Reflexivity of $\ell^p$

Question

I'm having bad difficulties in understanding how to prove that $\ell^p$ with $1<p<\infty$ are reflexive spaces. Every text I have consulted give that as a trivial result because "observing that $(\ell^p)^{\ast\ast} = ((\ell^p)^\ast)^\ast$, $\ell^q$ is isomorphic to $(\ell^p)^\ast$ and $\ell^p$ is isomorphic to $(\ell^q)^\ast$ then $(\ell^p)^{\ast\ast}$is isomorphic to $\ell^p$ and the result follows trivially".

Maybe trivially for you, books!! I want to show formally that the canonical application $J_{\ell^p}:\ell^p \rightarrow (\ell^p)^{\ast\ast}$ is surjective. I tried for hours but nothing, I'm blocked.

Let be $j_p: \ell^q \rightarrow (\ell^p)^\ast$ and $j_q: \ell^p \rightarrow (\ell^q)^\ast$ the isomorphisms that I have.

Let be $z \in (\ell^p)^{\ast\ast}$. I want to find an $x \in \ell^p$ such that $J_{\ell^p}(x)=z$, i.e. $\langle z,x'\rangle=\langle x',x\rangle$ for every $x' \in (\ell^p)^\ast$.

Probabily there will be many "$j_p, j_q, j_p^{-1}, j_q^{-1},J_{\ell^p}$" but I have no idea about how to choose them. Some help is greatly appreciated! Thank you!

Answer 1

This is the answer I've unravelled:

Let $z \in (\ell^p)^{\ast\ast}$. I want to prove that exists $x \in \ell^p$ such that $\langle z,f\rangle=\langle f,x \rangle$ for every $f \in (\ell^p)^\ast$.

I know that there are the isomorphisms:

$j_p: \ell^q \rightarrow (\ell^p)^\ast$ and $j_q: \ell^p \rightarrow (\ell^q)^\ast$

Now, fix $z \in (\ell^p)^{\ast\ast}$.

$\langle z, f\rangle = \langle z, j_p(y)\rangle$ for some $y$ in $\ell^q$.

I define $g(y)=\langle z, j_p(y)\rangle$. It's seen that $g$ is an element of $(\ell^q)^\ast$ then

$\langle z, f\rangle = \langle z, j_p(y)\rangle = \langle g, y \rangle$

Being an element of $(\ell^q)^\ast$, the number $\langle g, y \rangle$ can be represented as $\sum_{k=1}^\infty x_ky_k$ for some $x \in \ell^p$

Now fix that $x$. This sum, $\sum_{k=1}^\infty x_ky_k$ can be seen as a functional $u$ on $\ell^p$ described by $j_p(y)$.

Then $u=f$ and $\langle z, f\rangle = \langle z, j_p(y)\rangle = \langle g, y \rangle = \sum_{k=1}^\infty x_ky_k = \langle f,x \rangle$ as I wanted.

${}$

Now the answer seems correct to me, but still a bit confused because I would like to avoid to mix the $\langle \cdot, \cdot \rangle$ with the sum notation. And some suggestions to avoid that are appreciated.

After I would like to understand why the answer provided by Adam Hughes is rigorous: why does he claims that his $f_p: (\ell^q)^*\to (\ell^p)^{**}$ is the equality isomorphism? Indeed it's needed the isomorphism $j_p: \ell^q \rightarrow (\ell^p)^\ast$ to show that it is the equality isomorphism, but how? And then, acknowledged that his $f_p\circ f_q$ is an isomorphism from $l^p$ to $(l^p)^{\ast\ast}$, why it is said that it's equal to the canonical isomorphism?

I'm slow in understanding... I know! :-p

Answer 2

So you have the canonical maps:

$$\begin{cases} f_q: \ell^p\to (\ell^q)^* \\ f_p: (\ell^q)^*\to (\ell^p)^{**}\end{cases}$$

which are the isomorphisms between $\ell^p$ and $(\ell^q)^*$ and $\ell^q$ and $(\ell^p)^*$ respectively. Then just write $f_p\circ f_q=j_p:\ell^p\to (\ell^p)^{**}$.

Answer 3

This can be done with adjoints and I think it is more enlightening to do it this way.

Lemma: If $T:X\rightarrow Y$ is a surjective isometry between Banach spaces, then $T^*: Y^* \rightarrow X^*$ is also a surjective isometry.

Proof: We will only need to use that $T^*$ is surjective, but this is immediate: $T^*(x^* \circ T^{-1})=x^*$. I will prove that it preserves the norm because it is also straightforward, but we will not need this:

$$\lVert T^*(y^*)\rVert=\sup_{x\in B_X} \lVert y^*(T(x))\rVert=\sup_{y\in B_{Y}}\lVert y^*(y)\rVert=\lVert y^*\rVert\quad \quad\square$$

Now let us continue with our quest of proving reflexivity of $\ell^p$ for $1<p<\infty$.

If we consider the canonical surjective isometry $\phi: (\ell^p)^*\rightarrow l^q$, then its adjoint will satisfy:

$$\phi^*:(\ell^q)^*\rightarrow (\ell^p)^{**}$$

Next we consider $\psi: \ell^p \rightarrow (\ell^q)^*$ the canonical isometry. By using these maps we gain with our Lemma that composition is surjective:

$$\phi^*\circ \psi: \ell^p\rightarrow(\ell^p)^{**}$$

Furthermore let us denote $q$ and element of $\ell^q$ and $p$ an element of $\ell^p$ and $f_p$ the dual version of $p$ in $(\ell^q)^*$ and $f_q$ the dual version of $q$ in $(\ell^p)^*$:

$$\phi^* \circ \psi(p)(f_q)=\phi^*(f_p)(f_q)=f_p(\phi(f_q))=f_p (q)=\sum p_i q_i=f_q(p)=J_{\ell^p}(p)(f_q)$$

Hence, $J_{\ell^p}=\phi^* \circ \psi$ which we know to be surjective!