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## They are metrics

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Tolaso J Kos
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### They are metrics

Prove that $\displaystyle d\left ( a_n, b_n \right )=\sqrt{\sum_{n=1}^{\infty}\left ( a_n-b_n \right )^2}$ defines a metric in $\ell^2$.

In continuity prove that $\displaystyle d\left ( a_n, b_n \right )=\sum_{n=1}^{\infty}\frac{\left | a_n-b_n \right |}{2^n}$ defines a metric in $H^{\infty}$, where $H^{\infty}$ denotes the set of all sequences $a_n, \; n \in \mathbb{N}$ such that $\left | a_n \right |\leq 1$.
Imagination is much more important than knowledge.
Papapetros Vaggelis
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### Re: They are metrics

$\displaystyle{\ell^2=\left\{\left(a_{n}\right)_{n\in\mathbb{N}}: a_{n}\in\mathbb{R}\,,\forall\,n\in\mathbb{N}\,\land \sum_{n=1}^{\infty}a_{n}^2<\infty\right\}}$ .

Obviously, $\displaystyle{\mathbb{O}\in\ell^2\,\,,\left(\dfrac{1}{n}\right)_{n\in\mathbb{N}}\in \ell^2}$ and thus $\displaystyle{\ell^2\neq \varnothing}$ .

The $\displaystyle{\mathbb{R}}$ - vector space $\displaystyle{\left(\ell^2,+,\cdot\right)}$ , where:

$\displaystyle{+:\ell^2\times \ell^2\longrightarrow \ell^2 \left(\left(x_{n}\right)_{n\in\mathbb{N}},\left(y_{n}\right)_{n\in\mathbb{N}}\right)\mapsto \left(x_{n}+y_{n}\right)_{n\in\mathbb{N}}}$

$\displaystyle{\cdot:\mathbb{R}\times \ell^2\longrightarrow \ell^2 \left(k,\left(x_{n}\right)_{n\in\mathbb{N}}\right)\mapsto \left(k\,x_{n}\right)_{n\in\mathbb{N}}}$

equuiped with the function

$\displaystyle{||\cdot||_{2}:\ell^2\longrightarrow \mathbb{R}\,,x=\left(x_{n}\right)_{n\in\mathbb{N}} \mapsto ||x||_{2}=\sqrt{\sum_{n=1}^{\infty}\left|x_{n}\right|^2}}$

is a normed space. The triangular property is $\displaystyle{\rm{Minkowski}}$ 's inequality for $\displaystyle{p=2}$ and the other two are quite easy.

So, the function $\displaystyle{d:\ell^2\times \ell^2\longrightarrow \mathbb{R}}$ given by

$\displaystyle{d\,\left(\left(a_{n}\right)_{n\in\mathbb{N}},\left(b_{n}\right)_{n\in\mathbb{N}}\right)=\sqrt{\sum_{n=1}^{\infty}\left(a_{n}-b_{n}\right)^2}=||a-b||_{2}}$

where: $\displaystyle{a=\left(a_{n}\right)_{n\in\mathbb{N}}\,,b=\left(b_{n}\right)_{n\in\mathbb{N}}}$ is a metric in $\displaystyle{\ell^2}$ .

For the second part check here (http://www.mathimatikoi.org/forum/viewt ... ?f=4&t=272" onclick="window.open(this.href);return false;)
Tsakanikas Nickos
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### Re: They are metrics

A similar question to the second one suggested by Mr Apostolos is the following.

Let $\mathcal{S}$ be the set of all sequences of real numbers. Let $\displaystyle \left( m_{k} \right)_{k \in \mathbb{N}}$ be a sequence of positive real numbers such that $\displaystyle \sum_{k=1}^{\infty}m_{k} < +\infty$. If $\displaystyle x=\left( \xi_{k} \right)_{k \in \mathbb{N}} , y=\left( \eta_{k} \right)_{k \in \mathbb{N}} \in \mathcal{S}$, show that the function $\displaystyle d$ defined by
$\displaystyle d(x,y) = \sum_{k=1}^{\infty}m_{k}\frac{ |\xi_{k}-\eta_{k}| }{ 1 + |\xi_{k}-\eta_{k}| }$is a metric on $\mathcal{S}$.
Tolaso J Kos
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### Re: They are metrics

To continue Nickos's question what is the diameter of $(\mathcal{S}, d)$?
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Papapetros Vaggelis
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### Re: They are metrics

Hello everybody.

