about $l^p$ space and qoutient space

Question

Let $Y=\{x\in l^p:x_{2n}=0\}$, $1\leq p \leq \infty$. It can be proved that $Y$ is closed subspace of $l^p$. Define the quotient space $l^p/Y=\{x+Y:x\in l^p\}$. Then, by the fact that $Y$ closed, a norm on $l^p/Y$ can be defined by $||x+Y||=\inf\{\|x-y\|_p : y\in Y\}$.

prove that $l^p/Y$ is isometrically isomorphic $l^p$ .

I think we have :$\|x + Y\| = \left( \sum_{j} |x_{2j}|^p \right)^{1/p}$ but how we can show $l^p/Y$ is isometrically isomorphic $l^p$ ?

Answer 1

We start be defining the map $T:\ell^p\to\ell^p$ by $$T(x_1,x_2,\dots)=(x_2,x_4,x_6\dots)$$

This is a linear operator and obviously $\|Tx\|_p\leq\|x\|_p$. The kernel of this map is exactly $Y$ and it can easily be shown that its image is all of $\ell^p$: if $a=(a_n)\in\ell^p$, consider $b=(0,a_1,0,a_2,0,a_3,\dots)$. Then $b\in\ell^p$ and $Tb=a$. By elementary linear algebra, we have that

$$\ell^p/Y\cong\ell^p$$ as vector spaces via the isomorphism of vector spaces $\bar{T}$ that is defined by $\bar{T}(a+Y)=Ta$.

But $\bar{T}$ is isometric; as you observed, $\|a+Y\|=\bigg(\sum_{j}|a_{2j}|^p\bigg)^{1/p}=\|Ta\|_p=\|\bar{T}(a+Y)\|_p$.

The case $p=\infty$ works in a very similar way.

Answer 2

Consider the map $T : \ell^p \to \ell^p/Y$ given by $$Tx = (0,x_1,0,x_2,0,x_3\ldots) + Y, \quad x = (x_n)_n \in \ell^p.$$ Then $T$ is isometric: $$\|Tx\| = \|(0,x_1,0,x_2,0,x_3\ldots) + Y\| = \left(\sum_{n=1}^\infty |x_n|^p\right)^{\frac1p} = \|x\|_p.$$ Also $T$ is surjective since for every $y + Y \in \ell^p/Y$ we have $$T(y_{2n})_n = (0,y_2,0,y_4,0,y_6\ldots) + Y = y + Y.$$

Therefore $T$ is the desired isometric isomorphism.