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Oct 19, 2014 at 14:25 comment added Jochen Wengenroth It would be most natural to assume the continiuty of $X_{j+1} \hookrightarrow X_j$ so that you have a projective limit of Banach spaces. Then $X_\infty=\bigcap_j X_j$ is a Frechet space.
Oct 18, 2014 at 7:51 vote accept Hui Zhang
Oct 17, 2014 at 19:25 comment added Pietro Majer @HuiZhang It seems an interesting question to me, although a bit vague. Maybe you can add more details?
Oct 17, 2014 at 18:10 answer added Joonas Ilmavirta timeline score: 1
Oct 17, 2014 at 16:48 comment added Joonas Ilmavirta @HuiZhang, if the embeddings $X_j\to X_1$ are equicontinuous, then your condition implies that $x_j\to x$ in $X_1$. If they are not equicontinuous, I fear that limits may not be unique with your definition.
Oct 17, 2014 at 16:45 history edited Hui Zhang CC BY-SA 3.0
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Oct 17, 2014 at 16:42 comment added Hui Zhang @JoonasIlmavirta Suppose we have the continuity of the embeddings. Is $\|x_j-x\|_{X_j}$ a good quantity to indicate the error? You are right that I need more conditions for convergence. My question is: if it converges in that way, is the result a useful one? Or is there a better way to pose the convergence question such that we are confident that $x_j$ is a good approximation? (I find my question is really bad.)
Oct 17, 2014 at 16:29 comment added user10101 "This is perhaps a stupid question ... I have a sequence of Banach spaces". That made me LOL, thanks :)
Oct 17, 2014 at 16:00 comment added Joonas Ilmavirta If $y\in X_j$, do you know that $\|y\|_{X_j}\leq\|y\|_{X_1}$ or something else? You do need some kind of compatibility between the different norms. And even if you use the same norm on all spaces, the condition $x_j\in X_j$ allows $\|x_j\|\to\infty$ as $j\to\infty$, so you need more conditions to ensure convergence.
Oct 17, 2014 at 15:17 history asked Hui Zhang CC BY-SA 3.0