# Reflexivity of $\ell^p$

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!

• You really do want to use the definition of the double dual as an iterated dual. Then if you can show that $\ell^q=(\ell^p)^*$, it follows by dualizing $\ell^q$. Can you prove the $\ell^p=(\ell^q)^*$? – Adam Hughes Jul 17 '14 at 0:09
• Yes I can prove that $l^p$ and $(l^q)^\ast$ are isomorphic. The isomorphism is $j_q$. I just don't know how to use this isomorphism. – Benzio Jul 17 '14 at 0:16
• By definition $X^{**}$ is the dual space to $X^*$. Since $(\ell^p)^*=\ell^q$, by definition $(\ell^{p})^{**}=(\ell^{q})^{*}=\ell^p$ – Adam Hughes Jul 17 '14 at 0:17
• Yes, this is readily seen! My problem is to show that this equality is the same of the surjectivity of the canonical application $J_{l^p}$! – Benzio Jul 17 '14 at 0:20
• Benzio: in that case it's just a matter of writing it as a composition. See my answer below. – Adam Hughes Jul 17 '14 at 0:23

## 2 Answers

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 $f_p\circ f_q=j_p:\ell^p\to (\ell^p)^{**}$.

• shouldn't $j_p$ go from $l^q$ to $(l^p)^\ast$ and viceversa for $j_q$? Why you are saying $j_p: l^p \rightarrow l^q$? What operator do you intend with $j_p$? – Benzio Jul 17 '14 at 0:32
• @Benzio I got notation mixed up, I'll switch it. I guess that's what they mean when they say "mind your p's and q's." In any case your notation isn't standard, so really I could define my $j$ functions to be whatever. ;-) – Adam Hughes Jul 17 '14 at 0:34
• no problem for the different notation, but the isomorphisms I have are from $l^p$ to $(l^q)^\ast$, not to $(l^p)^\ast$. Or maybe you are not intending the same thing as mine. – Benzio Jul 17 '14 at 0:38
• @Benzio OHHHHH, I see. Yes, your notation confused me with all the different $J$s. One more edit, I'll just use new names, it's easier that way. – Adam Hughes Jul 17 '14 at 0:39
• @Benzio I've tried to clarify the language. See if it makes more sense now. – Adam Hughes Jul 17 '14 at 0:56

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