Legendre-Fenchel conjugate: $\dot x \in \operatorname{dom}(\Psi^*), -\operatorname{D}\mathcal E(x) \in \partial \Psi^*(\dot x)$

Question

Consider $\mathcal E: \Bbb R \to \Bbb R$ lower semi-continuous with $\mathcal E(x) \to \infty$ if $|x| \to \infty$ and the equation $\dot x = -\nabla \mathcal E(x)$. Apparently, this differential equation is equivalent to $\Psi(-\operatorname{D}\mathcal E(x)) + \Psi^*(\dot x) = - \langle \operatorname{D}\mathcal E(x), \dot x \rangle$ with $\Psi(x) = {1 \over 2}x^2$. This is the notion of gradient flows and this does hold if $\dot x \in \operatorname{dom}(\Psi^*), -\operatorname{D}\mathcal E(x) \in \partial \Psi^*(\dot x)$ where $\Psi^*$ is the Legendre-Fenchel conjugate. Please explain why this holds.

Answer 1

These are the main ideas, I might have skipped over some technical details.

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The first thing to notice is that $\Psi(x) = \frac{1}{2}\|x\|^2$ (I use the vector version but you can remove the bars if you'd like) is its own conjugate, $\Psi^* = \Psi$, and that its subdifferential is the identity. From this we can deduce,

$$\partial \Psi^*(\dot{x}) = \partial \Psi(\dot{x}) = \dot{x}.$$

If we assume the gradient flow differential equation holds then we have,

$$\dot{x} = -\nabla \mathcal{E}(x) = -D\mathcal{E}(x)$$

where I am assuming $D$ is the derivative operator. We have by Fenchel's inequality that,

$$\langle -D\mathcal{E}(x), \dot{x}\rangle \leq \Psi(-D\mathcal{E}(x)) + \Psi^*(\dot{x})$$

with equality iff $-D\mathcal{E}(x) \in \partial \Psi^*(\dot{x})$. But from our deduction we know that $\partial \Psi^*(\dot{x}) = \dot{x}$ and so we have equality in the Fenchel's inequality iff $-D\mathcal{E}(x) = \dot{x}$, which is exactly the gradient flow differential equation.