whether this set is closed subset of a Hilbert space

Question

Whether the set of sequences $(x_n \in l_2: \sum_{n=1}^{\infty}\frac{x_n}{n}=1 )$ is closed ? How do you find the limit point of a set of sequences? Moreover what is the complement of this set?, and whether that is open? (IIT-GATE 2015)

Answer 1

Define a linear functional $\varphi \colon \ell_2(\mathbb{N}) \rightarrow \mathbb{C}$ by

$$ \varphi(x) = \sum_{n=1}^{\infty} \frac{x_n}{n}. $$

By Cauchy-Schwartz, we have

$$ \sum_{n=1}^N \frac{|x_n|}{n} \leq \left( \sum_{n=1}^N |x_n|^2 \right)^{\frac{1}{2}} \left( \sum_{n=1}^N \frac{1}{n^2} \right)^{\frac{1}{2}} $$

and so $\varphi$ is well-defined and continuous with $||\varphi|| \leq \left( \sum_{n=1}^{\infty} \frac{1}{n^2} \right)^{\frac{1}{2}}$.

Your set is $\varphi^{-1}(\{1\})$ and since $\{ 1 \}$ is closed and $\varphi$ is continuous, the set is closed. It cannot be also open since $\ell_2(\mathbb{N})$ is connected (even path-connected).

Answer 2

We have $f=(1,{1 \over 2},..., {1 \over n},...) \in l_2$, hence $f^*(x)= \langle f,x\rangle$ is a continuous linear functional, hence the inverse image of a closed set is closed.

The complement is just the set $\{x | f^*(x) \neq 1 \}$ which is open because its complement is closed (or, indeed, because $\mathbb{R} \setminus \{1\}$ is open).

Another way is to note that $H=\{x | f^*(x) = 1 \} = \{e_1\}+\ker f^*$, that is, $H$ is just a translate of the kernel, which is closed.