How to formulate approximation from above? (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.
 A: 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.
