# Distance of the element from the subspace of $l_{1}.$

Let $$l_{1}$$ be as follows $$l_{1}=\Big\{\{x_{n}\}_{n=0}^{\infty}\subset \mathbb C\: : \: \sum_{n=0}^{\infty}|x_{n}|<\infty\Big\}$$ and its subspace be $$V=\Big\{\{x_{n}\}_{n=0}^{\infty}\in l_{1}\: : \: \sum_{n=0}^{\infty}\frac{n}{n+2}x_{n}=0\Big\}.$$ For every $$x\in l_{1}$$ I need to find its distance $$d_{v}(x)=\inf_{v\in V}||x-v||$$ from the subspace $$V,$$ where I suppose the norm is the p-th norm, that is $$||x||_{1}=\sum_{n=0}^{\infty}|x_{n}|$$ According to the theorem presented during lecture, I proceeded to find the linear form $$f$$ such that $$\ker{f}=V,$$ so that would obviously be $$\forall\: x\in l_{1}\: f(x)=\sum_{n=0}^{\infty}\frac{n}{n+2}x_{n}.$$ In such a case $$d_{v}(x)=\frac{|f(x)|}{||f||},$$ where $$||f||=\sup_{||x||_{1}=1}|f(x)|$$ the following conditions imply the value of the $$||f||$$: $$\begin{equation*} \begin{gathered} 1)\: \exists c : \forall x \in l_{1} |f(x)|\leq c||x||_{1}\\ 2)\: \exists x\in l_{1} : f(x)=c, \end{gathered} \end{equation*}$$ then $$||f||=c.$$ So to find $$c$$ I followed: $$\forall x\in l_{1}\: |f(x)|=\Big|\sum_{n=0}^{\infty}\frac{n}{n+2}x_{n}\Big|\leq\sum_{n=0}^{\infty}\frac{n}{n+2}|x_{n}|\leq\sum_{n=0}^{\infty}|x_{n}|,$$ so $$c$$ we are looking for would be $$c=1.$$ And now I came across the obstacle - how to find the sequence $$x=\{x_{n}\}_{n=0}^{\infty}$$ such that $$||x||_{1}=1 \: \wedge\: f(x)=c=1?$$ Is there any algorithm of finding such a vector or is the whole reasoning nonsense?

It is not necessary to find $$x$$ such that $$f(x)=1$$ and $$\|x\|=1$$. It is enough if we have $$x^{(n)}$$ such that $$\|x^{(n)}\|=1$$ and $$f(x^{(n)}) \to 1$$ because we then have $$\|f\| \geq f(x^{(n)})$$ for all $$n$$. Now just take $$x^{(n)}$$ to be the standard basis vector $$e_n$$ (with $$1$$ at the $$n-$$th place and $$0$$ elsewhere).
• In fact it is impossible to find $x$ such that $\|x\|=1$ and $f(x)=1$. – Kavi Rama Murthy May 8 '19 at 9:35