# The norm of an operator

Let $\rho(x)$ be a weight function in a unit sphere, such that $$\begin{array}{l} \displaystyle 1. \rho(x)\ge 0,\int_{\mathbb{R}^n}\rho(x)=1\\ \displaystyle 2. \rho(x)\in \mathrm{C}_{0}^{\infty}(\mathbb{R}^n) \\ \displaystyle 3. \rho(x)=0 \mbox{ if |x|\ge1} \end{array}$$ We define $\rho_{\varepsilon}(x)=\frac{1}{\varepsilon}\rho(\frac{x}{\epsilon})$,$u_{\varepsilon}(x)=(\rho_{\varepsilon}*u)(x)$.

$\Omega$ is the domain of $u$,when $x\not\in \Omega$, $u(x)=0$.

Is it true that the operator $A_{\epsilon}:L^{p}(\Omega)\to L^{p}(\Omega)$ $$A_{\epsilon}(u)= u_\varepsilon$$ satisfies: $$||A_{\epsilon}-I||\ge 1$$ Moreover, $$||A_{\epsilon}-I||=1$$

I know that $$\int_{B_{\epsilon}(0)}u(x-y)\rho_{\varepsilon}(y)dy=\int_{\Omega}u(z)\rho_{\varepsilon}(x-z)dz$$ But I am confused by the way to $\sup\frac{||Ax||}{||x|}$ , I am not familiar with convolution. Thanks for your help.

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