# If $X$ is a Banach space, is $\ell^{p}(X) \cong \ell^{q}(X^{*})$?

Here's the problem that I have: ($$\frac{1}{p}+\frac{1}{q} =1$$)

Let $$(X, ||\cdot||_{X})$$ be a Banach space, and let $$\ell^{p}(X) = \lbrace (x_{n})_{n=1}^{\infty} | \sum_{n=1}^{\infty} ||x_{n}||_{X}^p < +\infty \rbrace$$, for some $$1 \leq p < +\infty$$. Prove that:

1. $$\ell^{p}(X)$$ is a Banach space with respect to the norm $$||(x_{n})_{n=1}^{\infty}||_{p} = \sqrt[p]{\sum_{n=1}^{\infty}||x_{n}||_{X}^{p}}$$;

2. Every functional $$y^{*} \in \ell^{p}(X)$$ can be written as $$y^{*}((x_{n})_{n=1}^{\infty}) = \sum_{n=1}^{\infty}y_{n}^{*}(x_{n})$$, where $$y_{n}^{*} \in X^{*}$$, $$\sum_{n=1}^{\infty}||y_{n}^{*}||_{X^{*}}^{q} < +\infty$$ and $$||y^{*}|| = \sqrt[q]{\sum_{n=1}^{\infty} ||y_{n}^{*}||_{X^{*}}^{q}}$$.

A very similar question has been asked here: Banach valued sequence spaces $\ell^p(X)$. However, it's not enough for me to work out the details of my problem.

Here's what I have so far:

I did not have any trouble with part 1, and I would like to omit my proof here because it's pretty standard, but I will type it out if someone requests it.

For part 2:

Given a functional $$y^{*} \in \ell^{p}(X)^{*}$$, we define, for each $$n \in \mathbb{N}$$, a functional $$y_{n}^{*} \in X^{*}$$, $$y_{n}^{*}(x) = y^{*}(0,0,...,0,x,0,...)$$, where $$x$$ is at the $$n$$-th coordinate. $$y_{n}^{*}$$ are bounded because $$y^{*}$$ is bounded. Furthermore, for $$x = (x_{n})_{n \geq 1}$$, $$y^{*}(x) = y^{*}(\sum_{n=1}^{\infty} (0,...,0,x_{n},0,...))= \sum_{n=1}^{\infty}y^{*}(0,...,0,x_{n},0,...) = \sum_{n=1}^{\infty} y_{n}^{*}(x_{n})$$.

By part 1, and because $$X^{*}$$ is also a Banach space with the operator norm, it follows that $$\ell^{q}(X^{*})$$ is also a Banach space with the corresponding norm. Now, let us define $$Y = (y_{n}^{*})_{n \geq 1}$$. $$||Y||_{q} = \sqrt[q]{\sum_{n=1}^{\infty} ||y_{n}^{*}||^{q}} \in [0, +\infty]$$. I wish to prove that this norm cannot be infinity, and that $$||Y||_{q} = ||y^{*}||$$. By Hölder's inequality, $$|y^{*}(x)| = |\sum_{n=1}^{\infty}y_{n}^{*}(x_{n})| \leq \sum_{n=1}^{\infty}|y_{n}^{*}(x_{n})| \leq \sum_{n=1}^{\infty} ||y_{n}^{*}|| \hspace{1mm} ||x_{n}|| \leq \sqrt[q]{\sum_{n=1}^{\infty} ||y_{n}^{*}||^{q}} \sqrt[p]{\sum_{n=1}^{\infty} ||x_{n}||^{p}} = ||Y||_{q}||x||_{p},$$ so $$||y^{*}|| \leq ||Y||_{q}$$.

However, it seems (to me) pretty difficult to find an $$x$$ for which equality would hold in the previous inequality chain, and even more difficult to find a sequence of $$x$$'s, $$(x^{m})$$ for which I could let $$m \to \infty$$ and achieve equality. Therefore, I'm having trouble proving $$Y \in \ell^{q}(X^{*})$$ and $$||Y||_{q} \leq ||y^{*}||$$. The answer in the linked question says to proceed as in $$(\ell^{p})^{*} \cong \ell^{q}$$, however, the proof that I know uses a very specific construction of a sequence of complex numbers, a luxury that I don't have in this case.

Fix $\varepsilon>0$. For each $n$, there exists $z_n\in X$ with $\|z_n\|=1$ and $|y_n^*(z_n)|\geq\|y_n^*\|-\varepsilon/2^n$. Let $x_n=\alpha_n\|y_n^*\|^{q-1}\,z_n$, where $|\alpha|=1$ and $y_n^*(\alpha_nz_n)=|y_n^*(z_n)|$ . Then, for any $M\in\mathbb N$, $$\sum_{n=1}^M\|x_n\|^p=\sum_{n=1}^M\|y_n^*\|^{p(q-1)}=\sum_{n=1}^M\|y_n^*\|^q.$$

We have \begin{align} \sum_{n=1}^M\|y_n^*\|^q &=\sum_{n=1}^M\|y_n^*\|^{q-1}\|y_n^*\| \leq\sum_{n=1}^M\|y_n^*\|^{q-1}(|y_n^*(z_n)|+\varepsilon/2^n)\\ \ \\ &\leq\varepsilon \|y^*\|^{q-1}+ \sum_{n=1}^My_n^*(x_n)\\ \ \\ &=\varepsilon \|y^*\|^{q-1}+y^*(x_1,\ldots,x_M,0,\ldots)\\ \ \\ &\leq \varepsilon \|y^*\|^{q-1} +\|y^*\|\,\left(\sum_{n=1}^M\|x_n\|^p\right)^{1/p}\\ \ \\ &=\varepsilon \|y^*\|^{q-1}+\|y^*\|\,\left(\sum_{n=1}^M\|y_n^*\|^q\right)^{1/p}. \end{align} As we can do this for all $\varepsilon>0$, we get $$\sum_{n=1}^M\|y_n^*\|^q\leq \|y^*\|\,\left(\sum_{n=1}^M\|y_n^*\|^q\right)^{1/p}.$$ Since $1-1/p=q$, we obtain $$\left(\sum_{n=1}^M\|y_n^*\|^q\right)^{1/q}\leq \|y^*\|.$$ As $M$ is arbitrary, $$\|Y\|_q=\left(\sum_{n=1}^\infty\|y_n^*\|^q\right)^{1/q}\leq\|y^*\|.$$

As a byproduct we also obtain $x=(x_n)\in\ell^p(X)$ and $\|x\|_p^p=\|Y\|_q^q$.

• I haven't proven that $(y_{n}^{*}) \in \ell^{q}(X^{*})$. However, such a $c$ does exist, because for all $n$, $||y_{n}^{*}|| \leq ||y^{*}||$. May 22 '18 at 14:34
• Indeed, I have edited that. May 22 '18 at 14:39
• Can you explain how you get from $\sum\|y_n^*\|^q$ to $\epsilon\|y^*\|^q + \sum y_n^*(x_n)$? Somehow the exponents are not correct.
– daw
May 23 '18 at 6:06
• I understand everything you wrote, thanks for the great answer! However, I still don't understand why $\sum_{n=1}^{\infty} ||y_{n}^{*}||^{q}$ converges. More specifically, I don't understand the conclusion "so $x \in \ell^{p}(X)$...". We don't know that $\sum_{n=1}^{\infty} ||y_{n}^{*}||^{q}$ is finite, so we don't know if $x \in \ell^{p}(X)$ yet either. May 25 '18 at 12:47
• Yes. What needs to be done is, on the estimate, use finite sums. I'll edit the answer in a bit. May 25 '18 at 17:22