(Based on a question in University of Pittsburgh's Preliminary examination: Aug 2016, question 5.)
Let $f\colon \mathbb{R}^2 \to \mathbb{R}$ be a $C^2$ function. Let $B=\{x \in \mathbb{R}^2\colon |x|\leq 1\}$ be the closed unit ball. Suppose $f(o)=Df(o)=o$ and that for an $M\geq 0$, we have $$ \left \vert \frac{\partial^2 f}{\partial_{x_1}^2}(x)\right \vert^2 + 2\left \vert \frac{\partial^2 f}{\partial_{x_1}\partial_{x_2}}(x)\right \vert^2 +\left \vert \frac{\partial^2 f}{\partial_{x_2}^2}(x)\right \vert^2 \leq M^2 , \quad \text{for any $x=(x_1,x_2) \in B$}. $$
Give the sharpest bound on $|f(x)|$ for $x \in B$.
By that, I mean something in terms of $M$ and (powers of) $|x|$, or maybe a different norm, like, $|x_1|+|x_2|$.
Please include the details of your argument with clear reference to any lemma/theorem you use. Your time is appreciated!
Im assuming "$o$" means $0$. By Taylor's theorem with Lagrange remainder, for $x \in B$, $$f(x) = \frac{1}{2}x \cdot H(\theta x)x$$ where $\theta \in (0, 1)$ depends on $x$, and $H$ is the hessian of $f$. Note that the bound given in the problem statement is equivalent (by Clairaut's theorem since $f$ is $C^2$) to saying that for $x \in B$, $$||H(x)||_2^2 \leq M^2.$$ So $||H(x)||_2 \leq M$ for all $x \in B$. Let $x \in B$. Then \begin{align} |f(x)| &= \frac{1}{2}|x \cdot H(\theta x)x| \\ &\leq \frac{1}{2}||x|| \cdot ||H(\theta x)x|| \\ &\leq \frac{1}{2}||x|| \cdot ||H(\theta x)||_2 \cdot ||x|| \\ &\leq \frac{1}{2}M||x||^2. \end{align} The first inequality is by Cauchy Schwarz. The second inequality uses the result that $||AB||_2 \leq ||A||_2 \cdot ||B||_2$, which can be proved using Cauchy Schwarz. I believe the inequalities are sharp in the sense that they are achieved when $H(x) = MI$ on $B$.