If $u \in H^1(0,\infty)$, does $u_x(\infty) = 0$?

Question

Let $u \in H^1(0,\infty)$. Then is it true that $u'(x) \to 0$ as $x \to \infty$?

I am wondering by Green's formula/IBP in this setting; do I get something like $$\int_0^\infty u_{xx}v = -\int_0^\infty u_xv_x + u_x(0)v(0)?$$

Answer 1

For $H^1$- functions, this does not hold in general. It is basically the same argument that $L^2$- functions generally do not vanish at $\infty$, you just have to take such a function and integrate it, for example a bump function where the bumps get thinner when you go outside.

For $H^2$-functions however, it does hold: Take $$\int_0^a f'(x)f''(x)^*+f'(x)^*f''(x)^=|f'(a)|^2-|f'(0)|^2$$ As $f',f''$ are $L^2$, the product $f'f''$ is $L^1$, so the above expression has to be finite if you send $a$ to infinity. Because $f'$ is also $L^2$, it can only go to 0.