Limits of the Minkowski distance as related to the generalized mean Given that the Minkowski distance is $$d(X=(x_1,...,x_n),Y=(y_1,...,y_n))=(\sum_{i=1}^n|x_i−y_i|^p)^{1/p}$$ I understand that $$\lim_{p\to\infty}d(X,Y)=\max_{i=1}^n|x_i-y_i|$$ $$\lim_{p\to-\infty}d(X,Y)=\min_{i=1}^n|x_i-y_i|$$ However, I'm not sure what $\lim_{p\to0}d(X,Y)$ is. It seems like the relationship between the Minkowski distance and the generalized mean is $$d(X,Y)=n^{1/p}*mean(|x_1-y_1|,...,|x_n-y_n|)$$ Is this the case? If so, does that mean that $$\lim_{p\to0}d(X,Y)=n^{1/p}*\sqrt[n]{\prod_{i=1}^n|x_i-y_i|}$$ I'm not sure how to get rid of the $1/p$ in $n^{1/p}$.
 A: I'm pretty sure this "norm" is infinite for every set of vectors $\vec{x} = (x_1,...,x_n)$, $\vec{y} = (y_1,...,y_n)$ which have at least two distinct (non-equal dimensions). Here's what I mean: take cases.
1) If $|x_i - y_i| = 0$ for all $i$, then clearly $d(x,y)=0$.
2) If $|x_i - y_i| > 0$ for exactly one entry (say $i_1$), then the "norm" will be
$$d(x,y) = ( |x_{i_1} - y_{i_1}|^p )^{1/p} = |x_{i_1} - y_{i_1}|$$
On the other hand, if $|x_i - y_i| > 0$ for two or more entries, $i$, then we can examine the limit as $p\longrightarrow0$ using $p = 1/k$, for an increasing sequence of integers, $k$. Denote the sequence of norms thus obtained by "$D_k$". Then the sum
$$\sum_{i=1}^n|x_i - y_i|^{1/k}=\big[D_k(x,y)\big]^k$$
will approach the number of nonzero components, as $k\longrightarrow\infty$. In particular, there if there is more than one nonzero term in the sum, then there will be some finite $K > 0$ (notice, it's capital!) such that
$$\sum_{i=1}^n|x_i - y_i|^{1/k}\geq1.5\quad\forall k > K$$
Hence $D_k(x,y) \geq 1.5^k$ for all $k > K$. Taking the limit as $k\longrightarrow\infty$, get that $D_k(x,y)$ goes to infinity.
This has two basic consequences: First, if you don't take the 1/p power at the end, then the "norm" given by
$$d(\vec{x},\vec{y})=\lim_{p\longrightarrow0}\sum_{i=1}^n|x_n-y_n|^p$$
simply counts the number of nonzero elements in the vector difference $\vec{x} - \vec{y}$; but I believe it lacks certain essential properties associated to vector norms
(see http://en.wikipedia.org/wiki/P-norm)
However, if you do apply the $1/p$, then what you end up with is something which probably isn't very useful. It is null if the vectors are identical, equivalent to any other norm if they differ by one dimension; and infinite otherwise.
Best,
Dave Martin
