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If $u \in C_0^\infty (0,\infty)$, $v(x) := u(x) e^{-x/2}$, then I want to prove the equality $$ \int_0^\infty ( |u(x) |^2 + x | u'(x) |^2 ) e^{-x} dx = \int_0^\infty \left( \frac{x+2}{4} | v(x) |^2 + x | v'(x) |^2 \right) dx$$ holds. Here $C_0^\infty$ means that $C^\infty$ functions with a compact support.

And If I change the condition of $u$ as $\int_0^\infty ( |u(x) |^2 + x | u'(x) |^2 ) e^{-x} dx < \infty $ then does this still hold?

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Since $u(x)=v(x)e^{x/2}$, one has $$ u'(x)=(v'(x)+\frac{1}{2}v(x))e^{x/2}. $$ Therefore \begin{eqnarray} \int_0^\infty\left(|u(x)|^2+x|u'(x)|^2\right)e^{-x} dx &=& \int_0^\infty\left(|v(x)e^{x/2}|^2+x|(v'(x)+\frac{1}{2}v(x))e^{x/2}|^2\right)e^{-x} dx\cr &=& \int_0^\infty\left(|v(x)|^2+x|v'(x)+\frac{1}{2}v(x)|^2\right)dx\cr &=& \int_0^\infty\left(\frac{x+2}{4}|v(x)|^2++x|v'(x)|^2\right)dx+\int_0^\infty \frac{1}{2}|v(x)|^2dx\cr &+&\int_0^\infty xv(x)v'(x)dx. \end{eqnarray} If $u \in C_0^\infty(\mathbb{R})$, then there is some $R>0$ such that $u(x)=0$ for $x \ge R$, and $$ \int_0^\infty xv(x)v'(x)dx=\int_0^R xv(x)v'(x)dx=\frac{1}{2}xv^2(x)\big|_0^R-\frac{1}{2}\int_0^Rv^2(x)dx=-\frac{1}{2}\int_0^\infty v^2(x)dx. $$ Thus $$ \int_0^\infty\left(|u(x)|^2+x|u'(x)|^2\right)e^{-x} dx =\int_0^\infty\left(\frac{x+2}{4}|v(x)|^2++x|v'(x)|^2\right)dx. $$

Let's now replace the condition $u \in C_0^\infty(\mathbb{R})$ by $$ \int_0^\infty\left(|u(x)|^2+x|u'(x)|^2\right)e^{-x} dx<\infty. $$ Then $u \in L^2(\mathbb{R}, e^{-x}dx)$.

For every $R>0$ we have $$ \int_0^R\frac{1}{2}|v(x)|^2dx+\int_0^R xv(x)v'(x)dx=\frac{1}{2}Rv^2(R)=\frac{1}{2}Re^{-R}u^2(R). $$ Suppose $$ \lim_{x \to \infty}xe^{-x}u^2(x)=a>0. $$ Then there is an $r=r(a)>0$ such that $$ |xe^{-x}u^2(x)-a| \le \frac{a}{2} \quad \forall x \ge r. $$ It follows that $$ e^{-x}u^2(x)=\frac{xe^{-x}u^2(x)-a +a}{x}=\frac{xe^{-x}u^2(x)-a}{x}+\frac{a}{x} \ge -\frac{a}{2x}+\frac{a}{x}=\frac{a}{2x} \quad \forall x \ge r. $$ Thus $$ \int_0^\infty e^{-x}u^2(x)dx \ge \frac{a}{2}\int_r^\infty x^{-1}dx=\infty, $$ contradicting the fact that $u \in L^2(\mathbb{R},e^{-x}dx)$. Hence $$ \lim_{x \to \infty} xe^{-x}u^2(x)=0, $$ and the identity holds.

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Thank you very much! –  Ann Jul 31 '12 at 8:51

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