Let $f: \mathbb{R}^d \rightarrow \mathbb{R}$, and $g(x) = f(a) + \nabla f(a)^\top (x-a) + \frac{1}{2}(x-a)^\top \nabla^2f(a) (x-a)$ be its second-order expansion at $a$. We know that if $f$ is concave, then it can always upper bounded by its first-order expansion.
I was wondering, under what condition (i.e., we can impose additional assumptions on $f$), we can have
$f(x) \le g(x), \quad \forall x,a$?
A trivial solution will be $f(x) = x_0 + w^\top x + x^\top H x$, which gives $g(x) =f(x)$.
Is there any other solution?