1

By $H^2(\mathbb{R})$ denote the Sobolevspace $$ W^{2,2}(\mathbb{R}):=\left\{u\in L^2(\mathbb{R}): D^{\alpha}u\in L^2(\mathbb{R})~\forall\lvert\alpha\rvert\leq 2\right\} $$ which has an inner product $$ \langle u,v\rangle_{H^2}:=\sum_{i=0}^2\langle D^iu,D^iv\rangle_{L^2} $$ and norm $$ \lVert u\rVert_{H^2}:=\left(\sum_{\lvert\alpha\rvert\leq 2}\lVert D^{\alpha}u\rVert_{L^2}^2\right)^{1/2}. $$

Let $u\in H^2(\mathbb{R})$. Do we then have that $u\to 0$ and $Du\to 0$ as $x\to\pm\infty$?

idea

Suppose, $u$ is not tending to zero as $x\to\pm\infty$. Since $\lvert u(x)\rvert^2\geq 0$ for all $x\in\mathbb{R}$, I think this would imply that $\int_{\mathbb{R}}\lvert u(x)\rvert^2\, dx = \infty$, contradicting the assumption that $u\in H^2(\mathbb{R})$ since this means that $u\in L^2(\mathbb{R})$, i.e. $\int_{\mathbb{R}}\lvert u(x)\rvert^2\, dx <\infty$.

mathfemi
  • 2,631
  • I guess there is a typo in $u \to 0$ and $u_x \to 0$? Idea: Suppose that $u$ would not tend to zero and use the fact that $u \in L^2(\mathbb{R})$. – T'x Oct 06 '16 at 12:10
  • @T'x Where do you see a typo and why? – mathfemi Oct 06 '16 at 12:17
  • 1
    what do you mean with $u_x$? – T'x Oct 06 '16 at 12:18
  • I mean the derivative of u with respect to x. – mathfemi Oct 06 '16 at 12:18
  • 1
    Well, only the weak derivative exists. So you should write $u(x) \to 0$ and $Du(x) \to 0$ as $|x|\to \infty$. – T'x Oct 06 '16 at 12:21
  • Keep in mind that functions in $H^2$ are equivalence classes, so anything like $u(x)$ is not defined. Sorry for my misleading idea, I had the wrong picture in my mind. Maybe this helps you. – T'x Oct 06 '16 at 12:40
  • I am sorry, I do not understand. – mathfemi Oct 06 '16 at 12:55
  • Functions in $L^2$ (an so in $H^2$) are equal, if they are equal almost everywhere. That means for $f \in H^2$ the expression $f(x)$ is not well definied, since we could change the value of $f(x)$, but the integral would be the same. – T'x Oct 06 '16 at 14:25
  • Okay. Maybe I am just too stupid, but I still do not see how to prove that for $u\in H^2(\mathbb{R})$, we have that both u and its first weak derivative are vanishing for $x\to\pm\infty$. – mathfemi Oct 06 '16 at 14:32
  • The point is, that they are vanishing in $L^2$ sense. A possible proof is given here. As I said above $u(x) \to 0$ for $x \to \infty$ is not defined. (You first have to choose a smooth representative) – T'x Oct 06 '16 at 14:45
  • So let $u\in H^2(\mathbb{R})$. Then there exists some $w\in C^1(\mathbb{R})$ with $u=w$ almost everywhere. And then? – mathfemi Oct 06 '16 at 16:09

1 Answers1

1

We use the following result, which is classical and proven for example here in Theorem 3.37. See also this question.

Let $u \in W^{1,p}(\mathbb{R}^n)$ with $n<p<\infty$. Then $u \in C_0^{0,1-n/p}(\mathbb{R}^n)$ and $u(x) \to 0$ as $x \to \pm \infty$.

Let $u \in H^2(\mathbb{R})$. In particular $u \in W^{1,2}(\mathbb{R})$ and we see that $1=n<p=2$. Thus by applying above theorem we get that $u$ has zero limit as infinity.

Moreover, we have $Du \in W^{1,2}(\mathbb{R})$ and by applying above result on $Du$ gives us $Du(x) \to 0$ as $x\to \pm \infty$.

To clarify the comments: Yes, it is important that one says that one identifies $u$ and $Du$ with their continuous representations, so that $u(x)$ and $Du(x)$ are well-defined everywhere.

Cahn
  • 4,621