Jtdylktuy

My question is:

Let $ V $ be a vector space (over $ mathbb Kin{mathbb{R}, mathbb{C}} $), $ X,Ysubseteq V $ two subspaces equipped with norms $ |cdot|_X, |cdot|_Y $ such that $ (X,|cdot|_X) $ and $(Y,|cdot|_Y)$ are Banach spaces and $ Dsubseteq Xcap Y$. If $ D $ is dense in $ (X,|cdot|_X) $ and $(Y,|cdot|_Y)$, is $ D $ also dense in $ Xcap 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 $ Dsubseteq 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!

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.