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[[field|Here]] $\mathbb{F}$ denotes $\mathbb{R}$ or $\mathbb{C}$.
> [!proposition] Proposition. ([[the dual of the space of null sequences naturally identifies with l1]])
> [[Lp duality]] does not generally hold for $p=\infty,p'=1$. However, in the case [[Lp-norm|of]] $\ell^{\infty}$, the [[Lp duality|Hölder pairing]] $\langle -,- \rangle: \ell^{1} \times \ell^{\infty} \to \mathbb{F}$ restricts to a [[perfect pairing]] $\langle -,- \rangle : \ell^{1} \times c_{0} \to \mathbb{F},$where $c_{0} \leq \ell^{\infty}$ denotes the [[space of null sequences]] in $\mathbb{F}$, [[dual vector space|and]] the perfect [[isomorphism]] $ \begin{align}
\ell^{1} &\to c_{0}^{\vee} \\
(a_{n}) &\mapsto \langle (a_{n}), - \rangle
\end{align}$
is an [[Lipschitz continuous|isometry]].
^proposition
From [[Lp duality]] we know the [[Lp duality|Hölder pairing]] $\langle -,- \rangle: \ell^{\infty} \times \ell^{1}\to \mathbb{F}$ induces a [[Lipschitz continuous|isometric]] [[linear map|linear]] [[topological embedding]] $\begin{align}
\ell^{1} &\to (\ell^{\infty})^{\vee} \\
(x_{n}) & \mapsto \langle (x_{n}), - \rangle .
\end{align}$
The image of this embedding identifies isometrically [[dual vector space|with]] $(c_{0})^{\vee}$, where $c_{0}$ denotes the space of [[space of null sequences|null sequences]]. This gives an [[Lipschitz continuous|isometric]] identification $\ell^{1} \cong (c_{0})^{\vee}.$
> [!proof]- Proof. ([[the dual of the space of null sequences naturally identifies with l1]])
> ~
Define the '[[dual map|restriction map]]' $R:=(c_{0} \hookrightarrow \ell^{\infty})^{\vee}$; explicitly $\begin{align}
R:(\ell^{\infty})^{\vee} &\to (c_{0})^{\vee} \\
\varphi & \mapsto \varphi |_{c_0}.
\end{align}$
As the [[dual map|dual]] of the [[inclusion map]] $c_{0} \hookrightarrow \ell^{\infty}$, $R$ has [[norm]] $1$ ([[characterizing continuity of linear maps|in particular]] is [[continuous]]) and is [[surjection|surjective]] ([[Hahn-Banach Extension Theorem|Hahn-Banach]]). Labeling the natural map $J:\ell^{1} \to (\ell^{\infty})^{\vee}$, we claim $R \circ J$ is an isometric isomorphism of [[Banach space|Banach spaces]]. This requires showing that $R$ *restricted to the image of $J$* is still surjective, likewise injective, and that $R$ restricted to the image of $J$ is an isometry.
**Injectivity.** Easy.
**Surjectivity.** Let $\varphi \in (c_{0})^{\vee}$. We have to find $x \in \ell^{1}$ such that $\langle x, - \rangle=\varphi$, i.e., such that $\sum_{k=1}^{\infty} x_{k}y_{k}=\varphi(y) \text{ for all }y=(y_{k}) \in c_{0}.$
Let $e_{k}$ be the $k$th unit vector (in $c_{0 0} \subset c_{0}$). Set $x_{k}=\varphi(e_{k})$. Given $y \in c_{0}$, write $y^{(N)}:=\sum_{k=1}^{N} y_{k}e_{k}$. Then $y^{(N)} \to y$ in $\|\cdot\|_{\infty}$, so
Then $\begin{align}
\sum_{k=1}^{\infty} x_{k}y_{k}&= \sum_{k=1}^{\infty} \varphi (e_{k})y_{k} \\
&=\sum_{k=1}^{\infty} \varphi(y_{k}e_{k}) \\
&= \lim_{k \to \infty} \sum_{k=1}^{N} \varphi(y_{k} e_{k}) \\
&= \lim_{k \to \infty} \varphi\left( \overbrace{ \sum_{k=1}^{N} y_{k} e_{k} }^{ y^{(N)} } \right)
\end{align}$
and now the result follows because $\varphi$ is [[continuous]] (hence preserves limits).
**Isometry.** Let $x \in \ell^{1}$. We have to show that $\|(R \circ J)(x)\| = \sup_{\{ y \in c_{0} : \|y\|_{\infty}=1 \}}| \sum_{k=1}^{\infty} x_{k} y_{k}|$
equals $\|x\|_{1}=\sum_{k=1}^{\infty} |x_{k}|$. For $\leq \|x\|_{1}$, using [[Hölder's inequality]] $|\sum_{k=1}^{\infty} x_{k}y_{k}| \leq \sum_{k=1}^{\infty} |x_{k}y_{k}| \leq \|x_{1}\| \|y\|_{\infty} \leq \|x\|_{1}.$
For $\|x\|_{1} \leq$, we'll show that for all $\varepsilon>0$ there exists a sequence $(y_{k}) \in c_{0}$, $\|y\|_{\infty}=1$, such that $\|x_{1}\| < \varepsilon + |\sum_{k=1}^{\infty} x_{k}y_{k}|$. Indeed, fix $\varepsilon>0$. Choose $N$ large enough that $\|x\|_{1}-\sum_{k=1}^{N}|x_{k}|<\varepsilon$. Define $y^{(N)} \in c_{0}$ as $y^{(N)}_{k} := \begin{cases}
\text{sgn } x_{k} & k \leq N \\
0 & k > N
\end{cases}. $
Note $\|y^{(N)}\|_{\infty}=1$. Then $|\sum_{k=1}^{\infty} x_{k} y_{k}|=|\sum_{k=1}^{N} \overbrace{ x_{k} y_{k} }^{ = |x_{k}| }=\sum_{k=1}^{N} |x_{k}| .$
The result follows.
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#### References
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