So, I'm in a quest of understanding classical field theory on my own, and I'm interested in its rigorous construction. Here's the link for a previous post of mine on mathoverflow. The interesting discussions there led me to this new post here, which I will use some of the notations I used there.
Notation: If ${\bf{x}} = (x_{1},...,x_{n}) \in \mathbb{R}^{n}$ and $f=f({\bf{x}})$ is real-valued and differentiable, I'll denote: $$\frac{\partial f}{\partial \bf{x}} := \bigg{(}\frac{\partial f}{\partial x_{1}},...,\frac{\partial f}{\partial x_{n}}\bigg{)} \equiv \nabla f.$$
This notation is useful since, if $f$ is a function of more than one variable, e.g. $f=f(\bf{x},\bf{y},\bf{z})$, then $\partial f/{\partial \bf{x}}$ means the gradient with respect to the $\bf{x}$ variable.
Legendre Transforms for many variable functions
Here, I'm following Arnold. Let $f: \mathbb{R}^{n}\to \mathbb{R}$ be a twice-differentiable function such that its Hessian $\nabla^{2}f$ is positive-definite (so $f$ is strictly convex). Let $G=G({\bf{p}},{\bf{x}}) := \langle {\bf{p}},{\bf{x}}\rangle - f({\bf{x}})$, where $\langle \cdot, \cdot \rangle$ is the usual inner product on $\mathbb{R}^{n}$. Then, the Legendre transform of $f$ is defined to be the function $g=g({\bf{p}}) := \max_{{\bf{x}}}G({\bf{p}},{\bf{x}})$. Notice that $G$ attains its maximum iff $\frac{\partial G}{\partial \bf{x}} = 0$, so that the vector $\bf{x}$ which maximizes $G$ for a fixed $\bf{p}$ is the solution of: \begin{eqnarray} \frac{\partial f}{\partial \bf{x}} = \bf{p} \tag{1}\label{1} \end{eqnarray}
Let $L=L(t,{\bf{x}},\dot{{\bf{x}}})$ be a Lagragian on the phase space as studied in classical mechanics. Because the Hamiltonian $H=H(t,{\bf{x}},{\bf{p}})$ is the Legendre transform of $L$, equation (\ref{1}) becomes: \begin{eqnarray} \frac{\partial L}{\partial \dot{{\bf{x}}}} = {\bf{p}} \tag{2}\label{2} \end{eqnarray} which is one of the Hamilton's equations usually found in textbooks.
Classical Field Theory
As discussed in my previous question linked above, the Lagrangian and the Hamiltonian now become functions of fields, which are infinite-dimensional vectors indexed by space-time coordinates $(t,{\bf{x}})\in \mathbb{R}^{4}$. Let us denote $\mathcal{F}$ the space of fields, which we assume to be sufficiently smooth and regular at infinity so that the following integrals are always finite.
In textbooks, the Hamiltonian for a classical field theory is given by: \begin{eqnarray} H(t, \phi, \partial_{{\bf{x}}}\phi,\pi) := \int \pi(t,{\bf{x}})\dot{\phi}(t,{\bf{x}})d{\bf{x}} - L(t, \phi, \partial_{\mu}\phi) \tag{3}\label{3} \end{eqnarray}
Question 1: How does one define the Legendre transform in such infinite-dimensional space such that the Hamiltonian becomes (\ref{3})?
Question 2: Once question 1 is answered and the Hamiltonian is defined in this infinite-dimensional space, there should be an identity similar to (\ref{2}) so that the usual formula: \begin{eqnarray} \pi(t,{\bf{x}}) = \frac{\partial \mathscr{L}}{\partial \dot{\phi}(t,{\bf{x}})} \tag{4}\label{4} \end{eqnarray} holds. What is the meaning of the derivative in the right hand side of (\ref{4})? I'm assuming the space of fields $\mathcal{F}$ is a Banach space (actually, probably an inner product space) so that the above derivative is Fréchet?
ADD: As I stressed before, in classical mechanics one can define the Hamiltonian as: \begin{eqnarray} H(t,{\bf{p}},{\bf{x}}) = \langle {\bf{p}}, \dot{{\bf{x}}}\rangle - L(t,{\bf{x}},\dot{{\bf{x}}}) \tag{5}\label{5} \end{eqnarray} where, in (\ref{5}) it is understood that $\dot{{\bf{x}}}$ should be considered as a function of ${\bf{p}}$ by means of the solution of (\ref{2}). Thus, in classical field theory, we can define the Hamiltonian following the same recipe, by setting: \begin{eqnarray} H(t,\phi, \partial_{{\bf{x}}}\phi, \pi) := \int \pi(t,{\bf{x}})\dot{\phi}(t,{\bf{x}})d{\bf{x}} - L(t, \phi, \partial_{\mu}\phi). \tag{6}\label{6} \end{eqnarray}
However, in classical mechanics, the Hamiltonian (\ref{5}) is the Legendre transform of $L$ and (\ref{2}) follows naturally. So, the objective of my question is to check wether the infinite-dimensional case can also be defined by means of an appropriate infinite-dimensional Legendre transform analogous to the finite-dimensional case, so that the conjugate variable $\pi$ as defined by (\ref{4}) is naturally enherited from the maximality of this Legendre transform as it is the case for the finite-dimensional case.
