Suppose $a\in \mathbb{R}^p$ and $f$ is a real-valued function whose second-order partial derivatives all exist and are continuous on $B_r(a)$. Also, suppose that the operator norm $\|d^2f(x)\|$ of the matrix $d^2f(x)$ is bounded by $M$ on $B_r(a)$. Prove that $$|f(x)-f(a)-df(a)(x-a)|\le \frac{M}{2}\|x-a\|^2$$ for all $x\in B_r(a)$.
If I divide both sides by $\|x-a\|$, I get $$\frac{|f(x)-f(a)-df(a)(x-a)|}{\|x-a\|}\le \frac{M}{2}\|x-a\|.$$
What I know:
- The left side goes to $0$ as $x\to a$
- As a corollary to the Mean Value Theorem, $|df(x)-df(a)|\le M\|x-a\|$
- From the Mean Value Theorem, $f(x)-f(a) = df(c_1)(x-a)$ for some $c_1$ on the line segment from $x$ to $a$
- Less certain about this, but $df(x)-df(a)=d^2f(c_2)(x-a)$ for some $c_2$ on the line segment from $x$ to $a$
It seems like I have all the pieces I need, but I'm not sure how to put them together.
I imagine the $\frac{M}{2}$ comes from having $a$ be the center of the ball. The statements above hold even if $a$ were another arbitrary point in the ball, so it makes intuitive sense that we are bounded by $\frac{M}{2}$ because we only have half the distance to work with. But I'm not sure how to put this all together.
Possibly relevant:
