Question: Let (A: mathcal{C}_{c}^{infty}left(mathbb{R}^{d}ight) ightarrow mathcal{C}_{infty}left(mathbb{R}^{d}ight)) be a closed operator such that (|A u|_{infty} leqslant C|u|_{(2)}); here (|u|_{(2)}=|u|_{infty}+sum_{j}left|partial_{j} uight|_{infty}+sum_{j k}left|partial_{j} partial_{k} uight|_{infty}). Show that (mathcal{C}_{infty}^{2}left(mathbb{R}^{d}ight)=overline{mathcal{C}_{c}^{infty}left(mathbb{R}^{d}ight)}{ }^{|cdot|_{(2)}}
Let \(A: \mathcal{C}_{c}^{\infty}\left(\mathbb{R}^{d}ight) ightarrow \mathcal{C}_{\infty}\left(\mathbb{R}^{d}ight)\) be a closed operator such that \(\|A u\|_{\infty} \leqslant C\|u\|_{(2)}\); here \(\|u\|_{(2)}=\|u\|_{\infty}+\sum_{j}\left\|\partial_{j} uight\|_{\infty}+\sum_{j k}\left\|\partial_{j} \partial_{k} uight\|_{\infty}\). Show that \(\mathcal{C}_{\infty}^{2}\left(\mathbb{R}^{d}ight)=\overline{\mathcal{C}_{c}^{\infty}\left(\mathbb{R}^{d}ight)}{ }^{\|\cdot\|_{(2)}} \subset\) \(\mathfrak{D}(A)\).
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