Dense subsets in tensor products of Banach spaces

Question

Assume $B_1$ and $B_2$ are Banach spaces of univariate functions. Moreover, assume that the sets $D_1 \subset B_1$ and $D_2 \subset B_2$ are dense with respect to the respective norms $\|\cdot\|_{B_1}$ and $\|\cdot\|_{B_2}$.

Now we consider the algebraic tensor product $$B_1 \otimes_a B_2 := \{\sum_{i=1}^n f_i g_i \ :\ n \in \mathbb{N}, f_i \in B_1, g_i \in B_2\}$$ and define $B := \overline{B_1 \otimes_a B_2}$ as the completion with respect to some norm $\|\cdot\|_B$.

Now my question is: Is the tensor product $D:=D_1 \otimes D_2$ of the dense subsets also dense in $B$ with respect any choice of the norm $\|\cdot\|_B$? Or does $\|\cdot\|_B$ have to fulfill certain conditions?

I would argue: if $\|\cdot\|_B$ is a crossnorm, then $D \subset B_1 \otimes_a B_2$ is dense with respect to this crossnorm and since $B_1 \otimes_a B_2$ is dense in $\overline{B_1 \otimes_a B_2}$ the question holds true for all crossnorm? Is this correct?

Answer 1

If $\|\cdot\|_B$ is a cross norm, then yes, your argument shows that $D_1 \otimes_a D_2$ is dense in $B_1 \otimes_a B_2$ and therefore also dense in $B$.

Without this assumption or something similar, the answer is no, as then there is nothing at all relating the $B$ norm to the $B_1, B_2$ norms. This really has nothing to do with tensor products, so take $B_2$ to be one-dimensional, so that $B_1 \otimes_a B_2 = B_1$. Let's take $B_1 = \ell^2$ with its usual norm. Now the vector spaces $\ell^2$ and $\ell^\infty$ both have Hamel dimension $\mathfrak{c}$, so there is a linear isomorphism of vector spaces $T : \ell^2 \to \ell^\infty$. So define the $\|\cdot\|_B$-norm on $B_1 \otimes_a B_2 = B_1 = \ell^2$ by $\|f\|_B = \|Tf\|_{\ell^\infty}$. Then $B_1 \otimes_a B_2$ is already complete under the $\|\cdot\|_B$-norm and is isometrically isomorphic to $\ell^\infty$. Let $D_1, D_2$ be your favorite countable dense subsets of $B_1, B_2$. Then $D = D_1 \otimes_a D_2$ is countable, but $B$ is not separable, so $D$ is not dense in $B$.