Let $A$ and $B$ be Banach spaces with their own (possibly different) norms. Also, there is a non-empty subset $S \subset A \cap B$ such that $S$ is dense in $A$ and $B$ respectively.
Then, for $x \in A\cap B$, can we always extract a sequence $\{s_n\} \subset S$ such that $s_n \to x$ in $A$ and $s_n \to x$ in $B$?
This question is generalized from the situation $A = L^1(\mathbb{R}^n)$, $B = L^2(\mathbb{R}^n)$ and $S = \mathcal{S}(\mathbb{R}^n)$, in which case, we can find a sequence satisfying the above conditions.
I'd appreciate it if you'd help me!
Best Answer
Here is my proposed counterexample. It is inspired by considering the dual of the example given in Theorem 2 of http://faculty.missouri.edu/~stephen/preprints/interpolate.html.
Let $$ Z = L^1([0,1]) \oplus L^1([0,1]) \oplus L^1([0,1]). $$ Let $A$, $B$ be subspaces of $Z$ such that the following norms are finite: $$ {\|(f,g,h)\|}_{A} = {\|f-g\|}_\infty + {\|g\|}_1 + {\|h\|}_\infty ,$$ $$ {\|(f,g,h)\|}_{B} = {\|f-h\|}_\infty + {\|g\|}_\infty + {\|h\|}_1 .$$ Both spaces are isomorphic to $L^\infty([0,1]) \oplus L^\infty([0,1]) \oplus L^1([0,1])$, so they are Banach spaces.
We can calculate that $$ {\|(f,g,h)\|}_{A \cap B} := \max\{{\|(f,g,h)\|}_{A},{\|(f,g,h)\|}_{B}\}\approx {\|f\|}_\infty + {\|g\|}_\infty + {\|h\|}_\infty ,$$ because $$ {\|(f,g,h)\|}_{A \cap B} \le {\|(f,g,h)\|}_{A} + {\|(f,g,h)\|}_{B} \le 3 ({\|f\|}_\infty + {\|g\|}_\infty + {\|h\|}_\infty) ,$$ and \begin{align} {\|(f,g,h)\|}_{A \cap B} &\ge \tfrac12({\|(f,g,h)\|}_{A} + {\|(f,g,h)\|}_{B}) \\&\ge \tfrac14{\|f-g\|}_\infty + \tfrac14{\|f-h\|}_\infty + \tfrac12{\|g\|}_\infty + \tfrac12{\|h\|}_\infty \\&\ge \tfrac14({\|f\|}_\infty-{\|g\|}_\infty) + \tfrac14({\|f\|}_\infty-{\|h\|}_\infty) + \tfrac12{\|g\|}_\infty + \tfrac12{\|h\|}_\infty \\&\ge \tfrac14 ({\|f\|}_\infty + {\|g\|}_\infty + {\|h\|}_\infty) .\end{align} Hence $$ A \cap B = L^\infty([0,1]) \oplus L^\infty([0,1]) \oplus L^\infty([0,1]) .$$ Let $$ S = C([0,1]) \oplus L^\infty([0,1]) \oplus L^\infty([0,1]). $$ Clearly $S$ is not dense in $A \cap B$. We show $S$ is dense in $A$, as the argument for $S$ dense in $B$ is essentially identical.
Suppose $x = (f,g,h) \in A$ with ${\|x\|}_A \le 1$, that is, $$ {\|(f,g,h)\|}_A = {\|f - g\|}_\infty + {\|g\|}_1 + {\|h\|}_\infty \le 1.$$ Note that $f-g\in L^\infty \subset L^1$, and $g\in L^1$, which implies $f \in L^1$. Let $f_n \in C([0,1])$ be such that ${\|f-f_n\|}_1 \to 0$. Set $$ s_n = (f_n, g - f + f_n,h) .$$ Note $g - f + f_n = (g-f) + f_n \in L^\infty([0,1])$, so $s_n \in S$. Then as $n \to \infty$, $$ {\|x - s_n\|}_A = {\|(f-f_n, f-f_n, 0)\|}_A = {\|f-f_n\|}_1 \to 0. $$