Let $\displaystyle{x=\left(\xi_{k}\right)_{k\in\mathbb{N}}\,\,,y=\left(\eta_{k}\right)_{k\in\mathbb{N}}\in S}$ . We have that :

$\displaystyle{\forall\,k\in\mathbb{N}:\dfrac{\left|\xi_{k}-\eta_{k}\right|}{1+\left|\xi_{k}-\eta_{k}\right|} <1}$ and since

$\displaystyle{m_{k}>0\,,\forall\,k\in\mathbb{N}}$ we get:

$\displaystyle{m_{k}\,\dfrac{\left|\xi_{k}-\eta_{k}\right|}{1+\left|\xi_{k}-\eta_{k}\right|}<m_{k}(>0)\,,\forall\,k\in\mathbb{N}}$ .

Due to the fact that $\displaystyle{\sum_{k=1}^{\infty}m_{k}<\infty}$ ,

we conclude that $\displaystyle{\sum_{k=1}^{\infty}m_{k}\,\dfrac{\left|\xi_{k}-\eta_{k}\right|}{1+\left|\xi_{k}-\eta_{k}\right|}<\infty}$ .

Therefore, $\displaystyle{d\,(x,y)\in\mathbb{R}}$

and since $\displaystyle{m_{k}\,\dfrac{\left|\xi_{k}-\eta_{k}\right|}{1+\left|\xi_{k}-\eta_{k}\right|}\geq 0\,,\forall\,k\in\mathbb{N}}$

we have that $\displaystyle{d\,(x,y)\geq 0\,\,(I)}$ and thus the function $\displaystyle{d}$ is well defined.

Obviously, $\displaystyle{d(x,x)=0\,,\forall\,x=\left(\xi_{k}\right)_{k\in\mathbb{N}}\in S}$ .

Suppose that $\displaystyle{d\,(x,y)=0}$ for some $\displaystyle{x=\left(\xi_{k}\right)_{k\in\mathbb{N}}\,,y=\left(\eta_{k}\right)_{k\in\mathbb{N}}\in S}$ .

Then, $\displaystyle{\sum_{k=1}^{\infty}m_{k}\,\dfrac{\left|\xi_{k}-\eta_{k}\right|}{1+\left|\xi_{k}-\eta_{k}\right|}=0}$ . The sequence

$\displaystyle{b_{k}=\sum_{i=1}^{k}m_{i}\,\dfrac{\left|\xi_{i}-\eta_{i}\right|}{1+\left|\xi_{i}-\eta_{i}\right|}\,,k\in\mathbb{N}}$ is strictly

increasing, positive and bounded, so :

$\displaystyle{\sup\,\left\{b_{k}: k\in\mathbb{N}\right\}=\lim_{k\to \infty}b_{k}=\sum_{k=1}^{\infty}m_{k}\,\dfrac{\left|\xi_{k}-\eta_{k}\right|}{1+\left|\xi_{k}-\eta_{k}\right|}=0}$ .

If $\displaystyle{b_{i}>0}$ for some $\displaystyle{i\in\mathbb{N}}$ , then :

$\displaystyle{0=\sup\,\left\{b_{k}: k\in\mathbb{N}\right\}\geq b_{i}\implies b_{i}\leq 0}$, a contradiction, so:

\displaystyle{\begin{aligned} \forall\,k\in\mathbb{N}: b_{k}=0&\implies \forall\,k\in\mathbb{N}: m_{k}\cdot \left|\xi_{k}-\eta_{k}\right|=0\\&\implies \forall\,k\in\mathbb{N}: \left|\xi_{k}-\eta_{k}\right|=0\\&\implies \forall\,k\in\mathbb{N}: \xi_{k}=\eta_{k}\\&\implies x=y\end{aligned}} .

So, $\displaystyle{\forall\,x\,,y\in S: d\,(x,y)=0\iff x=y\,\,(II)}$ .

If $\displaystyle{x=\left(\xi_{k}\right)_{k\in\mathbb{N}}\,,y=\left(\eta_{k}\right)_{k\in\mathbb{N}}\in S}$ , then:

$\displaystyle{d\,(x,y)=\sum_{k=1}^{\infty}m_{k}\,\dfrac{\left|\xi_{k}-\eta_{k}\right|}{1+\left|\xi_{k}-\eta_{k}\right|}=\sum_{k=1}^{\infty}m_{k}\,\dfrac{\left|\eta_{k}-\xi_{k}\right|}{1+\left|\eta_{k}-\xi_{k}\right|}=d\,(y,x)\,\,(III)}$ .

The triangular property

Firstly, we define $\displaystyle{f:\left[0,+\infty\right)\longrightarrow \mathbb{R}}$ by $\displaystyle{f(x)=\dfrac{x}{1+x}}$.

