Existence of Distance Functional

Theorem
Let $\mathbb F \in \set {\R, \C}$.

Let $\struct {X, \norm \cdot_X}$ be a normed vector space over $\mathbb F$.

Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $X$.

Let $Y$ be a proper closed linear subspace of $X$.

Let $x \in X \setminus Y$.

Let:


 * $d = \map {\operatorname {dist} } {x, Y}$

where $\map {\operatorname {dist} } {x, Y}$ denotes the distance between $x$ and $Y$.

Then there exists $f \in X^\ast$ such that:


 * $(1): \quad$ $\norm f_{X^\ast} = 1$
 * $(2): \quad$ $\map f y = 0$ for each $y \in Y$
 * $(3): \quad$ $\map f x = d$.

That is:


 * there exists a distance functional for $x$.

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:


 * $\ds d \le \norm {x - \paren {-\frac y \lambda} } = \norm {x + \frac y \alpha}$

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.