I have to show that for all $1<p<\infty$ the space $\ell^p$ is reflexive using the adjoint operator.
Now, we say that the application $T\colon (\ell^p)'\to \ell^q$ defined as $$T(f(x))=(f(e_1),f(e_2),\dots)\quad\text{where}\quad f(x)=\sum_{k=1}^\infty\xi_kf(e_k)\quad \forall x=(\xi_k)\in \ell^p$$ is an isometric isomorphism. We remember that $e_n=(\delta_{nj})_j$.
Since also $1<q <\infty$ we have that $S\colon (\ell^q)'\to \ell^p$ defined as above is an isometric isomorphism. We consider the following composition $$\ell^p\color{RED}{\to}(\ell^q)'\color{BLUE}{\to}(\ell^p)'',$$ where $\color{RED}{S^{-1}\colon\ell^p\to (\ell^q)'}$ defined as $$(S^{-1}b)(x)=\sum_{k=1}^\infty\xi_k\beta_k\quad\forall x=(\xi_k)\in\ell^q\qquad (b=(\beta_k)\in\ell^p)$$ and $\color{BLUE}{{T}^\times\colon (\ell^q)'\to (\ell^p)^{''}}$ is the adjoint operator of $T$ that is defined as $$(T^\times g)(x)=g(Tx)\quad (g\in (\ell^q)', x\in \ell^p).$$
Now, we consider the natural embedding $J\colon \ell^p\to (\ell^p)^{''}$ defined as $$J(x)(f)=f(x)\quad x\in\ell^p, f\in(\ell^p)'.$$
I have tried so many times to show that
$J= T^\times\circ S^{-1}$
But I can't. I don't know if I'm on the right track. Could anyone give me some hints? Thank you.