Let $X$ be a normed $\Bbb{R}$-vector space. Let $Y \subset X$ be a closed subspace and $x_0 \in X \backslash Y$. We define $$d :=\text{dist}(x_0, Y) = \inf_{y \in Y}\,\lVert x_0 − y \rVert$$ I would like to prove that there exists $\lambda \in X^*$ with $\lVert \lambda \rVert = 1$, $\lambda|_Y = 0$, and $\lambda(x_0) = d$.
I attempted the following :
Let $f : Y+\Bbb Rx_0\rightarrow \Bbb{R}$ such that $\forall t \in \Bbb{R}, y \in Y$, $$f(y + tx_0) = t.d$$
One has that $f$ is linear and continuous. We also have $f(x_0) = d$ and $f|_Y =0$ as requested. We can compute that $\lVert f \rVert = \frac{d}{\lVert x_0 \rVert}$.
Now let's consider $\lambda$, the Hahn-Banach extension of $f$ to the whole space $X$ such that $\lVert \lambda \rVert = \lVert f \rVert$.
$\lambda$ almost corresponds to what I wanted but the norm of $\lambda$ is not equal to 1 and I am unable to fix that problem.
I would greatly appreciate any help. Thank you.
$$\begin{align}\|f\|&=\sup_{y\in Y,t\in\Bbb R\atop y+tx_0\ne0}\frac{|td|}{\|y+tx_0\|}\\&=\sup_{y\in Y,t\in\Bbb R^*\atop y+tx_0\ne0}\frac d{\|y/t+x_0\|}\\&=\frac d{\inf_{z\in Y}\|x_0-z\|}\\&=1.\end{align}$$