Invariant subspace problem: non-separable Hilbert space

Solution 1:

The space $\mathcal{M}_x$ is always $T$-invariant, whether $\mathcal{H}$ is separable or not. However, if $\mathcal{H}$ is separable then we may have $\mathcal{M}_x = \mathcal{H}$, so it isn't a nontrivial invariant subspace.

For example, take $\mathcal{H} = \ell^2(\mathbb{N})$ with its usual orthonormal basis $\{e_1, e_2, \dots\}$, and let $T$ be defined by $T e_n = e_1 + e_{n+1}$. Then if $x = e_1$, we have $T^n x = e_{n+1}$, so $\mathcal{M}_x = \operatorname{cl}(\operatorname{span}\{e_1, e_2, \dots\}) = \mathcal{H}$.

(Edit: In that example, I was thinking of $\mathcal{M}_x$ including $x$ itself, i.e. starting with $T^0 x$, but I just noticed that was not assumed. This $T$ can't work if $x$ is not included because its range is not dense. If I think of a better example I'll add it. Perhaps something like the left shift $T^*$ with $x$ something like $x = \sum \frac{1}{n} e_n$.)

Since $\mathcal{M}_x$ is always separable, if $\mathcal{H}$ is not separable then we indeed have $\mathcal{M}_x \subsetneq \mathcal{H}$ which is what we want.