Help proving that a Banach space is reflexive I'm having trouble in proving that the following space is reflexive:
$$E = \{ x= (x_n) : x_n \in \mathbb{R}^n \text{ and } \sum \|x_n\|^2_\infty < \infty\}$$
with the norm
$$ \|x\| = (\sum \|x_n\|^2_\infty)^\frac{1}{2}$$
I already tried an analogous of the proof that the $l_p$ spaces are reflexive, and showing that every element of $E^*$ is norm attaining.
I appreciate any help. Thanks.
 A: Consider the space
$$F = \left\lbrace y = (y_n) : y_n \in \mathbb{R}^n, \sum_{n=1}^\infty \lVert y_n\rVert_1^2 < \infty\right\rbrace$$
with the norm
$$\lVert y\rVert_F = \left(\sum_{n=1}^\infty \lVert y_n\rVert_1^2\right)^{\frac12}.$$
Since $(\mathbb{R}^n,\lVert\,\cdot\,\rVert_\infty)$ and $(\mathbb{R}^n,\lVert\,\cdot\,\rVert_1)$ are each isometrically isomorphic to the dual of the other, it is elementary to verify that
$$\begin{align}
\varphi \colon F \to E^\ast;&\quad \varphi(y)(x) = \sum_{n=1}^\infty \langle y_n,x_n\rangle_n,\\
\psi \colon E \to F^\ast;&\quad \psi(x)(y) = \sum_{n=1}^\infty \langle x_n,y_n\rangle_n,
\end{align}$$
where $\langle a,b\rangle_n = \sum_{k=1}^n a_k\cdot b_k$ is the standard inner product, are isometric embeddings.
Once one has shown that they are isomorphisms, one sees by the standard argument that
$$\left(\varphi^{-1}\right)^\ast \circ \psi \colon E \to E^{\ast\ast},$$
which is an isometric isomorphism, is the natural isometric embedding of $E$ into its bidual, so $E$ is reflexive.
It remains to see that $\varphi$ and $\psi$ are surjective. Let $\lambda \in F^\ast$. Restricting $\lambda$ to the subspace with all $y_k = 0$ except for $k = n$ gives a sequence $x = (x_n) \in \prod\limits_{n=1}^\infty \mathbb{R}^n$ with $\lVert x_n\rVert_\infty \leqslant \lVert \lambda\rVert$. Picking an $y_n \in \mathbb{R}^n$ with $\lVert y_n\rVert_1 = \lVert x_n\rVert_\infty$ and $\langle x_n,y_n\rangle_n = \lVert x_n\rVert_\infty^2$ for $1 \leqslant n \leqslant N$, setting $y_n = 0$ for $n > N$ and denoting that element of $F$ by $y^{(N)}$, it follows that
$$\sum_{n=1}^N \lVert x_n\rVert_\infty^2 = \lambda(y^{(N)}) \leqslant \lVert \lambda\rVert \cdot \lVert y^{(N)}\rVert_F = \lVert \lambda\rVert\cdot \left(\sum_{n=1}^N \lVert x_n\rVert_\infty^2\right)^{\frac12},$$
whence
$$\left(\sum_{n=1}^N \lVert x_n\rVert_\infty^2\right)^{\frac12} \leqslant \lVert\lambda\rVert,$$
and taking the limit $N\to\infty$, we see that $x \in E$ and $\lVert x\rVert \leqslant \lVert\lambda\rVert$. The two continuous linear forms $\lambda$ and $\psi(x)$ coincide on the dense subspace of elements of $F$ with only finitely many nonzero components, hence $\lambda = \psi(x)$ and $\psi$ is recognised as surjective, hence an isometric isomorphism. Showing the surjectivity of $\varphi$ is completely analogous.

