from $L^∞(μ)$ onto $L^1(μ)^∗$).

The dual of $L^∞$ is subtler. Elements of $(L^∞(μ))^∗$ can be identified with bounded signed finitely additive measures on $S$ that are absolutely continuous with respect to $μ$. See ba space for more details. If we assume the axiom of choice, this space is much bigger than $L^1(μ)$ except in some trivial cases. However, there are relatively consistent extensions of Zermelo-Fraenkel set theory in which the dual of $ℓ^∞$ is $ℓ^1$.

Note that $V^*$ above means the continuous dual of $V$, if I am correct.

there is always a naturally defined continuous linear operator $Ψ : V → V′′$ from a normed space $V$ into its continuous double dual $V′′$, defined by $$\Psi(x)(\varphi) = \varphi(x), \quad x \in V, \ \varphi \in V'. \,$$ As a consequence of the Hahn–Banach theorem, this map is in fact an isometry, meaning $||Ψ(x)|| = ||x||$ for all $x \in V$.

Note that $V'$ above means the continuous dual of $V$.

My questions are as following. The first quote says when the measure $\mu$ is sigma finite, the continuou dual of $L^1(\mu)$ is $L^\infty(\mu)$. Both $L^1(\mu)$ and $L^\infty(\mu)$ are normed spaces, according to the second quote, the double continuous dual of $L^1(\mu)$ should be $L^1(\mu)$ itself, but according to the first quote, the continuous dual of $L^\infty(\mu)$ may not be $L^1(\mu)$ and may be much bigger than $L^1(\mu)$. So I was wondering what I have misunderstood?

Thanks and regards!

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The second quote doesn't say that $\Psi$ is surjective. –  Martin Feb 21 at 3:21
@Martin: Thanks! (1) In the second quote, is $\Phi$ injective? (2) For a topological vector space, when is its continuous double dual same as itself? –  Tim Feb 21 at 3:27
$(1)$ $0 = \lVert \Psi(x) \rVert = \lVert x \rVert \implies x = 0$. (2) en.wikipedia.org/wiki/Reflexive_space // Where is weak and weak*-convergence in your question? –  Martin Feb 21 at 3:29
@Martin: I was planning to ask if weak convergence and weak* convergence for $L^p$ spaces are the same, when $p \in (1, \infty)$ (yes?), when $p=1$ (no?) and when $p=\infty$ (no?) ? –  Tim Feb 21 at 3:56

If the map $\Psi: V \rightarrow V^{\ast\ast}$ is surjective then $V$ is called reflexive. Notice that a non-reflexive vector space can still be isomorphic to its dual, just not via the canonical mapping, there's some discussion and a reference of this here. In general the map $\Psi$ is an linear embedding of $V$ into $V^{\ast\ast}$.

Here is a somewhat illustrative example. I will work with the sequence spaces $\newcommand{\N}{\mathbb N} \ell^1(\N)$, $\ell^\infty(\N)$ and $\ell^\infty(\N)^\ast$, where $\mathbb N$ is endowed with the counting measure. Consider the subspace $c(\mathbb N)\subset \ell^\infty(\N)$ of convergent sequences. Define a functional $f: c(\mathbb N) \rightarrow \mathbb C$ by

$$f(x_n)=\lim_{n\rightarrow \infty} x_n.$$ Notice that no element of $\ell^1(\N)$ induces this functional on $c(\mathbb N)$. To see this pick a sequence $y \in \ell^1(\N)$ then $$\hat{y}(x)=\sum_{n=1}^\infty y_nx_n,$$ in particular if $y\neq 0$ we may find some $y_n\neq 0$ and $\hat{y}(e_n) \neq 0$ but $f(e_n)=0$. Now we can extend $f$ to a functional $\tilde{f}$ on $\ell^\infty(\N)$ by Hahn Banach. We see that $\tilde{f} \notin \Psi(\ell^1(\N))$ so $\ell^1(\N)$ is not reflexive.

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Thanks! In your first paragraph, is $V$ assumed to be a normed space, a topological vector space, or a Banach space, or ...? –  Tim Feb 21 at 4:18
@Tim Sorry, yes it's a Banach space in this context. Although wikipedia tells me the definition is fine for locally convex topological vector spaces as well, whatever those are. –  JSchlather Feb 21 at 4:21
Thanks! "In general the map Ψ is an linear embedding of V into V∗∗", does a mapping being embedding mean that it is injective? Can "in general" be replaced with "always"? –  Tim Feb 21 at 23:30
@Tim I'm not sure in the most general context everything works out. I imagine whenever all the definitions make sense the map is injective. Whenever you have a normed linear space it should be the case that $\Psi$ is a linear isometric embedding. So it's injective and distance preserving. In general embedding says that it's injective, but also homemorphic onto its image. In the context of linear maps, this means we have an inequality of the form $c_1 || v|| \leq ||Tv|| \leq c_2 ||v||$ for some nonzero constants $c_i$ and any vector $v \in V$. –  JSchlather Feb 22 at 15:43
Thanks! Could you also take a look at my another similar question math.stackexchange.com/questions/309783/…? –  Tim Feb 22 at 15:48