Point-hyperplane distance in normed spaces

[gugo82]

Here's a generalization of the following classical formula from Linear Algebra:

If [tex]$x=(x_1,x_2,x_3)\in \mathbb{R}^3$[/tex] and [tex]$\Pi \subseteq \mathbb{R}^3$[/tex] is the plane of equation [tex]$ay_1+by_2+cy_3=\alpha$[/tex], then the point-plane distance from [tex]$x$[/tex] to [tex]$\Pi$[/tex] is:

[tex]$\text{dist} (x,\Pi )=\frac{|ax_1+bx_2+cx_3-\alpha|}{\sqrt{a^2+b^2+c^2}}$[/tex].

We're to work out the real case, but the theorem holds true also in the general case, i.e. for normed vector spaces over arbitrary fields.

***

Few things to remember:

Testo nascosto, fai click qui per vederlo

Let [tex]$(X,||\cdot ||)$[/tex] be an infinite-dimensional normed vector space over [tex]$\mathbb{R}$[/tex] and let [tex]$(X^*,||\cdot ||_*)$[/tex] be its normed topological dual space, i.e. the vector space [tex]$X^* := \{ u: X\to \mathbb{R} :\ u \text{ is linear and bounded} \} $[/tex] equipped with the dual norm:

[tex]$||u||_*:= \sup_{x\in X\setminus \{ \mathfrak{o}\}} \frac{|\langle u ,x\rangle |}{||x||} =\sup_{x\in \mathcal{B}} |\langle u,x\rangle |$[/tex].

Here and in what follows: [tex]$\mathfrak{o}$[/tex] is the null vector in [tex]$X$[/tex]; [tex]$\langle \cdot ,\cdot \rangle$[/tex] is the duality between [tex]$X^*$[/tex] and [tex]$X$[/tex], i.e. [tex]$\langle u , x\rangle := u(x) $[/tex]; and [tex]$\mathcal{B}$[/tex] is the open unit ball in [tex]$X$[/tex], i.e. [tex]$\mathcal{B} := \{ x\in X:\ ||x|| < 1 \}$[/tex].

We say that a subset [tex]$S\subseteq X$[/tex] is an affine subspace of codimension [tex]$N$[/tex] iff there exist exactly [tex]$N$[/tex] linearly independent functionals [tex]$u_1,\ldots ,u_N\in X^*$[/tex] and [tex]$N$[/tex] scalars [tex]$\alpha_1,\ldots ,\alpha_N \in \mathbb{R}$[/tex] s.t.:

[tex]$x\in S \Leftrightarrow \begin{cases} \langle u_1 ,x\rangle =\alpha_1 \\ \quad \vdots \\ \langle u_N ,x\rangle =\alpha_N \end{cases}$[/tex].

We refer to the [tex]$N$[/tex] equations in the previous formula as the affine equations of [tex]$S$[/tex]; the number [tex]$N$[/tex] is usually denoted [tex]$\text{codim} S$[/tex].

We call affine hyperplane each affine subspace [tex]$\Pi \subseteq X$[/tex] with [tex]$\text{codim} \Pi =1$[/tex].
If [tex]$\Pi$[/tex] is a hyperplane of equation [tex]$\langle u,x\rangle =\alpha$[/tex], we can also use the suggestive notation [tex]$\Pi_{u,\alpha}$[/tex].

Finally we remember that the distance of a point [tex]$x\in X$[/tex] from a nonempty subset [tex]$Y\subseteq X$[/tex] is the non negative number defined by:

[tex]$\text{dist} (x,Y):=\inf_{y\in Y} ||y-x||$[/tex].

***

Exercise:

Let [tex]$(X,||\cdot ||)$[/tex] an infinite-dimensional normed vector space and [tex]$\Pi=\Pi_{u,\alpha} \subseteq X$[/tex] an affine hyperplane.

1. Prove that for each [tex]$x\in X$[/tex] we have:

(*) [tex]$\text{dist} (x,\Pi) = \frac{|\langle u,x\rangle -\alpha|}{||u||_*}$[/tex].

