"Adding one dimension" to an infinite dimensional topological vector space

Question

Let $V$ be an infinite-dimensional Hausdorff topological vector space over $\mathbb{R}$. Is it true that $V\oplus\mathbb{R}$ is isomorphic (as a TVS) to $V$ itself? (How about $V\oplus V$?) If yes, I also wonder whether for every $v \ne 0$ in $V$ there exists an isomorphism $f : V \cong V \oplus \mathbb{R}$ such that the projection of $f(v)$ to $\mathbb{R}$ is nonzero.

If no:

1. What is the simplest counterexample?

2. What are some minimal conditions (complete/locally convex/normed/inner product etc.) that guarantee the existence of such isomorphisms? (If $V$ is a Hilbert space, desired isomorphisms exist in all three cases above.)

I looked briefly into Schaefer and Wolff's Topological Vector Spaces but did not find any clue.

Answer 1

Since this is a question in Functional Analysis, I will assume that isomorphism means "bicontinuous bijective linear map".

The answer is no in both cases. Not sure what the "simplest example is". The category "topological vector space", without further qualifications, allows for fairly nasty things to happen.

Let $V=L^p[0,1]$, with $0<p<1$. The "$p$-norm" in this case is not a norm, but $V$ is a complete metric topological vector space with the metric $$ d(f,g)=\int_0^1 |f-g|^p. $$ It turns out that $V^*=\{0\}$, that is $V$ has no nonzero continuous linear functionals. But $V\oplus\mathbb R$ (you are not saying what topology you consider there, but I will assume that it is the product topology) clearly has the continuous functional $\varphi(f,t)=t$, so its dual is nontrivial. So $V$ and $V\oplus\mathbb R$ cannot be isomorphic as topological vector spaces.

As for Banach spaces $V$ such that $V\not\simeq V\oplus V$, these exist but you will not find them among the classical examples of Banach spaces. Easiest examples (bear in mind that I'm not an expert) seem to be of the form infinite $\ell^2$-sums of spaces $\ell^{n_j}_{p_j}$ (this would be $\mathbb R^{n_j}$ with the $p_j$-norm). There even exists a Banach space that is not isomorphic to any square of any Banach space. The keyword to search for all this stuff is "Banach spaces nonisomorphic to their squares" (this is almost verbatim the title of a paper that has several references).

As far as I can tell, there is no characterization of Banach spaces isomorphic to their squares.

Answer 2

Sophie Morel on Zulip found this counterexample (by Gowers) to $V \cong V \oplus \mathbb{R}$ with $V$ a Banach space: A Solution to Banach's Hyperplane Problem, and there's a nice blog post about it; it turns out Banach asked both questions (about $V \oplus \mathbb{R}$ and $V \oplus V$) for Banach spaces.

The claim on Wikipedia that "a finite-codimensional subspace of a Banach space $X$ is always isomorphic to $X$" is apparently problematic, and Antoine Chambert-Loir offered an explanation how the mistake could have been made.

For the followup question, the answer should be true if $V^*$ (the continuous linear functionals) separates points in $V$ (which is true in any locally convex space by Hahn-Banach): from $V \cong V \oplus \mathbb{R}$ we know at least one closed hyperplane in $V$ is isomorphic to $V$, but all closed hyperplanes are isomorphic (I believe the proof works in any Hausdorff TVS, and so does this stronger result), so for any nonzero continuous linear functional $f \in V^*$ we get a decomposition $V \cong \text{ker }f \oplus \mathbb{R} \cong V \oplus \mathbb{R}$.