User:Dfeuer/General Euclidean Metrics are Topologically Equivalent

Proof
First we are going to show that:


 * $\ds \map {d_1} {x, y} \ge \map {d_2} {x, y} \ge \cdots \ge \map {d_r} {x, y} \ge \cdots \ge \map {d_\infty} {x, y} \ge n^{-1} \map {d_1} {x, y}$

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

Let $r \in \N: r \ge 1$.

Let $d_r$ be the metric defined as $\ds \map {d_r} {x, y} = \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^r}^{\frac 1 r}$.


 * First we wish to show that $\map {d_\infty} {x, y} \ge n^{-1} \map {d_1} {x, y}$.

By the definitions of $d_\infty$ and $d_1$, this is equivalent to showing that


 * $n \max_{i \mathop = 1}^n \size {x_i - y_i} \ge \sum_{i \mathop = 1}^n \size {x_i - y_i}$

But this holds trivially.


 * Now we wish to show that that $\forall r \in \N: \map {d_r} {x, y} \ge \map {d_{r + 1} } {x, y}$.

That is, that:
 * $\ds \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^r}^{\frac 1 r} \ge \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^{r + 1} }^{\frac 1 {r + 1} }$

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

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

Then the problem reduces to the equivalent one of showing that:
 * $\ds \paren {\sum_{i \mathop = 1}^{n - 1} \size {x_i - y_i}^r}^{\frac 1 r} \ge \paren {\sum_{i \mathop = 1}^{n - 1} \size {x_i - y_i}^{r + 1} }^{\frac 1 {r + 1} }$

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

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

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

Let $\ds \map f r = \paren {\sum_{i \mathop = 1}^n s_i^r}^{1/r}$.

Let $\ds u = \sum_{i \mathop = 1}^n s_i^r, v = \frac 1 r$.

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

So:

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

For the same reason, $\ds \forall j: \frac{s_j^r} {\sum_{i \mathop = 1}^n s_i^r} < 1$.

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

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

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.

Hence $\forall r \in \N: \map {d_r} {x, y} \ge \map {d_{r + 1} } {x, y}$.

Since $\forall r \in \N: \map {d_r} {x, y} \ge \map {d_{r + 1} } {x, y}$:
 * $\forall r \in \N: n^{-1} \map {d_r} {x, y} \ge n^{-1} \map {d_{r + 1} } {x, y}$

Since $\map {d_\infty} {x, y} \ge n^{-1} \map {d_1} {x, y}$, we see by transitivity that:
 * $\forall r \in \N: n^{-1} \map {d_\infty} {x, y} \ge n^{-1} \map {d_r} {x, y}$

Thus:
 * $\forall r \in \N: \map {d_r} {x, y} \ge n^{-1} \map {d_\infty} {x, y} \ge n^{-1} \map {d_r} {x, y}$

Therefore, $d_r$ and $d_\infty$ are Lipschitz equivalent for each $r$.