Are these two subspaces of $\ell^2$ closed?

Question

I have just met this exercise in functional analysis, asking us to determine if these two subspaces of the Hilbert space $\ell^2$ of square-summable complex sequences are closed:

1. The set of all sequences $\{x_n\}_{n=1}^{\infty}$ satisfying $$\sum_{n=1}^{\infty} \frac{1}{n} x_n = 0 $$

2. The set of all sequences $\{x_n\}_{n=1}^{\infty}$ satisfying $$\sum_{n=1}^{\infty} x_n = 0 $$

I know what I am supposed to do: to prove the subspace is closed I need to consider a general Cauchy sequence in the subspace and show its limit is also in the subspace and to prove it is not closed I only need to find one Cauchy sequence in the subspace whose limit is not in it. However, these two subspaces have me stuck, I do not know if they are closed or not so I have no idea on this. I thank all helpers.

Answer 1

For (1), as has been noted by past answers, we can use Cauchy-Schwarz to see that $f(x)= \sum_{n=1}^\infty \frac{1}{n}x_n$ is continuous $\ell^2 \to \mathbb{C}$, so the set $\{x \in \ell^2 \mid f(x) = 0\} = f^{-1}(0)$ is closed.

For (2), consider the sequence $x(k) \in \ell^2$ of the form $$x(k)_n = \begin{cases} 1 & n=1 \\ -\frac{1}{k} & 2 \leq n \leq k+1 \\ 0 & n\geq k+2 \end{cases}$$ Then for all $k\geq 1$, we have $\sum_{n=1}^\infty x(k)_n = 0$ and $\sum_{n=2}^\infty x(k)_n^2 = \frac{1}{k} \to 0,$ so $x(k) \to (1,0,0,0,\ldots)$ in $\ell^2$, but $(1,0,0,0,\ldots) \not\in \{x \in \ell^2 \mid \sum_{n=1}^\infty x_n = 0\}$, so the subspace is not closed.

Answer 2

Hints:

For $1).\ $ define the linear functional $x\mapsto \sum_{n=1}^{\infty} \frac{1}{n} x_n$, and show it is bounded, hence continuous.

For $2).\ $ Consider the sequence of sequences

$(1,-1,0,0,\cdots )$

$(1,-1/2,-1/2,0,0\cdots )$

$(1,-1/3,-1/3,-1/3,0,0\cdots )$

$\cdots$