More generally, let $A \neq \varnothing$ an index set, and $(X_\alpha)_{\alpha \in A}$ a family of normed spaces.
Let for $1 \leqslant p < \infty$ and $X \in \{E,F\}$
$$\ell^p(X,A) := \left\lbrace x \in \prod_{\alpha\in A} X_\alpha : \sum_{\alpha \in A} \lVert x_\alpha\rVert_{X_\alpha}^p < \infty\right\rbrace$$
and
$$\ell^\infty(X,A) := \left\lbrace x \in \prod_{\alpha\in A} X_\alpha : \sup_{\alpha\in A} \lVert x_\alpha\rVert_{X_\alpha} < \infty\right\rbrace.$$
We endow the spaces with the norms
$$\lVert x\rVert_p := \left(\sum_{\alpha\in A} \lVert x_\alpha\rVert_{X_\alpha}^p\right)^{1/p}\quad\text{resp.}\quad \lVert x\rVert_\infty := \sup_{\alpha \in A} \lVert x_\alpha\rVert_{X_\alpha}.$$
An element of $\ell^\infty(X,A)$ can have uncountably many nonzero components, but for $p < \infty$, each element of $\ell^p(X,A)$ has at most countably many nonzero components.
Before investigating the duality properties of these spaces, we need to collect a few basic facts:
Lemma 1: For each $\alpha$ and $1 \leqslant p \leqslant \infty$,
$$\iota_\alpha \colon X_\alpha \to \ell^p(X,A);\quad \pi_\beta(\iota_\alpha(v)) = \begin{cases} v &, \beta = \alpha\\ 0 &, \beta \neq \alpha \end{cases}$$
is an isometric embedding onto a closed subspace of $\ell^p(X,A)$.
Proof: $\lVert \iota_\alpha(v)\rVert_p = \left(\sum\limits_{\beta\in A}
\lVert \pi_\beta(\iota_\alpha(v))\rVert_{X_\beta}^p\right)^{1/p} =
\left(\lVert v\rVert_{X_\alpha}^p\right)^{1/p} = \lVert v\rVert_{X_\alpha}$,
so $\iota_\alpha$ is an isometry. The image $\mathcal{R}(\iota_\alpha) =
\left\lbrace x \in \ell^p(X,A) : (\beta \neq \alpha) \Rightarrow (\pi_\beta(x)
= 0)\right\rbrace$ is closed,
since the projections are continuous ($\lVert \pi_\beta(x)\rVert_{X_\beta} \leqslant \lVert x\rVert_p$).
Lemma 2: $\ell^p(X,A)$ is a Banach space if and only if each of the $X_\alpha$ is a Banach space.
Proof: If $\ell^p(X,A)$ is a Banach space, lemma $1$ tells us that each
$X_\alpha$
is a Banach space, since it is isometric to a closed subspace of
$\ell^p(X,A)$.
Conversely, let all $X_\alpha$ be Banach spaces, and $(x^{(n)})$ a Cauchy
sequence in $\ell^p(X,A)$. Then $\pi_\alpha(x^{(n)})$ is a Cauchy sequence
in $X_\alpha$, denote its limit by $x_\alpha$.
Let $x = (x_\alpha) \in \prod\limits_{\alpha\in A}X_\alpha$. We need to show
that $x \in \ell^p(X,A)$, and that $x = \lim\limits_{n\to\infty} x^{(n)}$.
Let $M = \sup_n \lVert x^{(n)}\rVert_p$. If $p = \infty$, we have
$\lVert x_\alpha\rVert_{X_\alpha} \leqslant M$ for all $\alpha \in A$,
hence $x \in \ell^\infty(X,A)$, and for $\varepsilon > 0$, let $n_\varepsilon$
such that $\lVert x^{(n)} - x^{(m)}\rVert_\infty \leqslant \varepsilon$ for
$m,n \geqslant n_\varepsilon$. Then in particular
$\lVert \pi_\alpha(x^{(n)} - x^{(m)})\rVert_{X_\alpha} \leqslant \varepsilon$
for
every $\alpha$, and letting $m\to\infty$ we see
$\lVert \pi_\alpha(x^{(n)}) - x_\alpha\rVert_{X_\alpha} \leqslant
\varepsilon$,
so $\lVert x - x^{(n)}\rVert_\infty \leqslant \varepsilon$ for
$n \geqslant n_\varepsilon$, hence $x^{(n)}\to x$ in $\ell^\infty(X,A)$.
If $p < \infty$, we have
$$\sum_{\alpha \in F} \lVert x_\alpha\rVert_{X_\alpha}^p \leqslant M^p$$
for every finite $F \subset A$ by the componentwise convergence, and taking
the supremum over all finite subsets $F \subset A$, we have
$$\sum_{\alpha \in A} \lVert x_\alpha\rVert_{X_\alpha}^p \leqslant M^p,$$
hence $x \in \ell^p(X,A)$. Once again, for $\varepsilon > 0$ let
$n_\varepsilon$ such that $\lVert x^{(n)} - x^{(m)}\rVert_p \leqslant
\varepsilon$
for $m,n \geqslant n_\varepsilon$. Then for every finite $F \subset A$,
we have
