P-Product Metrics on Real Vector Space are Topologically Equivalent

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Theorem

For $n \in \N$, let $\R^n$ be an Euclidean space.

Let $p \in \R_{\ge 1}$.

Let $d_p$ be the $p$-product metric on $\R^n$.

Let $d_\infty$ be the Chebyshev distance on $\R^n$.


Then $d_p$ and $d_\infty$ are topologically equivalent.


Proof

Let $r, t \in \R_{\ge 1}$.

Without loss of generality, assume that $r \le t$.

For all $x, y \in \R^n$, we are going to show that:

$\map {d_r} {x, y} \ge \map {d_\infty} {x, y} \ge n^{-1} \map {d_r} {x, y}$

Then we can demonstrate Lipschitz equivalence between all of these metrics, from which topological equivalence follows.

Let $d_r$ be the metric defined as:

$\ds \map {d_r} {x, y} = \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^r}^{1/r}$


Inequalities for Chebyshev Distance

By definition of the Chebyshev distance on $\R^n$, we have:

$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \size {x_i - y_i}$

where $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$.

Let $j$ be chosen so that:

$\ds \size {x_j - y_j} = \max_{i \mathop = 1}^n \size {x_i - y_i}$

Then:

\(\ds \map {d_\infty} {x, y}\) \(=\) \(\ds \paren {\size {x_j - y_j}^p}^{1/p}\)
\(\ds \) \(\le\) \(\ds \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^p}^{1/p}\) Power of Maximum is not Greater than Sum of Powers
\(\ds \) \(=\) \(\ds \map {d_p} {x, y}\)
\(\ds \) \(\le\) \(\ds \paren {n \size {x_j - y_j}^p}^{1/p}\) Sum of $r$ Powers is not Greater than $r$ times Power of Maximum
\(\ds \) \(=\) \(\ds n^{1/p} \size {d_\infty} {x, y}\)

$\Box$


Inequality for General Case

We show that $\ds \map {d_r} {x, y} \ge \map {d_t} {x, y}$, which is equivalent to proving that:

$\ds \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^r}^{1/r} \ge \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^t}^{1/t}$


Let $\forall i \in \closedint 1 n: s_i = \size {x_i - y_i}$.

Suppose $s_k = 0$ for some $k \in \closedint 1 n$.


Although this article appears correct, it's inelegant. There has to be a better way of doing it.
In particular: Putting off the analysis of s_k being 0 until near the end of the proof should shorten it without losing any clarity.
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Then the problem reduces to the equivalent one of showing that:

$\ds \paren {\sum_{i \mathop = 1}^{n - 1} \size {x_i - y_i}^r}^{1/r} \ge \paren {\sum_{i \mathop = 1}^{n - 1} \size {x_i - y_i}^t}^{1/t}$

that is, of reducing the index by $1$.

Note that when $n = 1$, from simple algebra $\map {d_r} {x, y} = \map {d_t} {x, y}$.


So, let us start with the assumption that $\forall i \in \closedint 1 n: s_i > 0$.

Let $\ds u = \sum_{i \mathop = 1}^n \size {x_i - y_i}^r = \sum_{i \mathop = 1}^n s_i^r$, and $v = \dfrac 1 r$.

From Derivative of Function to Power of Function‎, $\map {D_r} {u^v} = v u^{v - 1} \map {D_r} u + u^v \ln u \map {D_r} v$.

