Showing that sequences such that $\sum_{n=1}^\infty {x_n\over n} =1$ form a closed subset of $l^2$

Question

Let $H= \left\{(x_n)\in\ell^2 : \sum_{n=1}^\infty {x_n\over n} =1\right\}$

I need to show that H is closed in $l^2$.

then, it is sufficient to show that, closure of H $\subseteq$ H.

let x=($x_i$)$\in$closure of H,$\exists$ a sequence ($x^{(n)}$) with $x^{(n)}$=($x^{(n)}_1$,$x^{(n)}_2$,$x^{(n)}_3$,......)such that $\left\lVert x^{(n)}-x\right\rVert_2$ $\to$0 as n$\to$$\infty$

now, after this step, I am not able to complete the solution. please give a detailed solution.

thanks a lot...

Answer 1

Consider $x=(x_1,x_2,...,x_n,...)\in \bar {H}$. T.P. $x\in H$. as $x\in \bar H$ there exists seq say $x^n=(x^n_1,x^n_2,x^n_3,....)\in H$ converging to $x$. $\implies$ $\|x^n-x\|_2< \varepsilon$ $\implies$ on simplifying we will get $|x_k^n-x_k|< \varepsilon$ $\forall n \geq n_0$, $\forall k$. So $(x^n_k)_{n\in\mathbb{N}}\to x_k$ for each k. Now $\sum_{k=1}^{\infty}x^n_k/k=1 $ for each $n$. taking limit as n tends to infinity, $\sum_{k=1}^{\infty}x_k/k=1 $

Answer 2

Note that \begin{equation*} \mathbf{f}=(1,\frac{1}{2},\frac{1}{3},\cdots ) \end{equation*} is contained in $l^{2}$. Let \begin{equation*} \mathbf{x}=(x_{1},x_{2},\cdots )\in l^{2} \end{equation*} Then for $\mathbf{x}\in H$ \begin{equation*} (\mathbf{x},\mathbf{f})=1 \end{equation*} I think it is useful to switch to \begin{equation*} \mathbf{y}=\mathbf{x}-\frac{(\mathbf{x},\mathbf{f})}{\parallel \mathbf{f}% \parallel ^{2}}\mathbf{f} \end{equation*} (which becomes \begin{equation*} \mathbf{y}=\mathbf{x}-\frac{1}{\parallel \mathbf{f}\parallel ^{2}}\mathbf{f} \end{equation*} for $\mathbf{x}\in H$.) Now \begin{equation*} (\mathbf{y},\mathbf{f})=0 \end{equation*} which defines a closed subspace $K$ of $l^{2}$. \ Now suppose that $\{ \mathbf{x}^{j}\}\subset H$ is Cauchy, so it has a limit $\mathbf{x}\in l^{2}$ and so does $\{\mathbf{y}^{j}\}$ in $K$. But then it follows that \begin{equation*} (\mathbf{x},\mathbf{f})=\lim_{j\rightarrow \infty }(\mathbf{x}^{j},\mathbf{f} )=1 \end{equation*}

Answer 3

I was asked to look at this. Looking at it, I think it's possible to simplify Urgje's answer. Give $\bf{f}$ the same meaning as in Urgje's answer, i.e., $(1,1/2,1/3,\ldots,1/n,\ldots)$. We have ${\bf f}\in l^2$ because $||{\bf f}||^2 = \sum_{n=1}^\infty 1/n^2$ by definition of the norm and $\sum_{n=1}^\infty 1/n^2$ is a well-known convergent series, so $||{\bf f}||^2<\infty$.

Now, consider the function $g: {\bf x}\to ({\bf x},{\bf f})$ for ${\bf x}\in l^2$. The function $g$ takes $l^2$ to $\mathbb{R}$ or $\mathbb{C}$, according to what the scalar field for $l^2$ is. The function $g$ is continuous in ${\bf x}$ because $|({\bf x},{\bf f})|\le ||{\bf x}||\,||{\bf f}||$ (Rudin Functional Analysis 12.2) and so $|g({\bf x}_1)-g({\bf x}_2)|=|({\bf x}_1-{\bf x}_2,{\bf f})|\le ||{\bf x}_1-{\bf x}_2||\,||{\bf f}||\to 0$ when ${{\bf x}_1\to {\bf x}_2}$. Thus, the inverse image of a closed set under $g$ is closed (Baby Rudin 4.8).

For any ${\bf x}\in l^2$ we have ${\bf x} \in H$ iff $1=\sum_{n=1}^\infty x_n/n = ({\bf x},{\bf f})$. Thus, $H$ is the inverse image of $\{1\}$ under $g$, and we are done.