User:Anghel/Sandbox

Theorem
For $n \in \N_{>1}$, let $\R^n$ be an Euclidean space.

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

Let $d_r$, $d_t$ and $d_\infty$ be general Euclidean metrics on $\R^n$.

Then $d_r$, $d_t$ and $d_\infty$ are topologically equivalent.

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

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


 * $\displaystyle d_r \left({x, y}\right) \ge d_t \left({x, y}\right) \ge d_\infty \left({x, y}\right) \ge n^{-1} d_1 \left({x, y}\right) \ge n^{-1} d_r \left({x, y}\right)$

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

Let $d_r$ be the metric defined as $\displaystyle d_r \left({x, y}\right) = \left({\sum_{i \mathop = 1}^n \left|{x_i - y_i}\right|^r}\right)^{1/r}$.

Inequalities for Infinite Case
First, we show that $d_t \left({x, y}\right) \ge d_\infty \left({x, y}\right) \ge n^{-1} d_1 \left({x, y}\right)$.

By definition of $d_\infty$, we have $\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert}$.

With $i'$ chosen so $\left\vert{x_{i'} - y_{i'} }\right\vert = \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert}$, it follows that:


 * $\displaystyle d_t \left({x, y}\right) = \left({ \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert^t }\right)^{1/t} \ge \left({ \left\vert{x_{i'} - y_{i'} }\right\vert^t }\right)^{1/t} \ge = \left\vert{x_{i'} - y_{i'} }\right\vert = \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert} = d_\infty \left({x, y}\right)$

The second inequality holds trivially:


 * $\displaystyle n \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert} \ge  \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert$

Inequality for General Case
Next, we show that $\displaystyle d_r \left({x, y}\right) \ge d_t \left({x, y}\right)$, which is equivalent to proving that:


 * $\displaystyle \left({\sum_{i \mathop = 1}^n \left|{x_i - y_i}\right|^r}\right)^{1/r} \ge \left({\sum_{i \mathop = 1}^n \left|{x_i - y_i}\right|^t}\right)^{1/t}$

Let $\forall i \in \left[{1 \,.\,.\, n}\right]: s_i = \left|{x_i - y_i}\right|$.

Suppose $s_k = 0$ for some $k \in \left[{1 \,.\,.\, n}\right]$.

Then the problem reduces to the equivalent one of showing that:
 * $\displaystyle \left({\sum_{i \mathop = 1}^{n-1} \left|{x_i - y_i}\right|^r}\right)^{1/r} \ge \left({\sum_{i \mathop = 1}^{n-1} \left|{x_i - y_i}\right|^t }\right)^{1/t}$

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

Note that when $n = 1$, from simple algebra $d_r \left({x, y}\right) = d_t \left({x, y}\right)$.

So, let us start with the assumption that $\forall i \in \left[{1 \,.\,.\, n}\right]: s_i > 0$.

Let $\displaystyle u = \sum_{i \mathop = 1}^n \left|{x_i - y_i}\right|^r = \sum_{i \mathop = 1}^n s_i^r$, and $v = \dfrac 1 r$.

From Derivative of Powers of Functions‎, $D_r \left({u^v}\right) = v u^{v-1} D_r \left({u}\right) + u^v \ln u D_r \left({v}\right)$.

Here:
 * $\displaystyle D_r \left({u}\right) = \sum_{i \mathop = 1}^n s_i^r \ln s_i$ from Derivative of Exponential Function and Sum Rule for Derivatives
 * $\displaystyle D_r \left({v}\right) = - \frac 1 {r^2}$ from Power Rule for Derivatives

In the case where $r=1$, we have $D_r \left({u^v}\right) = 0$.

When $r > 1$, we have:

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

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

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

So for $r > 1$:


 * $\displaystyle D_r \left({\left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r}}\right) < 0$

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

As we assumed $r \le t$, we have $d_r \left({x, y}\right) \ge d_t \left({x, y}\right)$.

When we combine the inequalities, we have:


 * $d_r \left({x,y}\right) \ge d_t \left({x,y}\right) \ge d_\infty \left({x, y}\right) \ge n^{-1} d_1 \left({x,y}\right) \ge n^{-1} d_r \left({x, y}\right)$

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