Show that the determinant of the Hessian matrix of $f(x)=|x|^{\alpha}$, $\alpha>1$, is bounded away from zero when $1>|x|>c>0$

Question

Let $\alpha>1$ and let $f:\mathbb{R}^{n}\to \mathbb{R}$ be defined by $f(x)=|x|^{\alpha}$. Let $$H(x):=\partial_{x_{i}}\partial_{x_{j}}f (x).$$

How can one show that there exists some $C>0$ such that $$\text{det}(H(x))\geq C$$ whenever $c_{2}>|x|>c_{1}>0$.

Write $H(x)=(h_{ij}(x))$, $1\leq i,j\leq n$. Direct calculations give $$h_{ij}(x)=\alpha|x|^{\alpha-4}\begin{cases} (\alpha-2)x^2_{i}+|x|^2,& i=j,\\\\ (\alpha-2)x_{i}x_{j},& i\neq j. \end{cases}$$

So, $H=\alpha|x|^{\alpha-4}(S+|x|^2 I_{n})$ where $S$ is the symmetric matrix $(\alpha-2)(x_i x_j)$. It is easy to show that $\text{det}(S)=0$. Does this simplify the problem ?

Answer 1

$rank(S)=1$ so $0$ is an eigenvalue of order $n-1$.

The last eigen value is given by $\lambda + (n-1)\times 0 = Tr(S)$ and $$\lambda = |x|^2$$

$S$ is equivalent to $diag(|x|^2,0,0,...0)$

And $S+|x|^2I_n$ is equivalent to $diag(2|x|^2,|x|^2,...|x|^2)$

Therefore $$det(S+|x|^2I_n)=2|x|^{2n}$$

Answer 2

If $A$ and $B$ are $ n \times n$ matrices, then $\det(A+B)$ is the sum of the detrminants of all the matrices that can be formed by taking $m$ rows from $A$ and $ n-m $ rows from $B$ as $m$ ranges from $0$ to $n$. In the case of $\det(S+\vert x \vert^2I_n)$, if two or more rows are taken from $S$ the detrminant is $0$. If one row is taken from $S$, the determinant is $\vert x \vert^{2(n-1)}(x_1^2+...+x_n^2)=\vert x \vert^{2n}$. If no rows are taken from $S$ the determinant is $\vert x \vert^{2n}$. Thus the determinant of $S+\vert x \vert^2I_n$ is $2 \vert x \vert^{2n}$ so the determinant of the Hessian is $(\alpha \vert x \vert^{\alpha-4})^n2\vert x \vert^{2n}$. You can take it from there.