Existence of Distance Functional

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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 1

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.

Aiming for a contradiction, suppose 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:

\(\ds \map \phi {\lambda u_1 + \mu u_2}\) \(=\) \(\ds \map \phi {\lambda \paren {y_1 + \alpha_1 x} + \mu \paren {y_2 + \alpha_2 x} }\)
\(\ds \) \(=\) \(\ds \map \phi {\lambda y_1 + \lambda \alpha_1 x + \mu y_2 + \mu \alpha_2 x}\)
\(\ds \) \(=\) \(\ds \map \phi {\paren {\lambda y_1 + \mu y_2} + \paren {\lambda \alpha_1 + \mu \alpha_2} x}\)
\(\ds \) \(=\) \(\ds \paren {\lambda \alpha_1 + \mu \alpha_2} d\)
\(\ds \) \(=\) \(\ds \lambda \alpha_1 d + \mu \alpha_2 d\)
\(\ds \) \(=\) \(\ds \lambda \map \phi {y_1 + \alpha_1 x} + \mu \map \phi {y_2 + \alpha_2 x}\)

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:

\(\ds \cmod {\map \phi u}\) \(=\) \(\ds \cmod {\map \phi {y + \alpha x} }\)
\(\ds \) \(=\) \(\ds \cmod {\alpha d}\)
\(\ds \) \(=\) \(\ds \cmod \alpha d\)
\(\ds \) \(\le\) \(\ds \cmod \alpha \norm {x + \frac y \alpha}\)
\(\ds \) \(=\) \(\ds \norm {y + \alpha x}\) Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \) \(=\) \(\ds \norm u\)

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:

\(\ds \cmod {\map \phi {-y_n + x} }\) \(=\) \(\ds d\)
\(\ds \) \(\ge\) \(\ds \frac n {n + 1} \norm {x - y_n}\)

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.

$\blacksquare$


Proof 2

Consider the normed quotient vector space $X/Y$ with quotient mapping $\pi$.

From Kernel of Quotient Mapping, we have $\map \pi x \ne 0$.

So, from Existence of Support Functional, there exists $f \in \paren {X/Y}^\ast$ such that:

$\norm f_{\paren {X/Y}^\ast} = 1$

and:

$\map f {\map \pi x} = \norm {\map \pi x}_{X/Y}$

From the definition of the quotient norm, we have:

$\norm {\map \pi x}_{X/Y} = \map {\operatorname {dist} } {x, Y}$

From Normed Dual Space of Normed Quotient Vector Space is Isometrically Isomorphic to Annihilator, $g = f \circ \pi \in X^\ast$ and:

$\norm g_{X^\ast} = \norm f_{\paren {X/Y}^\ast} = 1$

with:

$\map g x = \map {\operatorname {dist} } {x, Y}$

So $g$ is a linear functional satisfying our requirements.

$\blacksquare$


Sources

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