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.
