Consider the function space Ω, consists of all functions $\phi(x)$ that satisfy the infinite set of conditions

Question

Consider the function space Ω, called the nuclear space, which consists of all functions $\phi(x)$ that satisfy the infinite set of conditions

$\int|\phi(x)|^2 (1+|x|)^n dx$ from -$\infty$ to $\infty$

for n = 0, 1, 2, . . . . The condition for n = 0 is precisely the one that defines H, so Ω is a subspace of H. Show that for any $\phi(x)$ in Ω, the function ˆx$\phi(x) ≡ $x$\phi(x)$ is also in Ω. This means the operator ˆx preserves Ω, even though it does not preserve all of H.

The attempt

$\int|\phi^*(x)ˆx\phi(x)| (1+|x|)^n dx$

$\int|\phi^*(x)x \phi(x)| (1+|x|)^n dx$

$\int|x\phi^*(x)\phi(x)| (1+|x|)^n dx$

$\int|x\phi^*(x)\phi(x)| (1+|x|)^n dx$

$\int|x\phi(x)|^2 (1+|x|)^n dx$

$\int|x\phi^*(x)\phi(x)|^2 (1+|x|)^n dx$

$\int|x^2$|$|\phi(x)|^2 (1+|x|)^n dx$

If this is the correct approach, I don't know how I can show that this is finite when phi is any function

Answer 1

\begin{align*} \int|x\phi(x)|^{2}(1+|x|)^{n}dx&=\int|\phi(x)|^{2}|x|^{2}(1+|x|)^{n}dx\\ &\leq\int|\phi(x)|^{2}(1+|x|)^{2}(1+|x|)^{n}dx\\ &=\int|\phi(x)|^{2}(1+|x|)^{n+2}dx\\ &<\infty. \end{align*}