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

Question

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?

Answer 1

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).