# Showing linear isometry

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.

• how'd u get $||\tilde{T}((\lambda_n)+(y_n)+Y|| = ||T((\lambda_n)+(y_n))||_1$? Apr 23, 2020 at 15:38
• This is wrong! It should be $\|\tilde{T}((\lambda_n)+(y_n)+Y)\|_1=\|T((\lambda_n)+(y_n))\|_1$ Apr 23, 2020 at 16:59
• 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) Apr 23, 2020 at 18:14
• have u seen the proof that projections onto a closed subspace have norm $1$ (context is general Banach space)? Apr 24, 2020 at 4:27
• Yes, I have seen that. Apr 24, 2020 at 5:01