I've found this result that states that super-reflexivity is separably determined, i.e., if every separable subspace $Y\subset X$ of a Banach space is superreflexive then $X$ itself is superreflexive. They always say that this is an easy consequence of the definition, but I'm not able to see it. I've tried proving it by contradiction. So, supposing that $X$ is not super-reflexive ensures that I can find a non-reflexive space, $Z$, which is finitely representable in $X$. Even more, since reflexivity is separably determined (easy consequence of James' characterization theorem for reflexivity), I can suppose that this $Z$ is separable. So I have $Z$ a separable, non-reflexive space that is finitely representable in $X$. My idea is to show that this implies that it is finitely representable in a separable subspace of $X$, which would give me a contradiction, but I couldn't be able. Could you guys lend me a hand?

Thank you so much! Bye!!

I've found this result that states that super-reflexivity is separably determined, i.e., if every separable subspace $Y\subset X$ of a Banach space is superreflexive then $X$ itself is superreflexive. They always say that this is an easy consequence of the definition, but I'm not able to see it. I've tried proving it by contradiction. So, supposing that $X$ is not super-reflexive ensures that I can find a non-reflexive space, $Z$, which is finitely representable in $X$. Even more, since reflexivity is separably determined (easy consequence of James' characterization theorem for reflexivity), I can suppose that this $Z$ is separable. So I have $Z$ a separable, non-reflexive space that is finitely representable in $X$. My idea is to show that this implies that it is finitely representable in a separable subspace of $X$, which would give me a contradiction, but I couldn't be able. Could you guys lend me a hand?

Maybe there is an easier approach, using renormings, for example.

Thank you so much! Bye!!