Does a convex function with a Lipschitz continuous gradient always have a strong convex conjugate?

Question

I got the answer is 'Yes' from a scribe.

But I am confused because:

Suppose there is a convex function $f(x)=x^THx$, where $x\in\mathbb{R}^N$ and $H\in\mathbb{R}^{M\times N}$ is positive semidefinite. Thus $f(x)$ is a convex with a Lipschitz continuous gradient $L=2\|H\|$.

$f^*(x)=\max\limits_y\langle x,y\rangle-y^THy$. Thus, $x=2Hy$. But $H$ can be not invertable.If it is invertable, then $f^*(x)=\frac{1}{4}x^TH^{-1}x$, which its gradient is $\frac{1}{2}H^{-1}x=\frac{1}{L}$.

Answer 1

If $Hy = 0$, then for any $x$ not perpendicular to $y$, you can see that $F^*(x) = \infty$. So you need $H$ to be strictly positive definite if you want to avoid $\infty$ as a possible answer.

The issue is related to sub-Riemannian manifolds, and whether every geodesic necessarily satisfies the Euler-Lagrange equations.