Metric Function

Real Analysis
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Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Metric Function

#1

Post by Papapetros Vaggelis »

Let \(\displaystyle{H^{\infty}:=\left\{a=\left(a_{n}\right)_{n\in\mathbb{N}}: a_{n}\in\mathbb{R}\ \land \left|a_{n}\right|\leq 1\ \forall n\in\mathbb{N}\right\}}\)

Prove that the set \(\displaystyle{H^{\infty}}\) with the function

\(\displaystyle{d\left(a_{n},b_{n}\right)=\sum_{n=1}^{\infty}\frac{\left|a_{n}-b_{n}\right|}{2^{n}}\,\,,\left(a_{n},b_{n}\right)\in H^{\infty}\times H^{\infty}}\)

is metric space.
tziaxri
Posts: 7
Joined: Mon Nov 09, 2015 4:32 pm

Re: Metric Function

#2

Post by tziaxri »

Let \(\displaystyle{H^{\infty}:=\left\{a=\left(a_{n}\right)_{n\in\mathbb{N}}: a_{n}\in\mathbb{R}\ \land \left|a_{n}\right|\leq 1\ \forall n\in\mathbb{N}\right\}}\)

and the function \(\displaystyle{d\left(a_{n},b_{n}\right)=\sum_{n=1}^{\infty}\frac{\left|a_{n}-b_{n}\right|}{2^{n}}\,\,,\left(a_{n},b_{n}\right)\in H^{\infty}\times H^{\infty}}\)

At first, \(\displaystyle\forall \left(a_{n}\right)_{n\in\mathbb{N}}\left(b_{n}\right)_{n\in\mathbb{N}}\in\mathbb{H^{\infty}}\) we have

\(\displaystyle{d\left(a_{n},b_{n}\right)=\sum_{n=1}^{\infty}\frac{\left|a_{n}-b_{n}\right|}{2^{n}}}\ \leq {\sum_{n=1}^{\infty}\frac{2}{2^{n}}}={2}<{\infty}\).

The properties \(\displaystyle{d\left(a_{n},b_{n}\right)=\sum_{n=1}^{\infty}\frac{\left|a_{n}-b_{n}\right|}{2^{n}}}={0}\Leftrightarrow \left(a_{n}\right)_{n\in\mathbb{N}}=(b_{n})_{n\in\mathbb{N}}\)

and \(\displaystyle{d\left(a_{n},b_{n}\right)}=\) \(\displaystyle{d\left(b_{n},a_{n}\right)}\) ,\({n\in\mathbb{N}}\) , are direct.

Now,for the triangular property of d , it is enough to show that for each \(\displaystyle\left(a_{n}\right)_{n\in\mathbb{N}},(b_{n})_{n\in\mathbb{N}}, (c_{n})_{n\in\mathbb{N}}\in\ H^{\infty}\) ,

holds: \(\displaystyle{d\left(a_{n},b_{n}\right)}\leq\) \(\displaystyle{d\left(a_{n},c_{n}\right)} +{d\left(c_{n},b_{n}\right)}\)

\(\displaystyle\Rightarrow {\sum_{n=1}^{\infty}\frac{\left|a_{n}-b_{n}\right|}{2^{n}}}\leq{\sum_{n=1}^{\infty}\frac{\left|a_{n}-c_{n}\right|}{2^{n}}} + {\sum_{n=1}^{\infty}\frac{\left|c_{n}-b_{n}\right|}{2^{n}}} \)

\(\displaystyle\Rightarrow {\sum_{n=1}^{\infty}\frac{\left|a_{n}-b_{n}\right|}{2^{n}}}\leq {\sum_{n=1}^{\infty}\frac{\left|a_{n}-c_{n}\right|+\left|c_{n}-b_{n}\right|}{2^{n}}}\)
\(\displaystyle\Rightarrow {\sum_{n=1}^{\infty}\frac{\left|a_{n}-c_{n}\right|+\left|c_{n}-b_{n}\right|-\left|a_{n}-b_{n}\right|}{2^{n}}}\geq{0}\)
Considering the triangle inequality for \(\displaystyle\left(a_{n}\right)_{n\in\mathbb{N}},(b_{n})_{n\in\mathbb{N}},(c_{n})_{n\in\mathbb{N}}\) we have
\(\displaystyle{\left|a_{n}-b_{n}\right|}\leq{\left|a_{n}-c_{n}\right|}+{\left|c_{n}-b_{n}\right|} \)

which proves the above relation.

So, the set \(\displaystyle{H^{\infty}:=\left\{a=\left(a_{n}\right)_{n\in\mathbb{N}}: a_{n}\in\mathbb{R}\ \land \left|a_{n}\right|\leq 1\ \forall n\in\mathbb{N}\right\}}\)

with the function \(\displaystyle{d\left(a_{n},b_{n}\right)}={\sum_{n=1}^{\infty}\frac{\left|a_{n}-b_{n}\right|}{2^{n}}}\) is metric space.
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