H={$(x_n)$ in $l_2$ | $\sum x_n/n = 1$} is closed and unbounded in $l_2$.

Question

Let H={$(x_n)$ in $l_2$ | $\sum x_n/n = 1$} Then how to prove H is closed and unbounded. Or give some example for its verification.

Answer 1

Unbounded :

Define $x^{(n)}$ by $x^{(n)}_k = n$ if $k=n$ and $0$ elsewhere.

We have :

• $x^{(n)} \in H$
• $\| x^{(n)} \| = n$

So $H$ is unbounded.

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Closed :

Notice that $\sum_{n=1}^\infty \frac{x_n}{n} = \langle x_n, v \rangle$ where $\langle , \rangle$ is the $l^2$ scalar product, and $v_k = \frac{1}{k}$

The functionnal $T_v : x\mapsto \langle x,v \rangle$ is a continuous linear form.

But $H = T_v^{-1}(\{1\})$, so it's the preimage of a closed set by a continuous function , hence closed.

Answer 2

Hint: $\sum x_n/n=\big\langle(x_n),(1/n)\big\rangle_{l_2}$.