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 Post subject: They are metrics
PostPosted: Thu Jul 14, 2016 1:30 pm 
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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 \( \left | a_n \right |\leq 1 \).

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 Post subject: Re: They are metrics
PostPosted: Thu Jul 14, 2016 1:30 pm 
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\(\displaystyle{\ell^2=\left\{\left(a_{n}\right)_{n\in\mathbb{N}}: a_{n}\in\mathbb{R}\,,\forall\,n\in\mathbb{N}\,\land \sum_{n=1}^{\infty}a_{n}^2<\infty\right\}}\) .

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 (viewtopic.php?f=4&t=272)


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 Post subject: Re: They are metrics
PostPosted: Thu Jul 14, 2016 1:32 pm 
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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 \( \mathcal{S} \).


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 Post subject: Re: They are metrics
PostPosted: Thu Jul 14, 2016 1:32 pm 
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To continue Nickos's question what is the diameter of \( (\mathcal{S}, d) \)?

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 Post subject: Re: They are metrics
PostPosted: Thu Jul 14, 2016 1:33 pm 
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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 \(\displaystyle{S}\) .


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