text: # $d(w, V) = 1$ and $U = V \oplus\{\lambda w :\lambda \in \mathbb{F}\}$.
  
  Let U be a normed space, $w \in U$ and $\|w\| = 1$. Show that $U$ has a closed
  subspace $V$ such that $d(w, V) = 1$ and $U = V \oplus\{\lambda w :\lambda \in \mathbb{F}\}$.
  
  #### Solutions Collecting From Web of "$d(w, V) = 1$ and $U = V \oplus\{\lambda w :\lambda \in \mathbb{F}\}$."
  
  As we know by Hahn-Banach theorem $\exists$ a bounded linear functional $f$ on $U$ such that $f(w)=||w||$ and $||f||=1$. Consider $V=Ker(f)$. As $f$ is continuous $V$ will be closed. It is easy to see that $V\cap L\{w\}=\{0\}$. As $Ker(f)$ is of co-dimensional $1$ and $w\notin V$, then $U=V\bigoplus L\{w\}$.
  
  Now it remains to show that $d(w,V)=1$.
  
  1. $d(w,V)\leq ||w-0||=1$ (as $0$ is always in $Ker(f)$).
  
  2. Now $1=||w||=f(w)=f(w-v)$, for any $v\in V$. As $f$ is bounded with $||f||=1$, so $1\leq ||w-v||$, for any $v\in V$. So $d(w,V)=inf_{v\in V}||w-v||\geq 1$.
  
  Note: $L\{w\}=\{\lambda w: \lambda \in \mathbb{F}\}$.