Let $u$ be a function on a domain $\Omega\subset R^n$, and $D^2u=\left(\frac{\partial^2 u}{\partial x_i\partial x_j}\right)_{n\times n}$ be the Hessian of $u$. If $D^2u$ is positively definite and $\det(D^2u)<\mu$ for some constant $\mu$, then $$ |D^2u|\leq (C(\varepsilon)+\varepsilon M)\sum_{i=1}^nu^{ii}, $$ holds for any $\varepsilon>0$, where $C(\varepsilon)$ is a constant depending only on $\varepsilon$, $(u^{ij})_{n\times n}$ is the inverse of $D^2u$, and $M=\sup_{x\in\Omega}|D^2u|$ .

I have tried the prove that in the case of $D^2u=\mathrm{diag}\{\lambda_1,\lambda_2,\ldots,\lambda_n\}$, but stacked for $n=3$. (in this case, n=1 and 2 are trivial.)


The question does not appear to have anything to do with $\Omega$ or with derivatives, since you are asking for a pointwise bound. It could be written as $\|A\|\le (C(\epsilon)+\epsilon \|A\|) \operatorname{tr} A^{-1}$.

Assume $A$ diagonal, as we can do because diagonalization preserves $\operatorname{tr} A^{-1}$. Let $\lambda_1\ge \dots \ge \lambda_n>0$ be the diagonal entries. Then $\operatorname{tr} A^{-1}\ge \lambda_n^{-1}$. Since $\mu>\lambda_1\cdots\lambda_n\ge \lambda_1\lambda_n^{n-1}$, it follows that $\lambda_n\le \lambda_1^{-1/(n-1)}\mu^{1/(n-1)}$. Hence $\operatorname{tr} A^{-1}\ge \lambda_1^{1/(n-1)}\mu^{-1/(n-1)}$. Let's simplify notation by writing $\operatorname{tr} A^{-1}\gtrsim \lambda_1^{1/(n-1)}$ where $\gtrsim$ indicates the presence of a multiplicative constant that may involve $\mu$ and $n$.

The matrix norm $\|A\|$ (whatever it is) is comparable to $\lambda_1$ in the sense that $\lambda_1 \lesssim \|A\|\lesssim \lambda_1$. Therefore, our task reduces to showing that $$\lambda_1 \lesssim C(\epsilon)\, \lambda_1^{1/(n-1)} + \epsilon\, \lambda_1^{n/(n-1)}$$ for all $\lambda_1>0$. But this is straightforward: no matter how small $\epsilon>0$, we have $\lambda_1 \le \epsilon\, \lambda_1^{n/(n-1)}$ for all sufficiently large $\lambda_1$. Then choose $C(\epsilon)$ so that $\lambda_1\le C(\epsilon) \, \lambda_1^{1/(n-1)}$ for all $\lambda_1$ that are not sufficiently large.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.