In the paper 'On the Holder continuity of solutions of second order elliptic equations in two variables' by Piccinini and Spagnolo, they prove the following estimate: $$ \begin{array}{ll} \left(\int_S p_{11} |u_T|^2 \right)^{\frac12}\left(\int_S \frac{\langle P(x) (u_N,u_T), (1,0) \rangle^2}{p_{11}} \right)^{\frac12} &\leq & \sqrt{\frac{\Lambda}{\lambda}} \left(\int_S \lambda |u_T|^2 \right)^{\frac12}\left(\int_S \frac{\langle P(x) (u_N,u_T), (1,0) \rangle^2}{p_{11}} \right)^{\frac12} \\ & \leq & \sqrt{\frac{\Lambda}{\lambda}}\frac12 \left(\int_S \lambda |u_T|^2 + \frac{(p_{11} u_N + p_{12}u_T)^2}{p_{11}} \right)\\ & \leq & \sqrt{\frac{\Lambda}{\lambda}}\frac12 \left(\int_S \left( p_{22} - \frac{p_{12}^2}{p_{11}}\right) |u_T|^2 + \frac{(p_{11} u_N + p_{12}u_T)^2}{p_{11}} \right)\\ & = & \sqrt{\frac{\Lambda}{\lambda}}\frac12 \left(\int_S \langle P(x) (u_N,u_T),(u_N,u_T)\rangle \right) \end{array} $$ Here $P(x)$ is a symmetric positive definite $2\times2$ matrix with entries $(p_{ij})$ and $\lambda |\xi|^2 \leq \langle P(x)\xi,\xi\rangle \leq \Lambda |\xi|^2$ and $(u_N,u_T)$ are the normal and tangential derivative with $S$ being the unit circle in $\mathbb{R}^2$.

The crucial part of the above calculation is the inequality $\lambda \leq \left( p_{22} - \frac{p_{12}^2}{p_{11}}\right)$.

My question is the following: Is there a different way to prove this estimate without explicitly writing down the expressions and instead proceed using only matrix inequalities?


1 Answer 1


Let $0<\lambda_1\le\lambda_2$ be the eigenvalues of the symmetric positive $2\times2$ matrix $P$. Then $$\lambda\le\lambda_1=\min_{|\xi|=1} \langle P\xi,\xi\rangle \le\langle Pe_1,e_1\rangle=p_{11}\le \lambda_2 =\max_{|\xi|=1} \langle P\xi,\xi\rangle,$$ so $\lambda p_{11}\le \lambda_1\lambda_2=\det P=p_{11}p_{22}-p_{12}^2$, whence the inequality, since $\langle Pe_1,e_1\rangle=p_{11}>0$.

  • $\begingroup$ (actually the middle $\le$ in the displayed inequalities is not needed, but it's also obvious) $\endgroup$ Dec 20, 2022 at 20:45
  • $\begingroup$ Thank you for the reply, this is very helpful. $\endgroup$
    – Adi
    Dec 28, 2022 at 16:48

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