I would like to get some help about the next problem:

I'm trying to understand the following example about metric in my book:

In the set $\mathbb{R}^n$, just as in any non-empty set, metric can be defined in more ways. Lets cite two examples of the generalisations of the previous example (My note: example is $\mathbf{d_2(x, y) = \sqrt{\sum\limits_{i = 1}^n (x_i - y_i)^2}, \,(x, y \in \mathbb{R}^n)}$):

i) Metric $d_p$, $p \ge 1$ is defined with $$d_p(x, y) = \sqrt[p]{\sum\limits_{i = 1}^n |x_i - y_i|^p}, \quad (x, y \in \mathbb{R}^n).$$ (My note: this part I understand and I proved it.)

ii) Metric $d_{\infty}$ is defined with $$d_{\infty} = \max_{1 \le i \le n}|x_i - y_i|, \quad (x, y \in \mathbb{R}^n).$$ (My note: I proved that this function is metric.)

From the inequality $$d_{\infty}(x, y) \le d_p(x, y) \le n^{\frac{1}{p}} d_{\infty}(x, y), \quad x, y \in \mathbb{R}^n, \quad p \ge 1,$$ we have that $d_{\infty}(x, y) = \lim\limits_{p \to \infty} d_p(x, y)$.

I guess that in the book they tried to use Squeeze theorem to prove the last equations, but I can't understand how the got the second part of the inequality, $d_p(x, y) \le n^{\frac{1}{p}} d_{\infty}(x, y)$. I only have this:

$$ d_p(x, y) \le n^{\frac{1}{p}} d_{\infty}(x, y) \iff \sqrt[p]{\sum\limits_{i = 1}^n |x_i - y_i|^p} \le \sqrt[p]{n} \cdot \max_{1 \le i \le n}|x_i - y_i| \iff $$ $$ |x_1 - y_1|^p + \cdot \cdot \cdot + |x_{a - 1} - y_{a - 1}|^p + |x_{a + 1} - y_{a + 1}|^p + \cdot \cdot \cdot + |x_n - y_n|^p \le (n - 1)|x_a - y_a|^p,$$

where $a \in \{1, ... ,n\}$ is such that $\max_{1 \le i \le n}|x_i - y_i| = |x_a - y_a|$.

Please, could you give me some advice about what should I do to prove this inequality?

P.S. I assume that limits will going to need their own question. :) (EDIT: Luckily, there is no need for another question.)

  • 1
    $\begingroup$ Your solution is perfect. Just observe that $|x_i-y_i|^p\le|x_a-y_a|^p$ for each of the $n-1$ summands on the left side, because of the assumption on the index $a$. $\endgroup$
    – Berci
    Mar 21 '19 at 11:24

Note that, if $x(x_1,\ldots,x_n)$ and $y=(y_1,\ldots,y_n)$, then, for each $i\in\{1,\ldots,n\}$, $\lvert x_i-y_i\rvert\leqslant d_\infty(x,y)$. Therefore\begin{align}d_p(x,y)&=\sqrt[p]{\sum_{i=1}^n\lvert x_i-y_i\rvert^p}\\&\leqslant\sqrt[p]{\sum_{i=1}^n\bigl(d_\infty(x,y)\bigr)^p}\\&=\sqrt[p]{n\times\bigl(d_\infty(x,y)\bigr)^p}\\&=\sqrt[p]n\times d_\infty(x,y).\end{align}


Let $|x_a - y_a|=\max_{1 \le i \le n}|x_i - y_i|$. Then

$$d_p(x, y) = \sqrt[p]{\sum_{i = 1}^n |x_i - y_i|^p}\leq \sqrt[p]{\sum_{i = 1}^n |x_a - y_a|^p}\\=\sqrt[p]{n\cdot|x_a - y_a|^p}=\sqrt[p]{n}|x_a - y_a|.$$


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