Adjoint of a bounded linear functional on normed space.

Question

The following is Exercise 3.1 in Megginson's An Introduction to Banach Space Theory:

Let $X$ be a normed space and let $x^* \in X^*$ be a bounded linear functional. Describe the adjoint of $x^*$.

I am not entirely sure what I am supposed to do here.

I know, by definition, that $x^* : X \to \mathbb{F}$ and $(x^*)^* : \mathbb{F} \to X^*$, where it is used that $\mathbb{F}^* \cong \mathbb{F}$. The action of $(x^*)^*$ is given by $\langle x, (x^*)^* \alpha \rangle = \langle x^* x, \alpha \rangle = \alpha x^* x = \langle x, \alpha x^* \rangle$, where $x \in X$ and $\alpha \in \mathbb{F}$. Thus $(x^*)^* \alpha = \alpha x^*$.

Answer 1

When you say $\mathbb F^*\simeq\mathbb F$, there is a duality to be considered (usually, multiplication).

Here $(x^*)^*(\alpha) \in X^*$, so given any $y\in X $, $$ (x^*)^*(\alpha )y=\alpha (x^*(y))=\alpha\,x^*(y). $$ So $(x^*)^*(\alpha)=\alpha x^*$.

Answer 2

Since $\mathbb{F}$ is 1-dimensional over itself, you can describe $(x^*)^*$ by its action on $1 \in \mathbb{F}$. So taking $\alpha = 1$ in your characterisation: $$\langle x, (x^*)^*(1) \rangle = \langle x^* x, 1 \rangle,$$ or in notation I prefer, $(x^*)^*(1)(x) = x^*(x)$. That is, $(x^*)^*(1) = x^*$.