# The Legendre Transform of a differentiable, strictly convex, and coercive function

I encountered the following unsubstantiated fact in a proof:

$$f$$ is differentiable, strictly convex, and coercive. Thus we may conclude $$(\nabla f)^{-1} = \nabla f^*$$ where $$f^*$$ is the Legendre transformation of $$f$$

Loosely speaking, I see how this might work: if a function is differentiable then Legendre trasformation will return the slope at some point $$x \in \text{dom} f$$. If $$f$$ is strictly coercive, then $$\nabla f$$ is monotonic, in the appropriate sense. So an inverse is defined. Loosely, I think the role of coercivity is to make the map between $$x \in \text{dom} f$$ and its slope at $$x$$ bijective. I can't understand how coercivity is causing this though.

My questions:

1. Can someone provide a heuristic description what coercivity is giving us?
2. Provide a proof of the fact above (or reference)? Assume $$f: \mathbb{R}^d \rightarrow \mathbb{R}$$.

Recall that $$f^*(y) = \sup \langle x,y\rangle -f(x)$$. Assuming there is a minimizer $$x^*$$ we have, $$f^*(y) = \langle x^*,y\rangle -f(x^*)$$. At the same time, $$x^*$$ satisfies the optimality conditions $$0=\nabla_x(\langle x,y\rangle -f(x))|_{x=x^*} = y-\nabla_x f(x^*)\implies y=\nabla_xf(x^*)$$. If we take the gradient of $$f^*(y)$$ w.r.t. $$y$$ to get $$\nabla_y f^*(y) = x^*(y)$$. So we know that $$\nabla f^*(y) = (\nabla f)^{-1}(x^*(y))$$. But $$x^*$$ might not be unique.

So you want,

$$\arg\sup_x \langle x,y\rangle - f(x)=\{x^*\}\neq\emptyset$$

a singleton for all $$y$$. Coercivity in $$x$$ will ensure this since $$f$$ is also strictly convex (and thus $$-f$$ is strictly concave).

The injectivity of $$\nabla f$$ needed to define an inverse is a consequence of strict convexity and coercivity. Since $$f$$ is strictly convex, $$\nabla f$$ is strictly monotone (and thus injective). Notice that

$$\arg\sup_x \langle x,y\rangle - f(x) = \arg\min_x \langle x,-y\rangle + f(x)$$

Fix $$y$$ and let $$\phi(x) = \langle x,-y \rangle + f(x)$$. The sublevel sets of $$\phi$$ are bounded by coercivity and closed since $$\phi$$ is continuous. Taking any sublevel set (e.g. the sublevel set for $$\phi(0)$$), we have that $$\phi$$ is strictly convex and differentiable and thus admits a unique minimum on a compact set. Then, for any $$y$$, $$x^*$$ exists and is unique so that $$\nabla f$$ is surjective.

1. You don't need coercivity. The result is true, e.g., for the exponential function which is not coercive.

2. Check Chapter 26 in Rockafellar's Convex Analysis about the Legendre transformation.