Existence of Distance Functional/Proof 1

== Theorem ==
== Proof ==
Since $x \not \in Y$, we have: 
:$d > 0$
from Point at Distance Zero from Closed Set is Element.
Let:
:$X_0 = \map \span {Y \cup \set x}$
From Linear Span is Linear Subspace, we have: 
:$X_0$ is a linear subspace of $X$.
Note that we can then write any $u \in X_0$ in the form: 
:$u = y + \alpha x$
for $y \in Y$ and $\alpha \in \mathbb F$. 
We want to define a map in terms of this representation, so we show that this representation is unique. 
Let: 
:$u = y_1 + \alpha_1 x = y_2 + \alpha_2 x$
Then: 
:$\paren {\alpha_2 - \alpha_1} x = y_1 - y_2$
If $\alpha_1 = \alpha_2$, then we have $y_1 = y_2$ as required.
 suppose that $\alpha_1 \ne \alpha_2$, then we would have: 
:$\ds x = \frac 1 {\alpha_2 - \alpha_1} \paren {y_1 - y_2}$
and so $x \in Y$, from the definition of a linear subspace, contradiction. 
So, we must have $\alpha_1 = \alpha_2$ and $y_1 = y_2$, and so the representation is unique. 
Now, define $\phi : X_0 \to \mathbb F$ by: 
:$\map \phi {y + \alpha x} = \alpha d$ 
for each $y \in Y$ and $\alpha \in \mathbb F$.
Clearly, we have: 
:$\map \phi x = d$
and:
:$\map \phi y = 0$ for all $y \in Y$. 
We will show that $\phi$ is a bounded linear functional and apply the Hahn-Banach Theorem. 
Let $u_1, u_2 \in X_0$ and $\lambda, \mu \in \mathbb F$. 
Write: 
:$u_1 = y_1 + \alpha_1 x$
and:
:$u_2 = y_2 + \alpha_2 x$
for $y_1, y_2 \in Y$ and $\alpha_1, \alpha_2 \in \mathbb F$. 
We then have: 
so $\phi$ is linear. 
We will now show that $\phi$ is bounded and that: 
:$\norm \phi_{\paren {X_0}^\ast} = 1$
To do this, we will show that: 
:$\norm \phi_{\paren {X_0}^\ast} \le 1$
and:
:$\norm \phi_{\paren {X_0}^\ast} \ge 1 - \epsilon$
for each $0 < \epsilon < 1$.
Let $u \in X_0$ and again write $u = y + \alpha x$. 
If $\alpha = 0$, then $u \in Y$, and we have: 
:$\cmod {\map \phi u} = 0 \le \norm u$ 
Now take $\alpha \ne 0$. 
Since $Y$ is a linear subspace, we have: 
:$\ds -\frac y \alpha \in Y$ 
Recall that, by the definition of the distance between $x$ and $Y$, we have: 
:$d = \inf \set {\norm {x - y} : y \in Y}$
So, we have: 
:$\ds d \le \norm {x - \paren {-\frac y \alpha} } = \norm {x + \frac y \alpha}$
Then, we have: 
So in any case, we have: 
:$\map \phi u \le \norm u$
So $\phi$ is bounded and we have: 
:$\norm \phi_{\paren {X_0}^\ast} \le 1$
from the definition of the dual norm.
Now fix $0 < \epsilon < 1$. 
Noting again that: 
:$d = \inf \set {\norm {x - y} : y \in Y}$
for each $n \in \N$ we can pick $y_n \in Y$ such that: 
:$\ds d \le \norm {x - y_n} \le d \paren {1 + \frac 1 n}$
That is: 
:$\ds \frac n {n + 1} \norm {x - y_n} \le d$ 
We then have: 
Now note that for $N \in \N$ with: 
:$\ds N \ge \frac 1 \epsilon - 1$ 
We have: 
:$\ds \frac 1 {N + 1} \le \epsilon$
and so: 
:$\ds \frac N {N + 1} = 1 - \frac 1 {N + 1} \ge 1 - \epsilon$
Then we have: 
:$\cmod {\map \phi {-y_N + x} } \ge \paren {1 - \epsilon} \norm {x - y_n}$
We then cannot have: 
:$\norm \phi_{\paren {X_0}^\ast} < 1 - \epsilon$
so we have: 
:$\norm \phi_{\paren {X_0}^\ast} \ge 1 - \epsilon$ 
Since $0 < \epsilon < 1$ was arbitrary, we have: 
:$\norm \phi_{\paren {X_0}^\ast} \ge 1$
and so: 
:$\norm \phi_{\paren {X_0}^\ast} = 1$
We apply: 
:Hahn-Banach Theorem: Real Vector Space: Corollary 2 if $\mathbb F = \R$
:Hahn-Banach Theorem: Complex Vector Space: Corollary if $\mathbb F = \C$
to find that there exists $f \in X^\ast$ such that:
:$f$ extends $\phi$ to $X$
with: 
:$\norm f_{X^\ast} = \norm \phi_{\paren {X_0}^\ast} = 1$
Since $f$ extends $\phi$, we have: 
:$\map f x = \map \phi x = d$
and:
:$\map f y = \map \phi y = 0$ for each $y \in Y$.
So $f$ is the required bounded linear functional.
Category:Existence of Distance Functional