This question came to me today, and I am now intrigued about it. For a system described by a Lagrangian $L$, the associated Hamiltonian is its Legendre transform. Suppose we consider a given Lagrangian function $L: \mathbb{R}^{n}\times \mathbb{R}^{n}\to \mathbb{R}$ (I am omitting any possible time-dependence here). This is, say, a $C^{k}$ function for some $k \ge 2$. Suppose its associated matrix: $$\bigg{[}\frac{\partial^{2}L}{\partial \dot{q}_{i}\partial \dot{q}_{j}}(q,\dot{q})\bigg{]}_{1\le i,j \le n}$$ is positive-definite, so that for each fixed $q \in \mathbb{R}^{n}$ the function $L(q,\cdot)$ is strictly convex.
In the literature, one is introduced by the Hamiltonian function by the formula: $$H(q,p) = \max_{\dot{q} \in \mathbb{R}^{n}}(\langle \dot{q},p\rangle - L(q,\dot{q})) \tag{1}\label{1}$$ where $\langle \cdot, \cdot \rangle$ is the usual Euclidean inner product. This is the Legendre transform.
My question is: what is the domain of definition of $H$? It is usually assumed to be $\mathbb{R}^{n}\times \mathbb{R}^{n}$ as well, but how can we assure that for every $q,p \in \mathbb{R}^{n}$, the maximum in (\ref{1}) is well-defined?
Let me be a little more specific. Suppose we define the function: $$\psi_{q,p}(\dot{q}) = \langle \dot{q},p\rangle - L(q,\dot{q}) \tag{2}\label{2}$$ for $q, p \in \mathbb{R}^{n}$ fixed. From the hypothesis on $L$, we know that:
- If $\psi_{q,p}$ has a critical point it is unique and
- If $\psi_{q,p}$ has a critical point it is also global maximum.
However, it is not clear that the maximum exists for every $q,p \in \mathbb{R}^{n}$.
What I can prove is the following. Because $L$ is strictly convex in the $\dot{q}$ variable, then for each fixed $q \in \mathbb{R}^{n}$ the function $$\dot{q} \mapsto f(\dot{q}):=\nabla_{\dot{q}}L(q,\dot{q}) \equiv (\frac{\partial L}{\partial \dot{q}_{1}}(q,\dot{q}),...,\frac{\partial L}{\partial \dot{q}_{n}}(q,\dot{q}))$$ is injective, so it is invertible over its range. Let $V_{q} := f(\mathbb{R}^{n})$ the range of $f$. Since $q$ is arbitrary, the maximum of $\psi_{q,p}$ is well-defined on $\bigcup_{q\in \mathbb{R}^{n}}\{q\}\times V_{q}$, so that the Hamiltonian $H$ is well-defined on this very same domain.
So, my question is: is it all? It seems like a very odd domain of definition. Am I missing something? Can we prove the domain of $H$ to be "larger" than this? Or do we need additional hypothesis?
