Hessian at a maximum point lying on the boundary

Question

Let $\Omega$ be a bounded domain of class $C^2$ in $\mathbb{R}^n$ and let $f: \overline{\Omega} \to \mathbb{R}$ be a smooth function. Assume $f$ attains its maximum at $x_0 \in \partial \Omega$. Can we say that the Hessian of $f$ at $x_0$ is negative semidefinite? I know this is true if $x_0$ lies in the interior of $\Omega$.

Answer 1

No. Take $\ \Omega\ $ to be the open unit ball in $\ \mathbb{R}^n\ $ and $\ f\left(x_1,x_2,\dots, x_n\right)=$$\sum_{i=1}^na_ix_i^2\ $ with $\ a_i>a_{i+1}>0\ $ for all $\ i=1,2,\dots, n-1\ $. Then the Hessian of $\ f\ $, $$ H=\pmatrix{2a_1&0&\dots&0\\ 0&2a_2&\dots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\dots&2a_n}\ , $$ is everywhere strictly positive definite, but $\ f\ $ attains a maximum of $\ a_1\ $ on $\ \overline\Omega\ $ at $\ x_0=\pmatrix{1,&0,&0,&\dots,&0}\ $.