why isomorphism preserves the function values. It is known that every functional $g \in l_{q}^{*}(\mathbb{K})$ can be identified
with functional $\varphi_{g} \in l_{p}^{**}(\mathbb{K})$ and that every
element $y \in l_{q}(\mathbb{K})$ can be identified with a functional $f_{y} \in l_{p}^{*}(\mathbb{K})$.
So we have that $g \simeq \varphi_{g}$ and $y \simeq f_{y}$.
I would like to ensure, reasoning in a symbolic fashion that
$g(y)$ is numerically equals that $\varphi_{g}(f_{y})$ for every $y \in l_{q}$, that is to say,
I need to ensure that
$$
  g(y) = \varphi_{g}(f_{y}), \quad \forall \, y \in l_{q} (\mathbb{K}).
$$
This is very important to show that $l_{p}(\mathbb{K})$ is reflexive, because to do it
I need to show that the function $\text{eval}: l_{p} \to l_{p}^{**}$ is surjective. So, given
a $\varphi \in l_{p}^{**}$ there is function $g \in l_{q}^{*}$ such that we can identify $g \simeq \varphi$ and we can write $\varphi = \varphi_{g}$. By Riesz Representation Theorem, there exists
$a = (a_{k})_{k \in \mathbb{N}} \in l_{p}(\mathbb{K})$ such that
$$
   g(y) = \sum_{k = 1}^{\infty} a_{k} y_{k}, \quad \forall \, y = (y_{k}) \in l_{q}(\mathbb{K}).
$$
Again by Riesz Representation Theorem, $y \in l_{q}$ can be identified with $f_{y} \in l_{p}^{*}(\mathbb{K})$ and
$$
  g(y) = \sum_{k = 1}^{k} a_{k} y_{k} = f_{y}(a) = \text{eval}(a) (f_{y})
$$
The only thing that I need to be clear is that $g(y) = \varphi_{g}(f_{y})$. With this clear I can stablish
the equality
$$
  \varphi_{g}(f_{y}) = \text{eval}(a) (f_{y}),\quad \forall f_{y} \in l_{p}^{*},
$$
so $\varphi_{g} = \text{eval}(a)$ and then $\text{eval}$ is surjective.
Please help me to understand the mentioned point.
 A: Let me introduce the following notation. Let $T_p\colon \ell_p\to\ell_q^*$, $1<p<\infty$, $p^{-1}+q^{-1}=1$, be the mapping given by the formula $$(T_p(a))(b)=\sum_{i=1}^\infty a_ib_i\text{ for }a\in\ell_p,\ b\in\ell_q.$$ We know it's an isomorphism and isometry, that is $$\|T_p(a)\|_{\ell_q^*}=\|a\|_p.$$ We see that
$$(T_p(a))(b)=(T_q(b))(a)\text{ for }a\in\ell_p,\ b\in\ell_q.\tag{1}$$
Let $T_q^*\colon \ell_p^{**}\to \ell_q^*$ be the dual operator to $T_q\colon \ell_q\to\ell_p^*$, that is $$T_q^*(\varphi)=\varphi\circ T_q\text{ for } \varphi\in \ell_p^{**}.\tag{2}$$
It's also an isomorphism.
Moreover let $\mathrm{eval}_p\colon \ell_p\to\ell_p^{**}$ be given by
$$\mathrm{eval}_p(a)(g)=g(a) \text{ for }a\in\ell_p,\ g\in \ell_p^*.\tag{3}$$
Our goal is to show that
$$\mathrm{eval}_p =  (T_q^*)^{-1}\circ T_p.$$ This is equivallent to
$$T_q^*\circ\mathrm{eval}_p =  T_p.$$
For $a\in\ell_p$ we have
$$(T_q^*\circ\mathrm{eval}_p)(a) = 
T_q^*(\mathrm{eval}_p(a)) \stackrel{(2)}= 
\mathrm{eval}_p(a)\circ T_q.
$$
Therefore, for $b\in\ell_q$ we have
$$((T_q^*\circ\mathrm{eval}_p)(a))(b) = 
(\mathrm{eval}_p(a))(T_q(b)) \stackrel{(3)}= 
(T_q(b))(a) \stackrel{(1)}= 
(T_p(a))(b).
$$
Since $b$ was arbitrary, we get
$$(T_q^*\circ\mathrm{eval}_p)(a) = 
T_p(a).
$$
Now, since $a$ was arbitrary, we get
$$T_q^*\circ\mathrm{eval}_p = 
T_p,
$$
which ends the proof.