We observe that $\displaystyle{f(x)=1-\dfrac{1}{1+x}\,,x\geq 0}$ and then for each $\displaystyle{x\,,y\geq 0}$ holds:

$\displaystyle{x<y\implies 1+x<1+y\implies \dfrac{1}{1+x}>\dfrac{1}{1+y}\implies 1-\dfrac{1}{1+x}=f(x)<1-\dfrac{1}{1+y}=f(y)}$,

which means that the function $\displaystyle{f}$ is strictly increasing at $\displaystyle{\left[0,+\infty\right)}$ .

Let $\displaystyle{x=\left(\xi_{k}\right)_{k\in\mathbb{N}}\,,y=\left(\eta_{k}\right)_{k\in\mathbb{N}}\,,z=\left(a_{k}\right)_{k\in\mathbb{N}}}$

be sequences of real numbers. Then, for every $\displaystyle{k\in\mathbb{N}}$ holds :

$\displaystyle{0\leq \left|\xi_{k}-\eta_{k}\right|\leq \left|\xi_{k}-a_{k}\right|+\left|a_{k}-\eta_{k}\right|}$, and thus:

$\displaystyle{f\,\left(\left|\xi_{k}-\eta_{k}\right|\right)\leq f\,\left(\left|\xi_{k}-a_{k}\right|+\left|a_{k}-\eta_{k}\right|\right)}$

or equivalently:

$\displaystyle{\dfrac{\left|\xi_{k}-\eta_{k}\right|}{1+\left|\xi_{k}-\eta_{k}\right|}\leq \dfrac{\left|\xi_{k}-a_{k}\right|+\left|a_{k}-\eta_{k}\right|}{1+\left|\xi_{k}-a_{k}\right|+\left|a_{k}-\eta_{k}\right|}}$

where :

$\displaystyle{\dfrac{\left|\xi_{k}-a_{k}\right|+\left|a_{k}-\eta_{k}\right|}{1+\left|\xi_{k}-a_{k}\right|+\left|a_{k}-\eta_{k}\right|}=}$

$\displaystyle{=\dfrac{\left|\xi_{k}-a_{k}\right|}{1+\left|\xi_{k}-a_{k}\right|+\left|a_{k}-\eta_{k}\right|}+\dfrac{\left|a_{k}-\eta_{k}\right|}{1+\left|\xi_{k}-a_{k}\right|+\left|a_{k}-\eta_{k}\right|}}$

$\displaystyle{\leq \dfrac{\left|\xi_{k}-a_{k}\right|}{1+\left|\xi_{k}-a_{k}\right|}+\dfrac{\left|a_{k}-\eta_{k}\right|}{1+\left|a_{k}-\eta_{k}\right|}}$

so:

$\displaystyle{\dfrac{\left|\xi_{k}-\eta_{k}\right|}{1+\left|\xi_{k}-\eta_{k}\right|}\leq \dfrac{\left|\xi_{k}-a_{k}\right|}{1+\left|\xi_{k}-a_{k}\right|}+\dfrac{\left|a_{k}-\eta_{k}\right|}{1+\left|a_{k}-\eta_{k}\right|}\implies}$

$\displaystyle{\implies m_{k}\,\dfrac{\left|\xi_{k}-\eta_{k}\right|}{1+\left|\xi_{k}-\eta_{k}\right|}\leq m_{k}\,\dfrac{\left|\xi_{k}-a_{k}\right|}{1+\left|\xi_{k}-a_{k}\right|}+m_{k}\,\dfrac{\left|a_{k}-\eta_{k}\right|}{1+\left|a_{k}-\eta_{k}\right|}}$

$\displaystyle{\implies \sum_{k=1}^{\infty}m_{k}\,\dfrac{\left|\xi_{k}-\eta_{k}\right|}{1+\left|\xi_{k}-\eta_{k}\right|}\leq \sum_{k=1}^{\infty}\left[m_{k}\,\dfrac{\left|\xi_{k}-a_{k}\right|}{1+\left|\xi_{k}-a_{k}\right|}+m_{k}\,\dfrac{\left|a_{k}-\eta_{k}\right|}{1+\left|a_{k}-\eta_{k}\right|}\right]}$

$\displaystyle{\implies d\,(x,y)\leq \sum_{k=1}^{\infty}m_{k}\,\dfrac{\left|\xi_{k}-a_{k}\right|}{1+\left|\xi_{k}-a_{k}\right|}+\sum_{k=1}^{\infty}m_{k}\,\dfrac{\left|a_{k}-\eta_{k}\right|}{1+\left|a_{k}-\eta_{k}\right|}}$

$\displaystyle{\implies d\,(x,y)\leq d\,(x,z)+d\,(z,y)\,\,\,(IV)}$ .

According to the relations $\displaystyle{(I)\,,(II)\,,(III)\,,(IV)}$ we have that the function $\displaystyle{d}$ is a metric on $\displaystyle{S}$ .