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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?

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I bet that you need to integrate by parts somewhere, which is just so pleasant on compact functions. – mixedmath Jul 29 '12 at 8:39
I think $C_0$ usually means 'goes to zero at $\infty$, $C_c$ is used for compact support? – copper.hat Jul 29 '12 at 9:07
up vote 2 down vote accepted

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 $\lim_{x\to\pm\infty}|u(x)|^2(\,x\,e^{-x^2})=0$.

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