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(This is perhaps a stupid question. If so, please give me a hint and a down vote.)

I have a sequence of Banach spaces $X_{1}\supset X_2\supset ...$ and a sequence of elements $x_j\in X_j$ ($j=1,2,..$). I also have a subspace $X\subset X_j$ for all $j=1,2,..$ and an element $x\in X$.

I want to prove $x_j\rightarrow x$ under some condition. But I find it is difficult to put the convergence question in a good space because $x_j-x$ is in $X_j$ but $X_j$ depends on $j$. Is it ok to ask $\lim_{j\rightarrow \infty}\|x_j-x\|_{X_j}=0$? This looks weird to me because I am not sure $\|\|_{X{j}}$ is a good measure of the approximation error. The situation in approximation from below sounds better to me.

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    $\begingroup$ 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. $\endgroup$
    – Joonas Ilmavirta
    Commented Oct 17, 2014 at 16:00
  • $\begingroup$ "This is perhaps a stupid question ... I have a sequence of Banach spaces". That made me LOL, thanks :) $\endgroup$
    – user10101
    Commented Oct 17, 2014 at 16:29
  • $\begingroup$ @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.) $\endgroup$
    – Hui Zhang
    Commented Oct 17, 2014 at 16:42
  • $\begingroup$ @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. $\endgroup$
    – Joonas Ilmavirta
    Commented Oct 17, 2014 at 16:48
  • $\begingroup$ @HuiZhang It seems an interesting question to me, although a bit vague. Maybe you can add more details? $\endgroup$
    – Pietro Majer
    Commented Oct 17, 2014 at 19:25

1 Answer 1

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Let $X_1\supset X_2\supset \dots$ be a nested sequence of Banach spaces, each equipped with a different norm $\|\cdot\|_j$, and suppose $X=\bigcap_jX_j\neq\{0\}$. Suppose the inclusions $X_j\to X_1$ are continuous. Let $(x_j)$ be a sequence such that $x_j\in X_j$ for all $j$. We define $x\in X$ to be a limit of the sequence $(x_j)$ if and only if $\|x_j-x\|_j\to0$ as $j\to\infty$.

Define $$ E=\{x\in X;\lim_{j\to\infty}\|x\|_j=0\}. $$ Clearly $E$ is a linear subspace of $X$.

Theorem: Suppose $x$ is a limit of $(x_j)$. Then $y$ is also a limit of $(x_j)$ if and only if $x-y\in E$. Consequently limits are unique if and only if $E=\{0\}$.

Proof: If a sequence $(x_j)$ has two limits $x,y\in X$, then $\|x-y\|_j\leq\|x-x_j\|+\|x_j-y\|_j\to0$ as $j\to\infty$ so $x-y\in E$. If $x$ is a limit of $(x_j)$ and $x-y\in E$, then $\|x_j-y\|_j\leq\|x_j-x\|_j+\|x-y\|_j\to0$ as $j\to\infty$ so also $y$ is a limit. $\square$

We have $E=\{0\}$, for example, when the inclusions $X_j\to X_1$ are equicontinuous. If you know what the space $E$ is, it might be of some use that the limit is unique in $X/E$.

I am not sure if this is exactly what you were after. Uniqueness is an important property of a concept of limits, so I focused on that one.

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