If $(u(x)v(x)), (u'v') \in L^1$ why are $u, v \in H^1$?

Question

Given functions $u(x), v(x)$ and given that $\int uv\:\mathrm{d}x < \infty$ and $\int u'v'\:\mathrm{d}x < \infty$ (that is, their product and the product of their derivatives are in $L^1$) why is it that $u, v \in H^1$?

Answer 1

This is not true in general.

Take $$ u(x)=\left\{\begin{array}{lll} \mathrm{e}^{-1/x^2} & \text{if} & x>0, \\ 0 & \text{if} & x\le 0,\end{array}\right. \quad\text{and}\quad v(x)=\left\{\begin{array}{lll} \mathrm{e}^{-1/x^2} & \text{if} & x<0, \\ 0 & \text{if} & x\ge 0.\end{array}\right. $$ Then $$ \int uv\,dx=\int u'v'\,dx=0, $$ but $u,v \not\in L^2(\mathbb R)$.