Re: an inequality from geometry


On Dec 13, 3:49 pm, ulfarsson <ulfars...@xxxxxxxxx> wrote:
On Dec 13, 4:51 pm, kar <k_yeriss...@xxxxxxxxx> wrote:





On Dec 13, 1:47 am, ulfarsson <ulfars...@xxxxxxxxx> wrote:

On Dec 11, 1:08 pm, kar <k_yeriss...@xxxxxxxxx> wrote:

Hello everybody!

(I write the mathematical formulas in LATEX format)

$n>=2$ is a natural number,

$H$ is the hyperplane consisting of all vectors in
$R^n$ the sum of
the components of which is zero, that is

$$H=\{ x\in R^n s.t. \sum_{i=1}^{n} x_i=0 \}$$

let's denote for ease of writing the cube

$$T=[-1,1]^n$$

I am interested if an inequality like the following holds

$$\forall x\in H$$
$$dist(x,T \cap H)\leq C dist(x,T)$$

for a constant $C$ not depending on $x$.

thanks in advance,
karen

Here are some thoughts about the case n=3:

If we restrict to points x = (x_0,y_0,z_0) on H with the properties

z_0 \geq 1,
x_0,y_0 \leq 0

then I claim that C=2 works. Here's why:

The distance from x to T is the distance from x to the "top" of the
box,
i.e., dist(x,T) = z_0 - 1.

The distance form x to T cap H is the distance from x to the line
segment
x+y+1=0, in the plane z=1, which is the intersection of the boundary
of the box
with the plane H.
If we draw a line from x to the origin then that line intersects the
line x+y+1=0,
z=1, in the point (x_0/z_0,y_0/z_0,1) so dist(x,T cap H) is the length
of the vector
(x_0-x_0/z_0,y_0-y_0/z_0,z_0-1).

We are interested in the fraction dist(x,T cap H)/dist(x,T), the
square of which is
[(x_0 - x_0/z_0)^2 + (y_0 - y_0/z_0)^2 + (z_0 - 1)^2]/(z_0 - 1)^2

which simplifies to (x_0/z_0)^2 + (y_0/z_0)^2 +1 =
(x_0/(x_0+y_0))^2 + (y_0/(x_0+y_0))^2 +1

which is bounded by 2.

Henning

Dear Henning, thank you for your consideration of my problem,
but there are some subtle things, for example
for the points that you take, that is

z_0 \geq 1,
x_0,y_0 \leq 0

the distance to $T$ is not always $z_0 - 1$, for example
take the point

x=(-2,-2,4)

the nearest point in $T$ is the point $(-1,-1,1)$!

so the distance dist(x,T)=\sqrt 11

thanks,
kar

Yes, you are right, I forgot to restrict to x_0, y_0 geq -1. Then I
think
what I said in my last message is correct. The case you mention above
can
be done similarly, but my guess is that C=2 will also work there.

Henning

in general you can prove that

$$ dist(x,T\cap H) \leq \sqrt{n^2-n} dist(x,T)$$ so we have $ C =
\sqrt{n^2-n} $

assume that $x \notin T\cap H$ then check for your self that $$
dist(x,T) \geq |max\{x_1,\ldots,x_n\}|-1 $$
and $$ dist(x,T\cap H) \leq (|max\{x_1,\ldots,x_n\}|-1)\sqrt{n^2-n} $$
mixing these last two inequalities you would get the first inequality.

you'll prove it, since this is not a good C and the inequality is not
sharp. However I think the sharpest C would be $\sqrt{\frac{n}{n-1}}$,
but it's a kind of hard to prove it. and it's sharp because the
inequality becomes an equality at the point $(-1,-1,...,-1,n-1)$.

Can I ask you triggered this question for you?

Saman
.

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