# Norm estimate of $(-\lambda A+ I)^{-1}$ for strictly elliptic operator

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.

• Do you know some form of the spectral theorem? That makes it relatively easy to compute such norms. – MaoWao Mar 10 at 9:54
• @MaoWao Thanks. I will check the theorem. – John Mar 10 at 10:36

## 1 Answer

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 $$\|u\|_{H^2} \le C \, (1 + 1/\lambda) \, \|f\|_{L^2}.$$