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!


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.

  • $\begingroup$ Nice answer. I think you want $\varphi(x)=-\varphi(Sx)$ (which of course doesn't change anything). $\endgroup$ Dec 1 '18 at 10:17
  • $\begingroup$ Thank you gerw, very enlightening. $\endgroup$
    – Nemesis
    Dec 1 '18 at 12:07
  • $\begingroup$ @SeverinSchraven: Yes, of course! Thank you. $\endgroup$
    – gerw
    Dec 1 '18 at 14:01

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