2. Show that the proof of (*) becomes easier if [tex]$X$[/tex] is a Hilbert space.

Testo nascosto, fai click qui per vederlo

Hints: 1. Use the definition of [tex]$||u||_*$[/tex] and homogeneity to determine which values of [tex]$r>0$[/tex] make [tex]$r\mathcal{B} \cap \Pi_{u,||u||_*} =\emptyset$[/tex] and which don't; use the translation invariance of the distance function to complete the proof.

2. Use a suitable version of the Orthogonal Projection Theorem.

[gugo82]

Exercise:

Let [tex]$(X,||\cdot ||)$[/tex] an infinite-dimensional normed vector space and [tex]$\Pi=\Pi_{u,\alpha} \subseteq X$[/tex] an affine hyperplane.

1. Prove that for each [tex]$x\in X$[/tex] we have:

(*) [tex]$\text{dist} (x,\Pi) = \frac{|\langle u,x\rangle -\alpha|}{||u||_*}$[/tex].

Testo nascosto, fai click qui per vederlo

Hints: 1. Use the definition of [tex]$||u||_*$[/tex] and homogeneity to determine which values of [tex]$r>0$[/tex] make [tex]$r\mathcal{B} \cap \Pi_{u,||u||_*} =\emptyset$[/tex] and which don't; use the translation invariance of the distance function to complete the proof.

I have to admit the hint was a little misleading because, when I wrote the text, I had in mind a different (and maybe subtler) proof of the formula... Recently I've found another proof (i.e. the one I present here) which is simpler than the previous; hope you like.

Let [tex]$\Pi_{u,\alpha} :=\{ y\in X:\ \langle u,y\rangle =\alpha\}$[/tex] and [tex]$x\in X$[/tex].

If [tex]$x\in \Pi_{u,\alpha}$[/tex] equality (*) trivially holds; hence let's choose [tex]$x\notin \Pi_{u,\alpha}$[/tex].

First we assume [tex]$x=\mathfrak{o}$[/tex] (hence [tex]$\alpha \neq 0$[/tex]): using homogeneity we find:

[tex]$\lVert u\rVert_* =\sup_{y\neq \mathfrak{o}} \frac{|\langle u,y\rangle|}{\lVert y\rVert} = \sup_{y\neq \mathfrak{o} ,\langle u,y\rangle \neq 0} \frac{\Big| \langle u,\frac{\alpha}{|\langle u,y\rangle|}\ y\rangle \Big|}{\left\lVert \frac{\alpha}{|\langle u,y \rangle|}\ y\right\rVert} $[/tex]
[tex]$=\sup_{z\in \Pi_{u,\alpha}} \frac{|\langle u,z\rangle|}{\lVert z\rVert} =\sup_{z\in \Pi_{u,\alpha}} \frac{|\alpha |}{\lVert z\rVert} =\frac{|\alpha|}{\inf_{z\in \Pi_{u,\alpha}} \lVert z\rVert} = \frac{|\alpha |}{\text{dist} (\mathfrak{o} ,\Pi_{u,\alpha})}$[/tex]

therefore (*) holds for [tex]$x=\mathfrak{o}$[/tex].

Now assume [tex]$x\neq \mathfrak{o}$[/tex]: using additivity we obtain [tex]$\Pi_{u,\alpha} -x:=\{ y-x,\ y\in \Pi_{u,\alpha}\} =\Pi_{u,\alpha -\langle u,x \rangle}$[/tex], hence:

[tex]$\text{dist} (x,\Pi_{u,\alpha}) =\inf_{y\in \Pi_{u,\alpha}} \lVert y-x\rVert =\inf_{z\in \Pi_{u,\alpha -\langle u,x\rangle}} \lVert z\rVert =\text{dist} (\mathfrak{o} ,\Pi_{u,\alpha -\langle u,x\rangle}) = \frac{|\langle u,x \rangle -\alpha|}{\lVert u\rVert_*}$[/tex]

so that (*) holds.