"Exotic" Banach spaces of sequences

Question

Does there exist a linear subspace of $\mathbb C ^{\mathbb N}$ that can be endowed a Banach space topology that is not finer than the locally convex topology of pointwise convergence?

Best, Martin

Answer 1

Martin, your question is at the level of a good homework problem in a course on functional analysis, so I will give only hints.

1. Show that there is NO Banach space topology on $\Bbb{C}^{\Bbb{N}}$ which is finer than the product topology.

2. Show that $\Bbb{C}^{\Bbb{N}}$ is algebraically isomorphic (that is, isomorphic as a vector space) to $\ell_2$.

Answer 2

Take any of the usual Banach spaces X of sequences. Let f be any unbounded linear functional on X such that $f({1,0,0,...})\neq 0$. Now consider the mapping $\Phi$ between sequences defined as follows: $$\Phi({x_1,x_2,x_3,...})=({f(x),x_2,x_3,...}).$$ Let $Y=\Phi(X)$, endowed with the topology $\|y\|=\|\Phi^{-1}(y)\|_X$. By construction, Y is a Banach space. However, since f was assumed unbounded, convergence in the norm of Y does not imply convergence of the first component.