2
$\begingroup$

Let $u \in C_0^\infty (\mathbb R)$, $v(x) := u(x) e^{-x^2 /2} $. And define the norm as $ \| u \|_1^2 = \int_{\mathbb R} | u' (x) |^2 e^{-x^2} dx + \int_{\mathbb R} | u(x) |^2 e^{-x^2} dx $ Then I want to prove that $ \| u \|_1^2 = \int_{\mathbb R} ( | v' (x) |^2 + x^2 | v(x)|^2 ) dx $

I think this is not trivial by just using the definition of the norm above. $C_0^\infty $ means that $C^\infty$ functions with a compact support. And I have one more question.

If the condition $u \in C_0^\infty ( \mathbb R) $ changes to "$\| u \|_1^2 < \infty$", then does this still hold?

  • 0
    I think $C_0$ usually means 'goes to zero at $\infty$, $C_c$ is used for compact support?2012-07-29

1 Answers 1

2

We have $ |v(x)|^2+x^2\,|v'(x)|^2=(|u'(x)|^2-2\,x\,u(x)\,u'(x)+2\,x^2|u(x)|^2)\,e^{-x^2}. $ The desired equality is thus equivalent to $ \int_\mathbb{R}(-2\,x\,u(x)\,u'(x)+2\,x^2|u(x)|^2)\,e^{-x^2}\,dx=\int_\mathbb{R}|u(x)|^2\,e^{-x^2}\,dx. $ Integration by parts gives $\begin{align*} \int_\mathbb{R}(-2\,x\,u(x)\,u'(x)\,e^{-x^2})\,dx&=-\int_\mathbb{R}(|u(x)|^2)'\,x\,e^{-x^2}\,dx\\ &=\int_\mathbb{R}|u(x)|^2(\,x\,e^{-x^2})'\,dx\\ &=\int_\mathbb{R}|u(x)|^2(1-2\,\,x^2)\,e^{-x^2}\,dx, \end{align*}$ and the result follows. The computation is valid as long as $\lim_{x\to\pm\infty}|u(x)|^2(\,x\,e^{-x^2})=0$.