$$\sum_{\alpha \in F} \lVert \pi_\alpha(x^{(n)}) - x_\alpha\rVert_{X_\alpha}^p
= \lim_{m\to\infty} \sum_{\alpha\in F} \lVert
\pi_\alpha(x^{(n)}-x^{(m)})\rVert_{X_\alpha}^p \leqslant \varepsilon^p,$$
so $\lVert x^{(n)}-x\rVert_p \leqslant \varepsilon$ and $x^{(n)}\to x$ in $\ell^p(X,A)$.
Now let $(E_\alpha)_{\alpha\in A};\; (F_\alpha)_{\alpha\in A}$ be two families of normed spaces, such that for every $\alpha \in A$ we have an isometric embedding $\delta_\alpha \colon F_\alpha \hookrightarrow E_\alpha^\ast$. Further, let $p,q\in [1,\infty]$ be conjugate exponents, i.e. $\frac1p + \frac1q = 1$.
Proposition: We have an isometric embedding
$$\varphi \colon \ell^q(F,A) \hookrightarrow \ell^p(E,A)^\ast;\quad \varphi(f) \colon e \mapsto \sum_{\alpha\in A} \langle\delta_\alpha(f_\alpha),e_\alpha\rangle\tag{1}$$ (where $\langle\,\cdot\,,\,\cdot\,\rangle$ denotes the natural pairing of $E_\alpha^\ast$ and $E_\alpha$).
Proof: Since $\lvert \langle \delta_\alpha(f_\alpha),e_\alpha\rangle \rvert \leqslant \lVert \delta_\alpha(f_\alpha)\rVert_{E_\alpha^\ast}\cdot \lVert e_\alpha\rVert_{E_\alpha} = \lVert f_\alpha\rVert_{F_\alpha}\cdot \lVert e_\alpha\rVert_{E_\alpha}$, Hölder's inequality assures the absolute convergence of the sum in $(1)$, and
$$\lVert \varphi(f)\rVert_{\ell^p(E,A)^\ast} \leqslant \lVert f\rVert_q.$$
Thus $\varphi$ is a continuous linear map of norm $\leqslant 1$. To see that
it is an isometry, consider $0 \neq f \in \ell^q(F,A)$ and let
$0 < \varepsilon < 1$. If $q = \infty$, choose an $\alpha \in A$ with
$\lVert f_\alpha \rVert_{F_\alpha} > (1-\varepsilon)\lVert f\rVert_\infty$,
and an $e_\alpha \in E_\alpha$ with $\lVert e_\alpha\rVert_{E_\alpha} = 1$
such that $\lvert \langle \delta_\alpha(f_\alpha),e_\alpha\rangle \rvert >
(1-\varepsilon)\lVert f_\alpha\rVert_{F_\alpha}$.
Then $$\lvert \varphi(f)(\iota_\alpha(e_\alpha))\rvert = \lvert \langle
\delta_\alpha(f_\alpha),e_\alpha\rangle\rvert > (1-\varepsilon)\lVert
f_\alpha\rVert_{F_\alpha} > (1-\varepsilon)^2 \lVert f\rVert_\infty \cdot
\lVert \iota_\alpha(e_\alpha)\rVert_1,$$
whence $\lVert \varphi(f)\rVert_{\ell^1(E,A)^\ast} > (1-\varepsilon)^2\lVert
f\rVert_\infty$.
As $\varepsilon$ can be chosen arbitrarily small, $\varphi$ is an isometry
in this case. If $q < \infty$, let $A_1 = \{ \alpha \in A : f_\alpha \neq
0\}$.
For $\alpha \in A_1$, choose $e_\alpha \in E_\alpha$ with
$\lVert e_\alpha \rVert_{E_\alpha} = \lVert f_\alpha\rVert_{F_\alpha}^{q-1}$ and
$\langle \delta_\alpha(f_\alpha),e_\alpha\rangle > (1-\varepsilon) \lVert
f_\alpha\rVert_{F_\alpha}^q$,
and for $\alpha \in A\setminus A_1$ let $e_\alpha = 0$. Thus an element
$e$ of $\ell^p(E,A)$ is defined. For $q = 1$, we have
$\lVert e\rVert_\infty = 1$, and for $1 < q < \infty$,
$$\lVert e\rVert_p = \left(\sum_{\alpha \in A} \lVert e_\alpha\rVert_{E_\alpha}^p\right)^{1/p} = \left( \sum_{\alpha\in A_1} \lVert f_\alpha\rVert_{F_\alpha}^{p(q-1)}\right)^{1/p} = \lVert f\rVert_q^{q/p} = \lVert f\rVert_q^{q-1}.$$
Further $$\varphi(f)(e) = \sum_{\alpha \in A_1} \langle \delta_\alpha(f_\alpha),e_\alpha\rangle > \sum_{\alpha\in A_1} (1-\varepsilon)\lVert f_\alpha\rVert_{F_\alpha}^q = (1-\varepsilon)\lVert f\rVert_q \cdot \lVert e\rVert_p,$$
so $\lVert \varphi(f)\rVert_{\ell^p(E,A)^\ast} > (1-\varepsilon)\lVert f\rVert_q$, and since $\varepsilon$ can be chosen arbitrarily small, $\varphi$ is seen to be an isometry also in this case.
The main result is the characterisation of the situations when $\varphi$ is surjective, hence an isometric isomorphism.
Theorem: The isometric embedding $\varphi$ is surjective if and only if


