Is $\ell^\infty({\mathbb N})\otimes\ell^\infty({\mathbb N})$ weakly dense in $\ell^\infty({\mathbb N}\times{\mathbb N})$?

Question

There is a natural embedding $\Phi$ of the (algebraic) tensor product $\ell^\infty({\mathbb N})\otimes\ell^\infty({\mathbb N})$ into the Banach space $\ell^\infty({\mathbb N}\times{\mathbb N})$ given by $$\big(\Phi(x\otimes y)\big)_{(n,m)}=x_n\cdot y_m$$ on elementary tensors and extended using linearity.

Question: Is the image of $\Phi$ dense with respect to the weak topology?

The fact that I can not answer it gives me the nagging suspicion that the answer is negative, but I'm hoping that it is positive.

It's easy to see that the image of $\Phi$ is not dense in norm; indeed its norm closure will not contain the vector $x_{n,m}=1_{n=m}$. On the other hand, as point out by Ben Grossmann in the comments, the image of $\Phi$ is dense in the weak$^*$ topology. I also believe that the restriction of $\Phi$ to $\ell^p({\mathbb N})\otimes\ell^p({\mathbb N})$ is dense in $\ell^p({\mathbb N}\times{\mathbb N})$ for every $1\leq p<\infty$, even in norm.

I assume the case $\ell^{\infty}$ case is hard because it's dual is not so nice. — Martin Brandenburg, Nov 01 '24 at 14:38
I don't know if this is at all a helpful observation, but if we identify $\ell^\infty$ with $(\ell^1)^*$, then we do have density relative to the weak* topology — Ben Grossmann, Nov 01 '24 at 14:41
@MartinBrandenburg Right, as Ben pointed out if the weak topology were the same as the weak$^*$ topology this would be relatively easy. — Joel Moreira, Nov 01 '24 at 15:11
If by weak topology you meant Banach weak topology, then no. The range of $\Phi$ is a subspace, so by Hahn-Banach, it is weakly dense iff it is norm-dense, and you’ve already observed it is not norm-dense. — David Gao, Nov 01 '24 at 16:46
@DavidGao Yes, that's exactly what I meant. Thank you for pointing that out; if you write it as an answer I'll acept it. — Joel Moreira, Nov 03 '24 at 16:05

score 4 · Accepted Answer · answered Nov 03 '24 at 17:19

As requested by the OP, here is my comment, re-organized into an answer:

By Hahn-Banach, the closure of a convex subset under the weak topology coincides with its closure under the norm topology. In particular, since the range of $\Phi$ is a linear subspace, it is weakly dense iff it is norm-dense. As the OP has already observed, the range of $\Phi$ is not norm-dense, so it is not weakly dense either.

Is $\ell^\infty({\mathbb N})\otimes\ell^\infty({\mathbb N})$ weakly dense in $\ell^\infty({\mathbb N}\times{\mathbb N})$?

1 Answers1