Infinite direct sum decomposition of the heat semigroup on $\mathbb R$

Question

This question is based on a very similar question posted by me yesterday. A very nice solution was provided by Aleksei Kulikov. Here I modify my question slightly.

Let $Q_t$ be the heat semigroup on $L^2(\mathbb R)$. Is there a direct sum decomposition $$\oplus_{k=1}^\infty \mathscr{H}_k\subseteq L^2(\mathbb R)$$ such that

1. Each $\mathscr H_k$ is an infinite dimensional Hilbert space.
2. $Q_t(\mathscr H_k)\subseteq \mathscr H_k$ for all $k$ and $t>0$.
3. If $Q^{(k)}_t$ denotes the restriction of $Q_t$ on $\mathscr H_k$ then $\sigma(Q^{(k)}_t)=\sigma(Q_t)=[0,1]$ for all $k$. Here $\sigma(A)$ means the spectrum of $A$.

For example, if $\mathscr{H}_1$ and $\mathscr{H}_2$ denote the Hilbert spaces of all $L^2(\mathbb R)$ functions that are even and odd respectively, it can be easily checked that $L^2(\mathbb R)=\mathscr{H}_1\oplus \mathscr{H}_2$, $Q_t(\mathscr H_k)\subseteq\mathscr{H}_k$ for $k=1,2$, and $\sigma(Q^{(k)}_t)=[0,1]$ for $k=1,2$. The above question asks whether we can have infinitely many such Hilbert spaces.

Note that the example provided by Aleksei Kulikov does not satisfy the property $\sigma(Q^{(k}_t)=[0,1]$.

Answer 1

Yes, this can be done. By spectral representation, $Q_t$ is unitarily equivalent to two copies of multiplication by $e^{-t\lambda}$ in $L^2(0,\infty)$, and now we only need to decompose $(0,\infty)=\bigcup A_k$ into disjoint Borel sets that are essentially dense in $(0,\infty)$, that is, $(x-\epsilon,x+\epsilon)\cap A_k$ has positive measure for all $x>0$, $\epsilon>0$, $k\ge 1$. Such sets can be obtained from positive measure Cantor sets (to construct $A_1$, start out with such a set $C_{11}$, then put smaller copies into each gap of $C_{11}$ etc.).

We can now take $H_k=L^2(A_k)$.