# 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.

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.

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