Chapter 6: Problem 6
Prove that \(H_{0}^{1, p}\left(\mathbf{R}^{n}\right)=H^{1, p}\left(\mathbf{R}^{n}\right)\)
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Chapter 6: Problem 6
Prove that \(H_{0}^{1, p}\left(\mathbf{R}^{n}\right)=H^{1, p}\left(\mathbf{R}^{n}\right)\)
These are the key concepts you need to understand to accurately answer the question.
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If \(p=n\), then we might suppose that \(H_{0}^{1, p}(\Omega)\) is contained in \(L^{\infty}(\Omega)\). However, consider \(u(x)=\log (\log (2 /|x|))-\log (\log 2)\) on \(\Omega=B_{1}(0)\) for \(n \geq 2\). Show that \(\|u\|_{1, n}<\infty\) but \(u \notin L^{\infty} .\) [Using mollification as in the next section, it is easy to show that \(\left.u \in H_{0}^{1, p}(\Omega) .\right]\)
For vector-valued functions \(\vec{u}=\left(u_{1}, \ldots, u_{N}\right)\) on a domain \(\Omega \subset \mathbf{R}^{n}\), it is natural to define \(L^{p}\left(\Omega, \mathbf{R}^{N}\right)\) as the functions \(\vec{u}: \Omega \rightarrow \mathbf{R}^{N}\) for which \(|\vec{u}| \in L^{p}(\Omega)\); that is, $$ \|\vec{u}\|_{p} \equiv\left(\int_{\Omega}|\vec{u}|^{p} d x\right)^{1 / p}<\infty, \quad \text { where }|\vec{u}|^{2}=\sum_{i=1}^{N} u_{i}^{2} . $$ Similarly, if each \(u_{i} \in C^{1}(\Omega)\), we may define $$ \|\nabla \vec{u}\|_{p} \equiv\left(\int_{\Omega}|\nabla \vec{u}|^{p} d x\right)^{1 / p}, \quad \text { where }|\nabla \vec{u}|^{2}=\sum_{i=1}^{N} \sum_{j=1}^{n}\left(\partial u_{i} / \partial x_{j}\right)^{2} . $$ (a) Define \(H_{0}^{1, p}\left(\Omega, \mathbf{R}^{N}\right)\). (b) If \(N=n\), show that div: \(H_{0}^{1, p}\left(\Omega, \mathbf{R}^{n}\right) \rightarrow L^{p}(\Omega)\) is a continuous linear operator. (c) Show that \(\tilde{H}_{0}^{1, p}\left(\Omega, \mathbf{R}^{n}\right) \equiv\left\\{\vec{u} \in H_{0}^{1, p}\left(\Omega, \mathbf{R}^{n}\right)\right.\) : div \(\left.\vec{u}=0\right\\}\) is a closed subspace of \(H_{0}^{1, p}\left(\Omega, \mathbf{R}^{n}\right)\).
If \(X\) is a separable Hilbert space, show that every bounded sequence contains a weakly convergent subsequence.
If \(T: D \rightarrow X\) is a densely defined operator on a complex Hilbert space \(X\), then \(T\) is symmetric if and only if \(D \subset D^{*}\) and \(\langle T x, x\rangle \in \mathbf{R}\) for all \(x \in D\).
If \(X, Y\), and \(Z\) are Banach spaces and \(T: X \rightarrow Y\) and \(S: Y \rightarrow Z\) are both bounded linear maps, with \(T\) or \(S\) (or both) compact, then \(S \circ T: X \rightarrow Z\) is compact.
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