Here:

$\ds \map {D_r} u = \sum_{i \mathop = 1}^n s_i^r \ln s_i$ from Derivative of Exponential Function and Sum Rule for Derivatives
$\map {D_r} v = - \dfrac 1 {r^2}$ from Power Rule for Derivatives

In the case where $r = 1$, we have:

$\map {D_r} {u^v} = 0$

When $r > 1$, we have:

\(\ds \map {D_r} {\paren {\sum_{i \mathop = 1}^n s_i^r}^{1 / r} }\) \(=\) \(\ds \dfrac 1 r \paren {\sum_{i \mathop = 1}^n s_i^r}^{1 / \paren {r - 1} } \paren {\sum_{i \mathop = 1}^n s_i^r \ln s_i} - \dfrac {\paren {\sum_{i \mathop = 1}^n s_i^r}^{1/r} \map \ln {\sum_{i \mathop = 1}^n s_i^r} } {r^2}\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {\sum_{i \mathop = 1}^n s_i^r}^{1 / r} } r \paren {\dfrac {\sum_{i \mathop = 1}^n s_i^r \ln s_i} {\sum_{i \mathop = 1}^n s_i^r} - \dfrac {\map \ln {\sum_{i \mathop = 1}^n s_i^r} } r}\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {\sum_{i \mathop = 1}^n s_i^r}^{1 / r} } r \paren {\dfrac {r \paren {\sum_{i \mathop = 1}^n s_i^r \ln s_i} - \paren {\sum_{i \mathop = 1}^n s_i^r} \map \ln {\sum_{i \mathop = 1}^n s_i^r} } {r \paren {\sum_{i \mathop = 1}^n s_i^r} } }\)
\(\ds \) \(=\) \(\ds K \paren {r \paren {\sum_{i \mathop = 1}^n s_i^r \ln s_i} - \paren {\sum_{i \mathop = 1}^n s_i^r} \map \ln {\sum_{i \mathop = 1}^n s_i^r} }\) where $\ds K = \dfrac {\paren {\sum_{i \mathop = 1}^n s_i^r}^{1 / r} } {r^2 \paren {\sum_{i \mathop = 1}^n s_i^r} } > 0$
\(\ds \) \(=\) \(\ds K \paren {\sum_{i \mathop = 1}^n s_i^r \map \ln {s_i^r} - \paren {\sum_{i \mathop = 1}^n s_i^r} \map \ln {\sum_{i \mathop = 1}^n s_i^r} }\) Logarithms of Powers
\(\ds \) \(=\) \(\ds K \paren {\sum_{j \mathop = 1}^n \paren {s_j^r \paren {\map \ln {s_j^r} - \map \ln {\sum_{i \mathop = 1}^n s_i^r} } } }\)
\(\ds \) \(=\) \(\ds K \paren {\sum_{j \mathop = 1}^n \paren {s_j^r \map \ln {\frac {s_j^r} {\sum_{i \mathop = 1}^n s_i^r} } } }\)

where $K > 0$ because all of $s_i, r > 0$.

For the same reason, $\ds \dfrac{s_j^r} {\sum_{i \mathop = 1}^n s_i^r} < 1$ for all $j \in \set {1, \ldots, n}$.

From Logarithm of 1 is 0 and Logarithm is Strictly Increasing, their logarithms are therefore negative.

So for $r > 1$:

$\ds \map {D_r} {\paren {\sum_{i \mathop = 1}^n s_i^r}^{1 / r} } < 0$

So, from Derivative of Monotone Function, it follows that (given the conditions on $r$ and $s_i$) $\ds \paren {\sum_{i \mathop = 1}^n s_i^r}^{1 / r}$ is decreasing.


As we assumed $r \le t$, we have $\map {d_r} {x, y} \ge \map {d_t} {x, y}$.

$\Box$


When we combine the inequalities, we have:


Although this article appears correct, it's inelegant. There has to be a better way of doing it.
In particular: In fact, this inequality (for real $r \ge 1$) can be established quite easily without the humongous calculation.
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$\map {d_r} {x, y} \ge \map {d_\infty} {x, y} \ge n^{-1} \map {d_1} {x, y} \ge n^{-1} \map {d_r} {x, y}$

Therefore, $d_r$ and $d_\infty$ are Lipschitz equivalent for all $r \in \R_{\ge 1}$.

$\blacksquare$


Sources

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