📂Partial Differential Equations

Hamiltonian and Lagrangian Convex Duality

Theorem¹

Legendre Transform

• $L$ is a convex function.
• $\lim \limits_{ |v|\to \infty} \dfrac{ L(v) }{ |v| }=+\infty$

Given the conditions, the Legendre transform $L^{\ast} : \mathbb{R}^{n} \to \mathbb{R}$ of $L$ for the Lagrangian $L : \mathbb{R}^{n} \to \mathbb{R}$ is defined as follows:

$$ L^{\ast} (p) := \sup \limits_{v \in \mathbb{R}^{n}} \big( p\cdot v -L(v) \big) \quad \forall \ p \in \mathbb{R}^{n} $$

Let’s assume the Lagrangian $L$ satisfies the conditions for the Legendre transform to be defined. Let the Hamiltonian $H$ be the Legendre transform of $L$. $$ H=L^{\ast} $$

Then, $H$ also satisfies the two conditions necessary for the definition of the Legendre transform, and $L=H^{\ast}$ holds.

• (a) $H$ is a convex function.

  $$ \lambda H(v_{1}) + (1-\lambda) H(v_{2}) \le H\big( \lambda v_{1} +(1-\lambda)v_{2} \big) \quad \forall\ v_{1},v_{2}\in \mathbb{R}^{n},\quad \forall\ 0\le \lambda \le 1 $$

• (b) $\lim \limits_{ |p|\to \infty} \dfrac{ H(p) }{ |p| }=+\infty$

• (c) $L=H^{\ast}$

The converse also holds. Moreover, if $H$ and $L$ are differentiable at $p$ and $v\in \mathbb{R}^{n}$, then all the following statements are equivalent.

• $p\cdot v=L(v) + H(p)$
• $p=DL(v)$
• $v=DH(p)$

Proof

(a)

Suppose $p,q \in \mathbb{R}^{n}$ and $0 \le \tau \le 1$. Then, the following holds.

$$ \begin{align*} H(\tau p+(1-\tau)q) &= \sup \limits_{v\in \mathbb{R}^{n}} \big( (\tau p + (1-\tau) q) \cdot v -L(v) \big) \\ &= \sup \limits_{v\in \mathbb{R}^{n}} \big( \tau ( p \cdot v -L(v) ) +(1-\tau)(q \cdot v -L(v)) \big) \\ &\le \tau \sup \limits_{v} (p\cdot v - L(v) ) +(1-\tau) \sup \limits_{v} (q\cdot v -L(v)) \\ &= \tau H(p) + (1-\tau)H(q) \end{align*} $$

The second equality follows by adding $\tau L(v) -\tau L(v)$. The third line holds because $L$ is a convex function. The final equality is due to the definition of the Legendre transform and the assumption $H=L^{\ast}$.

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(b)

Consider any fixed positive number $\lambda >0$ and suppose $p\ne 0 \in \mathbb{R}^{n}$. Then, by assumption, the following holds.

$$ H(p)= \sup \limits_{v \in \mathbb{R}^{n}}( p\cdot v - L(v)) $$

If we let $v=\lambda \frac{p}{|p|}$, then by the definition of $\sup$, if any $v$ is substituted into the above equation, it results in a value less than or equal, thus the following is obtained.

$$ \begin{align*} H(p) &= \sup \limits_{v \in \mathbb{R}^{n}}( p\cdot v -L(v)) \\ &\ge \lambda |p| -L \left( \lambda \frac{p}{|p|} \right) \\ & \ge \lambda |p| - \max \limits_{B(0,\lambda)} L \end{align*} $$

Dividing both sides by $|p|$ gives:

$$ \dfrac{H(p)}{|p|} \ge \lambda - \dfrac{\max L}{ |p| } $$

This implies that the following equation holds.

$$ \liminf \limits_{|p| \to \infty} \dfrac{ H(p) }{ |p| } \ge \lambda $$

Since the above equation holds for any $\lambda$, we obtain:

$$ \liminf \limits_{|p| \to \infty} \dfrac{ H(p) }{ |p| } = \infty $$

Therefore, it is as follows:

$$ \lim \limits_{|p| \to \infty} \dfrac{ H(p) }{ |p| } = \infty $$

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(c)

Since $H=L^{\ast}$ and $L^{\ast}=\sup (p \cdot v -L(v))$, for any $v$, $H$ is always greater or equal.

$$ H(p) \ge p \cdot v -L(v) \quad \forall \ v\in \mathbb{R}^{n} $$

Permuting $H$ and $L$ gives:

$$ L(v) \ge p \cdot v -H(p) \quad \forall \ v\in \mathbb{R}^{n} $$

Taking $\sup$ for $p \in \mathbb{R}^{n}$ on both sides, the right side equals $H^{\ast}$ by the definition of the Legendre transform.

$$ L(v) \ge \sup_{p \in \mathbb{R}^{n}} \big( p \cdot v -H(p) \big) =H^{\ast}(v) $$

Now, to prove that the opposite inequality $L(v) \le H^{\ast}(v)$ holds, $H^{\ast}$ is defined as follows by definition.

$$ H^{\ast}(v)= \sup_{p \in \mathbb{R}^{n}} \big( p \cdot v-H(p)\big) $$

However, since we assumed $H=L^{\ast}$, we obtain:

$$ \begin{equation} \begin{aligned} H^{\ast}(v) &= \sup_{p \in \mathbb{R}^{n}} \big( p \cdot v-\sup_{r \in \mathbb{R}^{n}}\big( p\cdot r - L(r) \big) \big) \\ &= \sup_{p} \inf_{r} \big( p\cdot(v-r) + L(r)\ \big) \end{aligned} \label{eq1} \end{equation} $$

Lemma

If a function $f : \mathbb{R}^{n} \to \mathbb{R}$ is convex, for all $x \in \mathbb{R}^{n}$, there exists $r \in \mathbb{R}^{n}$ satisfying the following:

$$ f(y) \ge f(x) + r\cdot(y-x) \quad \forall\ y\in \mathbb{R}^{n} $$

Since $L$ is convex, by the Lemma, for all $v \in \mathbb{R}^{n}$, there exists $s \in \mathbb{R}^{n}$ satisfying the following condition.

$$ L(r) \ge L(v)+s\cdot(r-v) \quad \forall \ r\in \mathbb{R}^{n} $$

Permuting $s\cdot (r-v)$ and substituting $p=s$ gives $\eqref{eq1}$ as follows.

$$ H^{\ast}(v) \ge \inf_{r} \big( s\cdot (v-r) \big) =L(v) $$

Therefore, $H^{\ast}(v) \ge L(v)$ and, $H^{\ast}(v)=L(v)$ holds.

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