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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.)

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1 Answer 1

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.

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