Case:
for which I was studying first the case with $X,Y$ as Hilbert spaces but then later with $X,Y$ as Banach spaces. I am interested on the rigorous proof. I think the linearity and continuity should be also proven for the equation.
My Proposal
Let $X,Y$ are Banach spaces, and let $A \in \mathscr{B}(X,Y)$ be fixed. For any $u \in X$ and $\psi \in Y'$ so $Au \in Y$: \begin{align*} \langle Au, \psi \rangle &= \psi(Au) \\ &= (\psi \circ A)(u) \\ &= \langle u,(\psi \circ A) \rangle \\ &= \langle u, \psi' \rangle, \end{align*} so $\psi' = \psi \circ A$. Because $\langle u, \psi \circ A \rangle$, $u \in X$ and $\psi \circ A \in X'$. So $(\psi \circ A)$ is the wanted mapping $A' : Y' \to X'$. So $\psi \in Y'$ and $A'(\psi) = (\psi \circ A)$. Then for any $u \in X$ holds: \begin{align*} \langle u, A'\psi \rangle (\in \mathbb{K}) &= \langle u, \psi \circ A \rangle \\ &= (\psi \circ A)(u) \\ &= \psi(Au) \\ &= \langle Au, \psi \rangle \in \mathbb{K} \end{align*} and because $u \in X$ and $(\psi \circ A) = A' \in X'$. Also $A': Y' \to X'$ because selecting any $\psi \in Y'$ gets $A'(\psi) = (\psi \circ A) \in X'$. So $A': \psi (\in Y') \to \psi \circ A (\in X')$ and $A' \in \mathscr{B}(Y',X')$.
We want to show that there exists a unique operator $A' \in \mathscr{B}(Y', X')$ that satisfies \begin{equation*} \forall x \in X, \, \forall \varphi \in Y', \, \langle Ax, \varphi \rangle = \langle x, A' \varphi \rangle. \end{equation*}
Fix $\varphi' \in Y'$. Let the adjoint map $A' \varphi : X \to \mathbb{F}$ such that $\langle x, A' \varphi \rangle = \langle A x, \varphi \rangle$ for $x \in X$.
The linearity and continuity. Let $\varphi_{1}, \varphi_{2}, \varphi \in Y$, $x_{1}, x_{2}, x \in X$, and $\lambda \in \mathbb{K}$: \begin{align*} \langle A' (\varphi_{1} + \varphi_{2}), x \rangle &= \langle \varphi_{1} + \varphi_{2}, Ax \rangle \\ &= \langle A' \varphi_{1} + A' \varphi_{2}, x \rangle \\ \end{align*} \begin{align*} \langle A'(\lambda \varphi), x \rangle &= \langle \lambda \varphi, Ax \rangle \\ &= \bar{\lambda} \langle \varphi, Ax \rangle \\ &= \langle \lambda A' \varphi, x \rangle \end{align*}
Let's prove that $\|A'\| = \|A\|$. If $Y = \{ 0 \}$, then $A = 0$ and $A' = 0$. Therefore, assume $Y \neq \{ 0 \}$. By Hahn-Banach, also $Y' \neq \{ 0 \}$. Choose any $x \in X$ with $\| x \| = 1$. Suppose that $Ax \neq 0$. Let $\varphi \in Y'$ such that \begin{equation*} \langle Ax, \varphi \rangle = \| Ax \|, \, \| \varphi \| = 1. \end{equation*} If $Ax = 0$, then $\varphi \in Y'$ such that $A'\varphi = 0$. Then, \begin{align*} \| Ax \| &= \langle x, A' y \rangle \\ &\leq \| A' y' \| \| x \| \\ &\leq \|A'\| \| x \| \\ &= \|A'\| \end{align*} So $\| Ax \| \leq \| A' \|$ and $\|A\| \leq \|A'\|$. Therefore, $\| A \| = \| A' \|$. $\square$
Sources
- Functional Analysis An introduction to Metric Spaces, Hilbert Spaces, and Banach Algebras by J. Muscat, 2014, p. 188, definition 10.19.
- Functional Analysis Lecture Notes: Adjoints in Banach Spaces, C. Heil, filename adjoint_banach.pdf
- Chapter 3: Duality in Banach Space, www4.ncsu.edu/~mtchu/Teaching/Lectures/MA719/chapter3.ps
How can you show the adjoint of an operator A on Banach spaces?