Reflexive Banach spaces and norms

Question

Let $(X,|.|)$ be a reflexive Banach space, and $Y\subset X$ such that $(Y,|.|_Y)$ is a Banach space with a norm $|.|_Y$ stronger than $|.|$, i.e. there's a constant $C$ such that $$|y|\leq C |y|_Y, \ \ \forall y\in Y$$ What can we say about $(Y,|.|_Y)$ ? is it reflexive ?

Answer 1

The answer is negative, a counterexample was given by David Mitra:

$Y=\ell_1$ sits inside $X=\ell_2$ and your inequality is satisfied with $C=1$.

Pushing this further, I'll show that every separable Banach space (and some nonseparable ones) can be realized as $Y$ inside of $X=\ell_2$. To begin with, every separable space is isometrically isomorphic to a subspace of $\ell_\infty$. Let $T:\ell_\infty\to\ell_2$ be the operator defined by $$T(\{x_n\}) = \{x_n/n\}$$ This is an injective bounded operator. Its image can be taken as $Y$ in your setting, with the norm $\|y\|_Y = \|T^{-1}y\|_{\infty}$.

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But if you assume in addition that $Y$ is closed in the norm of $X$, then (by the open mapping theorem) the two-sided inequality $$c\|y\|_Y\le \|y\|\le C\|y\|_Y \quad \forall y\in Y$$ holds, and therefore $Y$ is reflexive (being isomorphic to a subspace of a reflexive space).