(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.



How to formulate approximation from above?
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.


