### Point-hyperplane distance in normed spaces

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

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

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

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

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

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

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

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

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

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

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

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

***

Exercise:

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

1. Prove that for each $$x\in X$$ we have:

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

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

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

gugo82
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gugo82 ha scritto:Exercise:

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

1. Prove that for each $$x\in X$$ we have:

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

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

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

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

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

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

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

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

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

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)

gugo82
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Messaggio: 5375 di 17489
Iscritto il: 13/10/2007, 00:58
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