Let $x, y \in \mathbb{R}^d.$ Consider the functional $\Psi: Diffeo(\mathbb{R}^d) \to \mathbb{R}_{\ge 0}$ defined on the space $Diffeo(\mathbb{R}^d)$ of diffeomorphisms of $\mathbb{R}^d$ defined by:
$$\Psi[F]:= ||F(x)-F(y)||_2^{2},$$
where $||.||_2$ denotes the Euclidean $2$-norm.
Then, is $\Psi$ convex at all points of $ Diffeo(\mathbb{R}^d) ?$ We note that if we let $F$ be $F(x):= \lambda x$ for a scalar $\lambda \ne 0,$ then writing $\Psi = \Psi_{\lambda},$ we get $\frac{\partial\Psi_{\lambda}}{\partial \lambda^2}= 2 ||x-y||^2 > 0,$ implying $\Psi$ is convex on the subset of $Diffeo(\mathbb{R}^d)$ defined by scaling. But how do we calculate $\frac{\partial^{2}{\Psi}}{\partial F^2}$ to begin with? I need some help here!
To make the question a bit easy, we can perhaps replace the diffeomophism group in the domain by an appropriate vector space ($Diffeo(\mathbb{R}^d)$ is an infinite dimensional manifold!) and can take $d=1$, so now we consider instead:
$$F:\mathcal{C}^2(\mathbb{R}) \to \mathbb{R}, F[\psi]:=(\psi(x_1) - \psi(x_2)^2$$ How to calculate $\frac{\partial^{2}{\Psi}}{\partial {F^2}}$ as a bilinear form on $\mathcal{C}^2(\mathbb{R})$ and show that it's positive definite?