Hi,
i'm looking for the proof of minkowski metric in infinite that become the largest of the differences of the coordinates of x and y.
can anybody help me please?
I really don't understand what you are asking for either. However, for the record the Minkowski metric is eitherOriginally Posted by wosci
$\displaystyle \begin{bmatrix} 1 & 0 & 0 & 0\\0 & -1 & 0 & 0\\0 & 0 & -1 & 0\\0 & 0 & 0 & -1 \end{bmatrix}$
or
$\displaystyle \begin{bmatrix} -1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{bmatrix}$
depending on your choice of style. (The upper left corner is the $\displaystyle g_{00}$ entry.)
-Dan
The Minkowski distance between two points $\displaystyle x,\ y$ in $\displaystyle \mathbb{R}^N$ of order $\displaystyle p$ is:Originally Posted by wosci
$\displaystyle
d_p(x,y)=\left[ \sum_1^N |x_i-y_i|^p \right]^{1/p}
$
The problem here is to show that:
$\displaystyle
\lim_{p \to \infty} d_p(x,y)= \max_{i=1,..N} |x_i-y_i|
$
To prove this we observe that if we let $\displaystyle z_i=|x_i-y_i|$ then:
$\displaystyle
d_p(x,y)=D(z,p)=\left[ \sum_1^N z_i^p \right]^{1/p}
$
and if $\displaystyle z_j=\max_{i=1..N} z_i $
$\displaystyle
z_j \le D(z,p) = z_j \left[1+\sum_{i=1,..N;\ i \ne j} \left(\frac{z_i}{z_j}\right)^p \right]^{1/p}$$\displaystyle \le z_j\ \left[1+(N-1) \right]^{1/p}
$
that is:
$\displaystyle
z_j \le D(z,p) \le z_j\ \left[N \right]^{1/p}
$
But the limits of both ends of this chain of inequalities as $\displaystyle p \to \infty$ are the same and equal to $\displaystyle z_j$, hence:
$\displaystyle
\lim_{p \to \infty} D(z,p)=z_j=\max_{i=1..N} z_i
$,
which proves the required result.
RonL