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Let $\Omega$ be a smooth domain in $\mathbb{R}^n$, and $A$ be a strictly elliptic operator $$ Au=\partial_i(a^{ij}(x)\partial_j)u, $$ where $a^{ij}$ are bounded functions satisfying $$ a^{ij}(x)\xi_i\xi_j\ge \alpha |\xi|^2,\quad \forall \xi\in \mathbb{R}^n,\,x\in \mathbb{R}^n. $$ Then for any $f\in L^2(\Omega)$ and $\lambda>0$, there exists $u\in H^2(\Omega)$ such that $$ -\lambda Au+ u=f,\quad x\in \Omega; \quad x=0,\quad x\in \partial\Omega. $$ This implies $(-\lambda A+I)^{-1}:L^2(\Omega)\to H^2(\Omega)$ is a bounded operator (we assume zero boundary condition). Is it possible to get a precise estimate for the operator norm $|(-\lambda A+I)^{-1}|_{\mathcal{L}(L^2(\Omega),H^2(\Omega))}$ in terms of $\lambda$? In particular, I would like to know the behavior of the norm as $\lambda\to 0$.


Let $Dom(A)=H^2(\Omega)\cap H^1_0(\Omega)$, we know $A$ is a maximal monotone operator on $L^2(\Omega)$, hence $ |(-\lambda A+I)^{-1}|_{\mathcal{L}(L^2(\Omega))}\le 1$ for all $\lambda$. But it does not give the estimate in terms of $H^2$ norm.

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  • $\begingroup$ Do you know some form of the spectral theorem? That makes it relatively easy to compute such norms. $\endgroup$ – MaoWao Mar 10 at 9:54
  • $\begingroup$ @MaoWao Thanks. I will check the theorem. $\endgroup$ – John Mar 10 at 10:36
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One possibility is to rearrange the equation as $$ -A \, u + u = \frac1\lambda \, f + (1 - 1/\lambda) \, u.$$ The $L^2$-norm of the right-hand side can be bounded by your bound in $\mathcal L(L^2)$. Together with a bound for $(-A+I)^{-1}$, this should result in something like $$\|u\|_{H^2} \le C \, (1 + 1/\lambda) \, \|f\|_{L^2}.$$

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