1
$\begingroup$

Consider the heat semigroup $Q_t$ on $L^2(\mathbb{R}^n)$ generated by the Laplace operator $\Delta=\sum_{i=1}^n\frac{\partial^2}{\partial x^2_i}$. Does there exist a direct sum decomposition $$\oplus_{k=1}^\infty \mathscr{H}_k \subseteq L^2(\mathbb{R}^n)$$ such that

  1. $Q_t(\mathscr H_k)\subseteq \mathscr{H}_k$ for each $k$.
  2. $\mathscr{H}_k$ is isometric to $L^2(\mathbb{R}^m)$ for some $m$ and for each $k$.
Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

I might be misunderstanding the definitions, but isn't it kinda trivial to do with the Fourier transform? On the Fourier side $Q_t$ is a multiplication by $e^{-t |x|^2}$, and $L^2(A)$ for any $A\subset \mathbb{R}^n$ is invariant under all these guys (and, if $A$ has positive measure, it is isometric to all separable Hilbert spaces, so your second point is way more precise than it needs to be).

So, just chop $\mathbb{R}^n$ into countably many positive measure subsets and you are done. For example, an explicit choice is $\mathcal{H}_k = \mathcal{F}( L^2(\{x\in \mathbb{R}^n: k-1 < |x| \le k\}))$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .