Define a norm on $\ell^1$ by $$\|x\|=(\|x\|_1^2+\|x\|_2^2)^{\frac{1}{2}},$$ where $\|.\|_p$ denotes the canonical norm on $\ell^p$. Then $\|.\|$ is equivalent to $\|.\|_1$.

I want to show that a quotient space of $X=(\ell^1,\|.\|)$ is linearly isometric to $(\ell^1,\|.\|_1)$.

Let $D=\{x_n:n\in \mathbb{N}\}$ be a countable dense subset of $S_{\ell^1}$. Let $M,m>0$ be such that $M\|x\|_1\leq \|x\|\leq m\|x\|_1$ for all $x\in \ell^1$. Define $T:X\to (\ell^1,\|.\|_1)$ by $$T((\lambda_n))=\sum\limits_{n=1}^{\infty}M\lambda_n x_n \text{ for all }(\lambda_n)\in X.$$ Clearly, $T$ is linear. Also for all $(\lambda_n)\in X$, $$\|T((\lambda_n))\|_1=\|\sum\limits_{n=1}^{\infty}M\lambda_n x_n\|\leq M\|(\lambda_n)\|_1\leq \|(\lambda_n)\|.$$ Thus $T$ is continuous. \ Let $x\in (\ell^1,\|.\|_1)$. Choose $n_1\in \mathbb{N}$ such that $$\|x-\lambda_{n_1}x_{n_1}\|_1<\frac{\varepsilon}{2}, \text{ where }\lambda_{n_1}=\|x\|_1.$$ Choose $n_2\in \mathbb{N}$ with $n_2>n_1$ such that $$\|(x-\lambda_{n_1}x_{n_1})-\lambda_{n_2}x_{n_2}\|_1<\frac{\varepsilon}{2^2}, \text{ where }\lambda_{n_2}=\|x-\lambda_{n_1}x_{n_1}\|_1<\frac{\varepsilon}{2}.$$ Choose $n_3\in \mathbb{N}$ with $n_3>n_2>n_1$ such that $$\|(x-\lambda_{n_1}x_{n_1}-\lambda_{n_2}x_{n_2})-\lambda_{n_3}x_{n_3}\|_1<\frac{\varepsilon}{2^3}, \text{ where }\lambda_{n_3}=\|x-\lambda_{n_1}x_{n_1}-\lambda_{n_2}x_{n_2}\|_1<\frac{\varepsilon}{2^2}.$$ Proceeding in this way we get a sequence $(\lambda_{n_k})$ such that $$\lambda_{n_{k+1}}=\|x-(\lambda_{n_1}x_{n_1}+\ldots+\lambda_{n_k}x_{n_k})\|_1<\frac{\varepsilon}{2^k}.$$ It follows that $x=\sum\limits_{k=1}^{\infty}\lambda_{n_k}x_{n_k}$. Let $\alpha_{n_k}=\frac{1}{M}\lambda_{n_k}$ for all $k\in \mathbb{N}$ and $\alpha_n=0$ for all $n\notin \{n_1,n_2,\ldots\}$. Then $$\sum\limits_{n=1}^{\infty}|\alpha_n|\leq \frac{1}{M}(\|x\|+\sum\limits_{k=1}^{\infty}\frac{\varepsilon}{2^k})=\frac{1}{M}(\|x\|_1+\varepsilon)<\infty.$$ Consequently $(\alpha_n)\in X$ and $T((\alpha_n))=\sum\limits_{k=1}^{\infty}M\alpha_{n_k}x_{n_k}=\sum\limits_{k=1}^{\infty}\lambda_{n_k}x_{n_k}=x$. Hence $T$ is onto. Since $T$ is continuous, $Y=\ker T$ is a closed subspace of $X$ and so $X/Y$ is a Banach space. Therefore by the first law of isomorphism of Banach spaces, $T$ induces a linear isomorphism $\tilde{T}$ from $X/Y$ onto $(\ell^1,\|.\|_1)$ given by $$\tilde{T}((\lambda_n)+Y)=T(\lambda_n)\text { for all }(\lambda_n)\in X.$$ We prove that $\tilde{T}$ is an isometry.

For all $(y_n)\in Y$, $$\|\tilde{T}((\lambda_n)+Y)\|_1=\|\tilde{T}((\lambda_n)+(y_n)+Y)\|_1=\|T((\lambda_n)+(y_n))\|_1 \leq \|(\lambda_n)+(y_n)\|.$$ Consequently $$\|\tilde{T}((\lambda_n)+Y)\|_1\leq \|(\lambda_n)+Y\|.$$

I got stuck here. How to show the reverse inequality? Any help will be appreciated.

  • $\begingroup$ how'd u get $||\tilde{T}((\lambda_n)+(y_n)+Y|| = ||T((\lambda_n)+(y_n))||_1$? $\endgroup$ Apr 23, 2020 at 15:38
  • $\begingroup$ This is wrong! It should be $\|\tilde{T}((\lambda_n)+(y_n)+Y)\|_1=\|T((\lambda_n)+(y_n))\|_1$ $\endgroup$
    – Anupam
    Apr 23, 2020 at 16:59
  • $\begingroup$ Why is $||\tilde{T}((\lambda_n)+Y)|| = ||\tilde{T}((\lambda_n)+(y_n)+Y)||_1$? You should edit your question to clear all of this up... (even after my first comment) $\endgroup$ Apr 23, 2020 at 18:14
  • $\begingroup$ have u seen the proof that projections onto a closed subspace have norm $1$ (context is general Banach space)? $\endgroup$ Apr 24, 2020 at 4:27
  • $\begingroup$ Yes, I have seen that. $\endgroup$
    – Anupam
    Apr 24, 2020 at 5:01


You must log in to answer this question.