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.

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Few things to remember:

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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.


Risposte
gugo82
"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].


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. :-D

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.

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