*

*all $\delta_\alpha$ are surjective, and

*$p < \infty$ or $\bigl\lbrace\alpha \in A : E_\alpha \neq \{0\}\bigr\rbrace$ is finite.


Proof: If for some $\alpha$ there is a $\lambda_\alpha \notin \mathcal{R}(\delta_\alpha)$, then $\lambda_\alpha \circ \pi_\alpha \notin \mathcal{R}(\varphi)$.
For if $\lambda_\alpha\circ\pi_\alpha = \varphi(f)$, then for all $e_\alpha \in E_\alpha$
$$\lambda_\alpha(e_\alpha) = \varphi(f)( \iota_\alpha(e_\alpha)) = \langle\delta_\alpha(f_\alpha),e_\alpha\rangle,$$
so $\lambda_\alpha = \delta_\alpha(f_\alpha) \in \mathcal{R}(\delta_\alpha)$.
Thus the surjectivity of all $\delta_\alpha$ is necessary for the surjectivity of $\varphi$.
Also, if $p = \infty$ and infinitely many $E_\alpha$ are nontrivial,
choose countably many such, call them $E_1,\, E_2,\, \dotsc$, and for
each $n$, a $\lambda_n \in E_n^\ast$ with $\lVert \lambda_n\rVert_{E_n^\ast} =
1$.
Let $C$ be the subspace of $e \in\ell^\infty(E,A)$ such that
$\Lambda_0(e) := \lim\limits_{n\to\infty} \lambda_n(e_n)$ exists.
Since $\Lambda_0 \in C^\ast$, the Hahn-Banach theorems assert the existence
of a continuous extension $\Lambda \in \ell^\infty(E,A)^\ast$. Such a
$\Lambda$ is not in the range of $\varphi$: For all $n$, choose $e_n \in E_n$
with $\lVert e_n\rVert_{E_n} \leqslant 1+2^{-n}$ and $\lambda_n(e_n) = 1$.
For $k \geqslant 1$, let $\eta_k = \sum\limits_{n=k}^\infty \iota_n(e_n)$.
Then $\Lambda(\eta_k) = 1$ for all $k$, and if we had $\Lambda = \varphi(f)$,
then $$0 = \Lambda(\eta_k-\eta_{k+1}) = \Lambda(\iota_k(e_k)) = \langle\delta_k(f_k),e_k\rangle,$$ hence $$0 = \sum_{n=1}^\infty \langle \delta_k(f_k),e_k\rangle = \varphi(f)(\eta_1) \neq \Lambda(\eta_1) = 1.$$
If $p = \infty$ and only finitely many $E_\alpha$ are nontrivial, without loss
of generality suppose $A$ finite and all $E_\alpha$ nontrivial. If all
$\delta_\alpha$ are surjective, for $\lambda \in \ell^\infty(E,A)$ consider
$$f = \sum_{\alpha \in A} \iota_\alpha^F \left(\delta_\alpha^{-1}(\lambda\circ
\iota_\alpha^E)\right) \in \ell^1(F,A).$$
We claim that $\lambda = \varphi(f)$:
$$\lambda(e) = \lambda\left(\sum_{\alpha \in A} \iota_\alpha^E(e_\alpha)\right) = \sum_{\alpha\in A} (\lambda\circ \iota_\alpha^E)(e_\alpha) = \sum_{\alpha\in A} \langle\delta_\alpha\left(\pi_\alpha(f) \right),e_\alpha\rangle = \varphi(f)(e).$$
Suppose now that $p < \infty$ and all $\delta_\alpha$ are surjective. Let $0 \neq\lambda \in \ell^p(E,A)^\ast$, and
$$f = (\delta_\alpha^{-1}(\lambda\circ \iota_\alpha))_{\alpha\in A} \in \prod_{\alpha\in A} F_\alpha.$$
For each $\alpha$ choose $e_\alpha$ with
$\lVert e_\alpha\rVert_{E_\alpha} = \lVert f_\alpha\rVert^{q-1}$
and $\langle \delta_\alpha(f_\alpha), e_\alpha\rangle \geqslant (1-\varepsilon) \lVert f_\alpha\rVert_{F_\alpha}^q$. If $B \subset A$ is finite, then
$$\begin{align}
\sum_{\alpha\in B} \lVert f_\alpha\rVert_{F_\alpha}^q &\leqslant \frac{1}{1-\varepsilon} \sum_{\alpha\in B} \langle \delta_\alpha(f_\alpha), e_\alpha\rangle\\
&= \frac{1}{1-\varepsilon}\sum_{\alpha\in B} (\lambda\circ \iota_\alpha)(e_\alpha)\\
&= \frac{1}{1-\varepsilon} \lambda\left( \sum_{\alpha\in B} \iota_\alpha(e_\alpha)\right)\\
&\leqslant \frac{1}{1-\varepsilon} \lVert\lambda\rVert\cdot \left\lVert \sum_{\alpha\in B} \iota_\alpha (e_\alpha)\right\rVert\\
&= \frac{1}{1-\varepsilon} \lVert\lambda\rVert\cdot \left(\sum_{\alpha\in B} \lVert f_\alpha\rVert_{F_\alpha}^{q}\right)^{1/p},
\end{align}$$
hence
$$\sum_{\alpha\in B} \lVert f_\alpha\rVert_{F_\alpha}^q \leqslant \frac{1}{(1-\varepsilon)^q} \lVert \lambda\rVert^q.$$
Thus $f \in \ell^q(F,A)$, and we have $\lambda = \varphi(f)$:
$$\begin{align}
\lambda(e) &= \lambda\left(\sum_{\alpha\in A} \iota_\alpha(e_\alpha)\right)\\
&= \sum_{\alpha\in A} \lambda(\iota_\alpha(e_\alpha))\\
&= \sum_{\alpha\in A} \langle \delta_\alpha(f_\alpha), e_\alpha\rangle\\
&= \varphi(f)(e).
\end{align}$$
An easy consequence is the
Corollary: $\ell^p(E,A)$ is reflexive if and only if


