Why is $1=\ell (\frac{x_{0}-y_{n}}{\operatorname{dist}(x_{0},Y)})$

Question

I am on the way to proving that for $Y\subsetneq X$ closed subspace in a normed space $(X, \vert \vert \cdot \vert \vert)$ that:

$\exists \ell \in (\operatorname{span}(\{x_{0}\} \cup Y))^{*}$ where $\ell\vert_{Y}=0$ and $\ell(x_{0})={\operatorname{dist}(x_{0},Y)} $ and $\vert \vert \ell \vert \vert_{*}=1$. I used $\ell(\alpha x_{0})=\alpha {\operatorname{dist}(x_{0},Y)}$. It is rather simple to show that $\vert \vert \ell \vert \vert_{*}\leq1$ but on the issue of $\vert \vert \ell \vert \vert_{*}=1$, I have stumbled into trouble. When I look at the solutions, they do not make sense to me either.

The supposed solution: Let $(y_{n})_{n}\subseteq Y: \lim\limits_{n \to \infty} \vert\vert x_{0} - y_{n}\vert\vert={\operatorname{dist}(x_{0},Y)}$ , then it follows that $1=\ell (\frac{x_{0}-y_{n}}{\operatorname{dist}(x_{0},Y)})\leq\vert\vert \ell\vert\vert_{*}\vert\vert\frac{x_{0}-y_{n}}{\operatorname{dist}(x_{0},Y)}\vert\vert\Rightarrow \vert \vert \ell \vert \vert_{*}=1$.

Is this solution even correct, I do not understand where the first equality comes from and how would the entire inequality imply $\vert \vert \ell \vert \vert_{*}=1$. Appreciate the help.

Answer 1

I believe your $\ell$ is actually defined as $$\ell(\alpha x_0 + y) = \alpha \operatorname{dist}(x_0, Y)$$ for all scalars $\alpha$ and $y \in Y$.

Therefore we have

$$1 = \ell\left(\frac{x_0}{\operatorname{dist}(x_0, Y)}\right) = \ell\left(\frac{x_0}{\operatorname{dist}(x_0, Y)} - \frac{y_n}{\operatorname{dist}(x_0, Y)}\right) = \ell\left(\frac{x_0-y_n}{\operatorname{dist}(x_0, Y)}\right)$$

so $$1 = \left|\ell\left(\frac{x_0-y_n}{\operatorname{dist}(x_0, Y)}\right)\right| \le \|\ell\|_*\frac{\|x_0-y_n\|}{\operatorname{dist}(x_0, Y)} \xrightarrow{n\to\infty} \|\ell\|_*$$

It follows $\|\ell\|_* \ge 1$.