User:Dfeuer/General Euclidean Metrics are Topologically Equivalent

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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:

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

$K > 0$ 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 $r$.

$\blacksquare$