Exists $l\in X^*$ such that $\|l\|=1, l|_Y=0$ and $l(x_0)=\operatorname{dist}(x_0, Y)$?

Question

Let $(X, \|.\|)$ be a normed space and $Y\subsetneq X$.
How can you prove that for $x_0 \in X\setminus Y$ there exists $l\in X^*$ such that $\|l\|=1$, $l|_Y=0$ and $l(x_0)=\operatorname{dist}(x_0, Y)$ ?

Answer 1

I assumed that $Y$ is a subspace and $x_0$ doesn't belong to the closure of $Y$ (hence we may furthermore assumed that $Y$ is closed).

Define on $V:=Y+\operatorname{Vect}(x_0)$ the linear functional $f$ by $$f(y+\lambda x_0):=\lambda d( x_0,Y).$$ This is well defined, linear and for each $v\in V$, $|f(v)|\leqslant d(v,Y)$. The map $v\mapsto d(v,Y)$ is sublinear. Indeed, $$d(v_1+v_2,Y)=\inf_{y_1,y_2\in Y} \lVert v_1-y_1+v_2-y_2\rVert.$$ Therefore we may extend $f$ to $X$: there exists $g\in X^*$ such that $f(v)=g(v)$ for each $v\in V$ and $|g(x)|\leqslant d(x,Y)$ for any $x\in X$. We have $f(x_0)=d(x_0,Y)$ and $|g(x)|\leqslant \lVert x\rVert$ hence $\lVert g\rVert\leqslant 1$. To conclude that the norm of $g$ is indeed $1$, we actually prove it for $f$ using Riesz lemma.