Discussion
Let \(H = \{(x_{n}) \in \ell_{2} :\) \(\sum_{n=1}^{\infty} \frac{x_{n}}{n} = 1\}\). Then \(H\)
Answer: B
\(H\) is closed.
Define \(\phi(x) = \sum \frac{x_{n}}{n}\) on \(\ell_{2}\). Since \((\frac{1}{n}) \in \ell_{2}\), the Cauchy-Schwarz inequality gives \(|\phi(x)| \le \|x\|_{2} \left(\sum \frac{1}{n^{2}}\right)^{1/2}\), so \(\phi\) is a bounded, hence continuous, linear functional. \(H = \phi^{-1}(\{1\})\) is the preimage of a closed set, so it is closed.
H is a translate of the hyperplane \(\ker\phi\): it is not a subspace (it does not contain 0), it is not bounded (add any multiple of a vector in \(\ker\phi\)), and a proper closed hyperplane has no interior points.
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