The infinite-dimensional separable Hilbert space $H$ has the unusual property that it is diffeomorphic to its unit sphere. Therefore, $H$ admits the round metric as a complete and bounded Riemannian structure, which only applies to compact Riemannian manifolds in finite-dimensional case.

According to corollary 4 in this article, every infinite-dimensional separable Hilbert manifold is homeomorphic to $\lvert K \rvert \times H$, where $K$ is a locally finite simplicial complex. Since "homotopic" equals "diffeomorphic" for Hilbert manifolds (corollary 2 and 3 in the same article), the product Riemannian structure is complete and bounded if $\lvert K \rvert$ is homotopic to a closed smooth manifold.

Is it true that every infinite-dimensional, separable Hilbert manifold admits a complete and bounded Riemannian structure?