Point-hyperplane distance in normed spaces

Messaggioda gugo82 » 24/03/2010, 01:41

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

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

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

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

\( \displaystyle ||u||_*:= \sup_{x\in X\setminus \{ \mathfrak{o}\}} \frac{|\langle u ,x\rangle |}{||x||} =\sup_{x\in \mathcal{B}} |\langle u,x\rangle | \) .

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

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

\( \displaystyle x\in S \Leftrightarrow \begin{cases} \langle u_1 ,x\rangle =\alpha_1 \\ \quad \vdots \\ \langle u_N ,x\rangle =\alpha_N \end{cases} \) .

We refer to the \( \displaystyle N \) equations in the previous formula as the affine equations of \( \displaystyle S \) ; the number \( \displaystyle N \) is usually denoted \( \displaystyle \text{codim} S \) .

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

Finally we remember that the distance of a point \( \displaystyle x\in X \) from a nonempty subset \( \displaystyle Y\subseteq X \) is the non negative number defined by:

\( \displaystyle \text{dist} (x,Y):=\inf_{y\in Y} ||y-x|| \) .

***

Exercise:

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

1. Prove that for each \( \displaystyle x\in X \) we have:

(*) \( \displaystyle \text{dist} (x,\Pi) = \frac{|\langle u,x\rangle -\alpha|}{||u||_*} \) .

2. Show that the proof of (*) becomes easier if \( \displaystyle X \) is a Hilbert space.

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Hints: 1. Use the definition of \( \displaystyle ||u||_* \) and homogeneity to determine which values of \( \displaystyle r>0 \) make \( \displaystyle r\mathcal{B} \cap \Pi_{u,||u||_*} =\emptyset \) 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.
Sono sempre stato, e mi ritengo ancora un dilettante. Cioè una persona che si diletta, che cerca sempre di provare piacere e di regalare il piacere agli altri, che scopre ogni volta quello che fa come se fosse la prima volta. (Freak Antoni)
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Messaggioda gugo82 » 18/04/2010, 02:52

gugo82 ha scritto:Exercise:

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

1. Prove that for each \( \displaystyle x\in X \) we have:

(*) \( \displaystyle \text{dist} (x,\Pi) = \frac{|\langle u,x\rangle -\alpha|}{||u||_*} \) .

Testo nascosto, fai click qui per vederlo
Hints: 1. Use the definition of \( \displaystyle ||u||_* \) and homogeneity to determine which values of \( \displaystyle r>0 \) make \( \displaystyle r\mathcal{B} \cap \Pi_{u,||u||_*} =\emptyset \) 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. :-D

Let \( \displaystyle \Pi_{u,\alpha} :=\{ y\in X:\ \langle u,y\rangle =\alpha\} \) and \( \displaystyle x\in X \) .

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

First we assume \( \displaystyle x=\mathfrak{o} \) (hence \( \displaystyle \alpha \neq 0 \) ): using homogeneity we find:

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

therefore (*) holds for \( \displaystyle x=\mathfrak{o} \) .

Now assume \( \displaystyle x\neq \mathfrak{o} \) : using additivity we obtain \( \displaystyle \Pi_{u,\alpha} -x:=\{ y-x,\ y\in \Pi_{u,\alpha}\} =\Pi_{u,\alpha -\langle u,x \rangle} \) , hence:

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

so that (*) holds.
Sono sempre stato, e mi ritengo ancora un dilettante. Cioè una persona che si diletta, che cerca sempre di provare piacere e di regalare il piacere agli altri, che scopre ogni volta quello che fa come se fosse la prima volta. (Freak Antoni)
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