# Definition of the Hamiltonian via Legendre transform.

In my book of classical mechanics (Mathematical methods of classical mechanics by V.I. Arnold), the Hamiltonian is introduced in this way (my translation):

Let us consider the system of equations $$\dot p = \partial L /\partial \dot q$$ ($$p\in \mathbb R^n$$, $$q\in \mathbb R ^n$$, the second member is the gradient of the Lagrangian with respect to $$\dot q$$), defined by a Lagrangian that we will suppose convex with respect to the second argument $$\dot q$$.

[...]

By definition, the Legendre transform in $$\dot q$$ of $$L(q,\dot q ,t)$$ is a function $$H(p)=p\dot q-L(\dot q)$$, where $$\dot q$$ is given by the relation: $$p=\dfrac{\partial L}{\partial \dot q}.$$

Now, my definition of Legendre transform of a function $$f:\mathbb R ^n \to \mathbb R$$ is: $$g(p)=\sup _{x\in \mathbb R^n} (\langle p,x\rangle-f(x)).$$ I can see that the quoted definition coincides with mine if for example, we suppose that $$f$$ is a quadratic form $$f(x)=x^T A x,$$ for a positive definite symmetric matrix. In the general case of a convex $$f$$ (that, to clarify, here means “definite positive Hessian matrix”), however, I don't see how are we granted that:

1. The maximum is attained at a point $$x\in \mathbb R ^n$$ (we should at least require that $$f$$ is coercive right? Counterexample: $$f(x)=-\ln x$$)
2. The equation $$p=\partial f / \partial x$$ has a unique solution.

What are sufficient (or maybe necessary and sufficient) conditions for the above formulas to properly define a function that coincides with the Legendre transform of $$L$$?

Suppose that $$f:\mathbb R ^n \to \mathbb R$$ has a positive definite hessian matrix $$f''(x)$$ for all $$x$$. Also suppose that $$\lim _{|x|\to \infty } \frac{f(x)}{|x|} = \infty .$$ Then the transform $$f^*$$ exists and is given by $$f^*(p)=\left\langle p,\xi (p)\right\rangle-f(\xi (p)),$$ where $$\xi$$ is the inverse of $$f'$$.

In fact, if the limit holds, one can easily see that $$G_p (x) = f(x)-\left\langle x,p\right\rangle$$ is coercitive and admits a minimum in $$\mathbb R ^n$$, that corresponds to $$-f^*(p)$$. At this point, the derivative vanishes, so $$f'(x)=p$$ (this also proves that $$f'$$ is surjective). Finally, since $$f''>0$$, $$f'$$ is injective and has an inverse $$\xi$$ and the Legendre transform is as said above.

• note $g(p)=\sup_{x\in\Bbb R}\left(px-f(x)\right)$ Dec 4, 2013 at 8:04
• I don't understand this comment. If it's referred to my counterexample $f=-\ln$, note that $f$ is convex and $\sup _{x \in \mathbb R} px - f(x)=\infty$. Dec 4, 2013 at 18:42
• @pppqqq Just want to throw out a comment saying that I think your questions are valid. I'll think about it, and I might get back to you later. I do think the equation has a unique solution (although that requires a proof). I think you are correct when you say a solution might not exist. Dec 4, 2013 at 22:18
• Notice two facts: 1) In physics, kinetic energy (i.e. $L$ considered as a function of $\dot q$) is always assumed to be a quadratic form with respect to $\dot q$; and 2) in §14 (italian version) Arnold introduces Legendre transform and remarks that the point $x$ we are looking for doesn't need to exist. However, in §15 a footnote says that we "often" assume that kinetic energy will be a quadratic form, so I think your work is necessary and useful as a mathematical one. Jan 6, 2014 at 12:14
• @Federico Thank you, I was really more concerned in some sufficient hypothesis to make the above argument meaningful, without making further assumptions on the transformed function. Jan 6, 2014 at 18:48

It is clear from your examples that the supremum might not exist.

However, the solution to the equation $\frac{\partial L}{\partial \dot q} = p$, if it exists, is unique. Because the Hessian of $L(x,y)$ with respect to $y$ is positive definite, this means that $f:\mathbb R \to \mathbb R$, $t\mapsto L(x,y_0+t(y_1-y_0))$ satisfies $f''>0$ whenever $y_0 \ne y_1$. This means that $f'(y_0) \ne f'(y_1)$, and $f'(y) = (y_1-y_0) \cdot \frac{\partial L}{\partial y}$, hence $\frac{\partial L}{\partial y}(x,y_0) \ne \frac{\partial L}{\partial y}(x,y_1)$.

You have shown coercivity implies existence. What remains is the converse.

Since the Hessian is positive definite, by the implicit function theorem, the map $y \mapsto \frac{\partial L}{\partial x}(x,y)$ is locally invertible with a continuous local inverse. Since the map is injective, it is a continuous map from $\mathbb R^n$ onto its image.

For now, write $G(y) = \frac{\partial L}{\partial x}(x,y)$

Pick $M > 0$. Then for every $z \in \mathbb R^n$ with $|z|\le M$, there exists a unique $y_z$ such that $G(y_z) = z$. The map $z \mapsto y_z$ is well defined and continuous. Hence $\sup_{|z|=1} |y_z| = N$ exists. Now $G^{-1}(\{|z|\le M\}$ is compact and hence bounded. Hence $G^{-1}(\{|z|>M\}$ is unbounded. Therefore if $|y|>N$, then $|G(y)| \ge M$.

Now suppose that $|y| > N$. Create a path along the ODE $\eta(0) = y$, $\eta'(t) = G(\eta(t))/|G(\eta(t))|$. Let $T = \inf\{t:|G(\eta(t))| = M$ (and its OK if $T = \infty\}$, but as it happens it won't). Then $T \ge |y|-N$. Then it can be seen that $L(x,y) \ge L(x,\eta(T)) + T M \ge T M + \inf_{\xi} L(x,\xi)$. Hence if $|y|$ is large enough, the $L(x,y) \ge \frac12M |y|$. So $L(x,y)$ is coercive in $y$.

Kind of a complicated argument. Maybe there is something simpler.

• thank you for your answer, your first paragraph answers point #2 that was the most interesting part. It'd be nice to have some precise statement about the existence, though. Dec 5, 2013 at 11:51
• I am thinking about it, but I don't have a rigorous answer yet. Dec 5, 2013 at 16:53
• Thank you. I've updated the question with a first result, try to give a look. Dec 5, 2013 at 17:37