3

Let $H=\left\{(x_n)\in \ell^2(\mathbb{N})\mid\sum \frac{x_n}{n}=1\right\}$.

To check which one is true:

  • (a) $H$ is bounded
  • (b) $H$ is closed
  • (c) $H$ is a subspace
  • (d) $H$ has interior points

My try: (c) is not true as $x_n\in H$ does not imply that $cx_n\in H$ for $c\in \mathbb R$. (d) is also not true as $x_n\in H$ does not imply $x_n+t\in H$ for $t$ however small it is.

I am not sure about (a),(b). How to proceed?

Silvia Ghinassi
  • 4,707
Learnmore
  • 31,062

2 Answers2

3

(a) is false as the sequence $(0,\dots,0,n,0,\dots)$ is in $H$.

(b) is true due the continuity of inner product of $l_2$ and that the sequence $(1/n)$ is in $l_2$.

Quang Hoang
  • 15,854
  • 1
    please explain b ;i did not get that – Learnmore Aug 26 '15 at 06:03
  • $l_2$ is a Hilbert space with the inner product $\mathbf{x}\cdot\mathbf{y}=\sum x_iy_i$. The continuity of the dot product follows from the Cauchy-Schwarz inequality. The argument should be available in standard Function Analysis textbook. – Quang Hoang Aug 26 '15 at 06:07
0

For (b),

$1 = \sum_{n=1}^{\infty} \frac {x_n}{n} \le \sum_{n=1}^{\infty} |\frac {x_n}{n}| \le (\sum_{n=1}^{\infty} |x_n|^2)^{\frac 12}(\sum_{n=1}^{\infty}|\frac 1n|^2)^{\frac 12}=t \frac {\pi}{\sqrt 6}$, Where $t=||(x_n)||=(\sum_{n=1}^{\infty}|x_n|^2)^{\frac 12}$.

$\Rightarrow t=||(x_n)|| \ge \frac {\sqrt 6}{\pi}...(*)$

Hence $H$ contains all those elements with property ($*$).

Now it might be easy for you to check why it's closed. :)

Error 404
  • 6,006