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:

*

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


*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.
 A: 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 $\|x\|_p^p=\|Y\|_q^q$. 
