Proving that there exists $c \in S$ such that $f(x) = f(a) + \nabla f(a)\cdot (x − a) + \frac{1}{2} H_{f}(c)(x − a)\cdot (x − a)$

Question

$S \subseteq \mathbb{R}^{n}$ is open and convex

Let $f : S \to \mathbb{R}$ be $C^{2}$ on $S$. Prove that for every $x, a \in S$, there exists $c \in S$ such that $f(x) = f(a) + \nabla f(a)\cdot (x − a) + \frac{1}{2} H_{f}(c)(x − a)\cdot (x − a)$

I don't quite know where to start the proof on this question. I know that I need to show that there exists some c such that the equality holds, but I do not know how to define c or how to go about proving that it is equal to f(x).

I have a result that the second order Taylor Polynomial of f at $a \in S$ is:

$P_{a,2}(x) = f(a) + \nabla f(a) \cdot (x-a) + \frac{1}{2}h_{f}(a)(x-a) \cdot (x-a)$

But I don't know if I could apply it here. Would letting c be a and claiming the proposition true by the result satisfy the proof? Please help/point me in the right direction for the proof

Answer 1

Consider the function $$ g(t):= f(x + t(x-a)) \quad t\in [0,1]. $$ Now apply the Taylor theorem for functions from $\mathbb R$ to $\mathbb R$. You will end up with evaluating the second-derivative at some intermediate value $\xi\in(0,1)$, which will give the $c$.