Is $\ell^\infty$ a Compact Metric Space?

[ozkan12]

Let $X={\ell}^{\infty}:=\left\{u\in{\ell}^{2}\left(R\right):\left| {u}_{k} \right|\le\frac{1}{k}\right\}$ and $T:{\ell}^{\infty}\to{\ell}^{\infty}$, defined by $T{u}_{k}=\frac{k}{k+1}{u}_{k}.$. Then

1) ${\ell}^{\infty}$ is a compact metric space,

2) $T$ is not a contraction,

3) $T$ is not a contractive,

4) Fix(T)=0, the null sequence,

5) The Picard İteration converges (uniformly) to the null sequence, i.e.,

${T}^{n}{u}_{k}^{(0)}=({\frac{k}{k+1}})^n {u}^\left(0\right)_{k}\to 0$

for any ${u}^\left(0\right)_{k}\in {\ell}^{\infty}$

Please prove 1, 2, 4 and 5...And please can you give definition of null sequence ? Thank you for your attention...Best wishes :)

 

[Jhoneine]

1) ${\ell}^{\infty}$ is a compact metric space,Proof: Let $\left\{ {u^n} \right\}_{n=1}^{\infty}$ be a sequence in ${\ell}^{\infty}$, then for each $u^n$, there exists a natural number $N_n$ such that $\left| {u^n_k} \right|\le\frac{1}{N_n}$ for all $k\ge N_n$. Thus,$||u^n-u^m||_2=\sqrt{\sum_{k=1}^{\infty}\left| {u^n_k-u^m_k} \right|^2}\le\sqrt{\sum_{k=N_n}^{\infty}\left| {u^n_k-u^m_k} \right|^2}\le\sqrt{\sum_{k=N_n}^{\infty}\left(\frac{1}{N_n}+\frac{1}{N_m}\right)^2}\le\sqrt{\sum_{k=N_n}^{\infty}\frac{2}{N_n N_m}}$Thus, the sequence $\left\{ {u^n} \right\}_{n=1}^{\infty}$ is Cauchy and hence converges since ${\ell}^{\infty}$ is complete. Therefore, ${\ell}^{\infty}$ is a compact metric space. 2) $T$ is not a contraction,Proof: Let $u^1,u^2\in{\ell}^{\infty}$ be two arbitrary elements. Then,$||T(u^1)-T(u^2)||_2=\sqrt{\sum_{k=1}^{\infty}\left| \frac{k}{k+1}{u^1_k}-\frac{k}{k+1}{u^2_k} \right|^2}=\sqrt{\sum_{k=1}^{\infty}\frac{k^2}{(k+1)^2

 

[math771]

1) To prove that ${\ell}^{\infty}$ is a compact metric space, we need to show that it is complete and totally bounded.

First, we will show that ${\ell}^{\infty}$ is complete. Let $\{u_n\}$ be a Cauchy sequence in ${\ell}^{\infty}$. This means that for any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $m,n \geq N$, we have $||u_n - u_m|| < \epsilon$. Since ${\ell}^{\infty}$ is a subset of ${\ell}^{2}$, which is a complete metric space, this means that $\{u_n\}$ is also a Cauchy sequence in ${\ell}^{2}$. Therefore, it converges to some $u \in {\ell}^{2}$.

Next, we need to show that $u \in {\ell}^{\infty}$. Since $u_n \in {\ell}^{\infty}$ for all $n$, we have $|u_{n,k}| \leq \frac{1}{k}$ for all $n$ and $k$. Taking the limit as $n \to \infty$, we get $|u_{k}| \leq \frac{1}{k}$ for all $k$, which means that $u \in {\ell}^{\infty}$. Therefore, ${\ell}^{\infty}$ is complete.

To show that ${\ell}^{\infty}$ is totally bounded, we will use the fact that ${\ell}^{2}$ is totally bounded. Let $\epsilon > 0$ be given. Since ${\ell}^{2}$ is totally bounded, there exists a finite subset $F \subset {\ell}^{2}$ such that for any $u \in {\ell}^{2}$, there exists $v \in F$ such that $||u - v|| < \epsilon$. Now, for any $u \in {\ell}^{\infty}$, we can choose $v \in F$ such that $||u - v|| < \epsilon$. This means that $|u_{k} - v_{k}| < \epsilon$ for all $k$, which implies that $|u