Hessian of a $C^{\infty}$ function $f: \mathbb R^n \to \mathbb R$ restricted to a subspace

Question

Let $f: \mathbb R^n \to \mathbb R$ be an infinitely differentiable function. Let $\mathcal V \subset \mathbb R^n$ be a vector subspace. Let $g = f|_{\mathcal V}$ be the restriction of $f$. Then we can conclude $g$ is also infinitely differentiable and gradient of $g$ is $\nabla g(x) = P_{\mathcal V}(\nabla f(x))$ where $P_{\mathcal V}$ is the orthogonal projection. I am wondering if there is some similar relationship between the Hessian $\nabla^2 f(x) : \mathbb R^n \to \mathbb R^n$ and $\nabla^2 g(x)$. For example, I was tempting to think $\nabla^2 g(x) = \nabla^2 f(x)|_{\mathcal V}$ but it doesn't make sense since $\mathcal V$ might not be invariant under the map.

Answer 1

We can define $\tilde g:\mathbb R^n \to \mathbb R$ via $$ \tilde g(x) := g( Px) = f(Px)$$ where $P$ is the linear orthogonal(thus symmetric) projection map $\mathbb R^n \to V$. From here, the formulae are easier to derive. (One can return from $\tilde g$ to $g$ using the restriction map again.) Using the basic chain rule formulae for Frechet derivatives,

$$ d \tilde g(x)h = df(Px) Ph $$

In terms of e.g. the matrix representative in the standard basis, $$ \nabla\tilde g(x) \cdot h = \nabla f(Px)^T Ph = (P\nabla f(Px))^Th =(P\nabla f(Px)) \cdot h $$ recreating your quoted result. Computing the second derivative,

$$ d^2\tilde g(x) (u,v) = d_x(df(Px)Pu)(v) = d(df(Px))(Pu,v)= d^2f(Px)(Pu,Pv)$$

In terms of the matrix representative, $$ u^T H\tilde g(x)v = (Pu)^T Hf(Px)Pv = u^T \ (PHf(Px) P) \ v$$