Dense subset of two Banach spaces also dense in the intersection My question is:
Let $ V $ be a vector space (over $ \mathbb K\in\{\mathbb{R}, \mathbb{C}\} $), $ X,Y\subseteq V $ two subspaces equipped with norms $ \|\cdot\|_X, \|\cdot\|_Y $ such that $ (X,\|\cdot\|_X) $ and $(Y,\|\cdot\|_Y)$ are Banach spaces and $ D\subseteq X\cap Y$. If $ D $ is dense in $ (X,\|\cdot\|_X) $ and $(Y,\|\cdot\|_Y)$, is $ D $ also dense in $ X\cap Y $ equipped with $ \|\cdot\|:=\|\cdot\|_X + \|\cdot\|_Y $?
At first sight, it seemed very clear to me that this should be true. But I even fail to answer the following (possibly) easier question:
Let $ X $ be a vector space over $ \mathbb K $ equipped with two norms $ \|\cdot\|_1, \|\cdot\|_2 $ such that $ (X,\|\cdot\|_1) $ and $(X,\|\cdot\|_1)$ are Banach spaces and $ D\subseteq X$. If $ D $ is dense in $ (X,\|\cdot\|_1) $ and $(X,\|\cdot\|_2)$, is $ D $ also dense in $ X $ equipped with $ \|\cdot\|:=\|\cdot\|_1 + \|\cdot\|_2 $?
The answer is yes, if $ \|\cdot\|_1, \|\cdot\|_2 $ are equivalent, so I tried to think about a counterexample using nonequivalent norms on a specific space and I also found a nice paper about nonisomorphic complete norms (https://www.researchgate.net/publication/226200984_Equivalent_complete_norms_and_positivity) but it didn't helped me so far to construct anything useful for my question.
Thanks for your help!
 A: I have a negative answer for your "easier" question based on the construction of https://mathoverflow.net/a/184471.
Let $X_1 := (X, \|\cdot\|_1)$ be an infinite dimensional Banach space and let $\varphi$ be an unbounded linear functional on $X_1$. We fix $y \in X$ with $\varphi(y) = 1$ and define
$$ S x := x - 2 \, \varphi(x) \, y.$$
It is easily checked that $S^2 x := S S x = x$.
The norm
$$ \|x \|_2 := \| S x\|_1$$
gives rise to the normed space $X_2 := (X, \|\cdot\|_2)$. Since $S  :X_2 \to X_1$ is an isometric isomorphism (by definition), $X_2$ is complete.
From $\varphi(x) = -\varphi(Sx)$ one can check that $\varphi$ is also unbounded on $X_2$. Indeed, we find $x_n \in X$ with $\varphi(x_n) \ge n$ and $\|x_n\|_1=1$. Hence, $\varphi( S x_n) \ge n$ and $\|S x_n\|_2 = \|x_n\|_1 = 1$.
Thus, the kernel of $\varphi$ is dense in $X_1$ and $X_2$.
However, we can check that $\varphi$ is bounded w.r.t. $\|\cdot\|=\|\cdot\|_1+\|\cdot\|_2$:
$$
2 \, \|y\|_1 \, |\varphi(x)|
=
\| 2 \, \varphi(x) \, y \|_1
\le
\|x\|_1 + \| x - 2 \, \varphi(x) \, y \|_1 
=
\|x\|_1 + \| S x\|_1
=
\|x\|.$$
Hence, the kernel of $\varphi$ is closed and therefore not dense w.r.t. the norm $\|\cdot\|$ in $X$.
I would imagine that your original question would be also interesting if we add the following (reasonable) assumption:
if $\{z_n\} \subset X \cap Y$ satisfies $z_n \to x$ in $X$ and $z_n \to y$ in $Y$
then $x = y$.
Note that this is not satisfied in my counterexample.
