# Is the space of maps between Hilbert spaces that have at most polynomial growth of order m a separable Banach space?

If $U$ and $H$ are two real Hilbert spaces (they are infinite dimensional), let $C_{m}(U,H)$ denote the space of continuous maps from $U$ to $H$ which have at most polynomial growth of order $m$. It is known that for any $u \in U$, if we have $|u|_{C_{m}}$= $sup_{h \in H} \frac {|u(h)|} {1+|h|^{m}}$, then $C_{m}(U,H)$ is Banach space with this norm.
I would like to know
1. Is $C_{m}(U,H)$ also separable?
2. If not, is it possible to give it a norm so that $C_{m}(U,H)$ is a separable Banach space?
Thank you very much!

• Obviously the answer will depend on the dimensions of $U$ and $H$. Do you mean to assume they are separable and infinite-dimensional? Also, what do you mean by "maps"? Arbitrary maps? Continuous maps? Something else? If you mean arbitrary maps, then $C_m(U,H)$ is absolutely gigantic (assuming $U$ and $H$ are nontrivial) and cannot be separable merely for cardinality reasons. – Eric Wofsey Aug 27 '18 at 23:17
• @Eric Wofsey Thank you for the comments! Let me make it clearer. By Hilbert space, I mean infinite dimensions, and by maps I mean continuous and bounded maps. – misakaczy Aug 27 '18 at 23:33

No and no. First, one can not ask if $C_{m}(U,H)$ is separable until it has some topology on it, e.g. the one defined by the norm. Second, even the space of bounded linear operators, i.e. continuous linear polynomials, on a separable Hilbert space is non-separable. It contains an isometric copy of $\ell_\infty$ represented by the "diagonal" operators, see Separability of the space of bounded operators on a Hilbert space. Spaces of continuous maps between two infinite-dimensional spaces are just too large to expect separability.