Let $Y,Z$ be subspaces of a Banach space $X$ such that $Y$ is isomorphic to $Z.$ Are $X/Y$ and $X/Z$ isomorphic?

Question

The following question is taken from 'Banach Space Theory; The Basis for Linear and Nonlinear Theory', page $35,$ question $29.$

Question: Let $Y,Z$ be subspaces of a Banach space $X$ such that $Y$ is isomorphic to $Z.$ Are $X/Y$ and $X/Z$ isomorphic?

Hint given states that it is no by letting $X = \ell_2, Y=\{(0,x_2,x_3,...)\}$ and $Z = \{(0,0,x_3,x_4,...)\}.$

I am trying to use the hint to solve the question. Let $T:Y\to Z.$ Clearly $T$ is a bounded bijection. But I am not able to prove that $T^{-1}$ is also bounded, that is, I could not find a constant $D>0$ such that $$\sum_{k=2}^\infty |x_k|^2\leq D \sum_{k=3}^\infty |x_k|^2.$$

Another doubt is that what is a motivation of considering the spaces given. In other words, how to 'see' that the question has negative answer.

Answer 1

You first need to show that $Y$ and $Z$ as defined as actually isomorphic. Intuitively that's definitely the case, $Z$ is just $Y$ with an extra zero in front of all the elements. If you wish to show this formally take $T: Y \to Z$ given by $$ T(0, x_2, x_2, \ldots) = (0, 0, x_2, x_3, \ldots) $$ This is clearly linear, and also it's bijective since we can immediately give an inverse function $T^{-1}:Z \to Y$ where $$ T^{-1}(0, 0, 0, x_3, x_4, \ldots) = (0, 0, x_3, x_4, \ldots) $$ It's not difficult to show in fact $T$ and $T^{-1}$ are both isometries and hence continuous.

You could appeal to the inverse function theorem to show $T^{-1}$ is continuous once you've shown $T$ is continuous by showing $Y$ and $Z$ are closed subspace of $\ell^2$ and hence Banach spaces.

Can you see why $X/Y$ and $X/Z$ are not isomorphic? (Hint: Consider dimensions)

Answer 2

I think the following reasoning works.

$(X/Y)^{\ast}\cong Y^{\perp}\cong \left<(1,0,...)\right>$, and $(X/Z)^{\ast}\cong Z^{\perp}\cong\left<(1,0,...),(0,1,...)\right>$. We cannot have $X/Y\cong X/Z$, if not, then $(X/Y)^{\ast}\cong(X/Z)^{\ast}$, then $\left<(1,0,...)\right>\cong\left<(1,0,...),(0,1,...)\right>$, this is clearly invalid because of the different dimensions.