*

*all $E_\alpha$ are reflexive, and

*$1 < p < \infty$ or all but finitely many $E_\alpha$ are trivial.


Proof: If $\ell^p(E,A)$ is reflexive, the $E_\alpha$ must be reflexive, since they
are isometrically isomorphic to closed subspaces of $\ell^p(E,A)$.
If $p \in \{1,\infty\}$ and infinitely many $E_\alpha$ are nontrivial,
$\ell^p(E,A)$ contains an isometric copy of $\ell^1$ resp. $\ell^\infty$ and can
therefore not be reflexive.
If all $E_\alpha$ are reflexive and $1 < p < \infty$ or only finitely many
$E_\alpha$ are nontrivial, by the theorem we have isometric isomorphisms
$$\varphi_1 \colon \ell^p(E,A) \xrightarrow{\cong} \left(\ell^q(E^\ast,A)\right)^\ast;\quad
\varphi_2 \colon \ell^q(E^\ast,A) \xrightarrow{\cong} \left(\ell^p(E,A)\right)^\ast.$$
Then the transpose of $\varphi_2$ is an isometric isomorphism
$\varphi_2^\ast \colon \left(\ell^p(E,A)\right)^{\ast\ast} \to \left(\ell^q(E^\ast,A)\right)^\ast$ and with the natural embedding
$J\colon \ell^p(E,A) \hookrightarrow \left(\ell^p(E,A)\right)^{\ast\ast}$ we have
$\varphi_1 = \varphi_2^\ast\circ J$: For all $e\in \ell^p(E,A)$ and $\lambda \in \ell^q(E^\ast,A)$
$$\begin{align}
(\varphi_2^\ast \circ J)(e)(\lambda) &= J(e)(\varphi_2(\lambda))\\
&= \varphi_2(\lambda)(e)\\
&= \sum_{\alpha\in A} \langle \lambda_\alpha,e_\alpha\rangle\\
&= \sum_{\alpha\in A} \langle J_\alpha(e_\alpha), \lambda_\alpha\rangle\\
&= \varphi_1(e)(\lambda).
\end{align}$$
Thus $J = (\varphi_2^\ast)^{-1} \circ \varphi_1$ is an isomorphism, i.e. $\ell^p(E,A)$ is reflexive.
