Is this set closed in $H^1(\mathbb{R}^2)$?

Question

Only hints please no full answers!

Let $$E=\{u\in H^1(\mathbb{R}^2):\lim_{y\to\infty}u(x,y)=0,\forall x\in \mathbb{R}\}.$$

We have that $E$ is a subspace. Is $E$ closed?

Suppose that $\{u_n\}$ is a sequence in $E$ such that $u_n\to u\in H^1(\mathbb{R}^2)$. We have that $u_n\to u\in L^2(\mathbb{R^2})$ and $\nabla u_n|_\omega\to \nabla u|_\omega\in L^2(\mathbb{R}^2)\times L^2(\mathbb{R}^2)$ for all compact sets $\omega\subset\mathbb{R}^2$. I'm not sure if this is enough though to show that $u\in E$.

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Since I don't have an idea on how to solve the above, I am tempted to ask that in general, can we find integers $m,p>0$ such that $$E=\{u\in W^{m,p}(\mathbb{R}^2):\lim_{y\to\infty}u(x,y)=0,\forall x\in \mathbb{R}\},$$ a closed set?

Answer 1

$\newcommand{\R}{\mathbb{R}}$ As pointed out by Raoul, already the definition of $E$ is troublesome, since $u \in H^1(\R^2)$ is not really a function. Even if we say that it's a function up to equality a.e., limits along lines still don't make sense.

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To make things meaningful, recall the trace operator, or rather the whole family of traces: $$ T_x \colon H^1(\R^2) \to L^1(\R) \quad \text{for all } x \in \R. $$ The fundamental property of $T_x$ is that

• $T_x \varphi(y) = \varphi(x,y)$ for $\varphi \in C_c^\infty(\R^2)$ (i.e., for such functions $T_x$ is really the restriction to $\{ x \} \times \R$),
• $T_x$ is a continuous linear operator from $H^1(\R^2)$ to $L^1(\R)$.

With $T_x$ you can now meaningfully refer to $u(x,y)$ as a function of $y$ (for each fixed $x$). But what about the limit?

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$\newcommand{\ACL}{\mathrm{ACL}}$ For this, recall the $\ACL^2$-characterization of $H^1(\R^2)$ ($\ACL$ stands for absolutely continuous on (almost all) lines). That is, the following theorem:

ACL characterization.

• If $u \in H^1(\R^2)$ then $T_x u \in H^1(\R)$ for almost every $x \in \R$. Moreover, for such $x$ we have $(T_x u)'(y) = \frac{\partial u}{\partial y}(x,y)$ a.e.
• Any function $v \in H^1(\R)$ is absolutely continuous: its derivative satisfies the fundamental theorem of calculus. Moreover, since $$ |v(a)-v(b)| \le \int_a^b |v'(t)| dt \le |a-b|^{1/2} \left( \int_a^b |v'(t)|^2 dt \right)^{1/2}, $$ $v$ is even $\frac 12$-Holder continuous.

Asking for the limit $\lim_{y \to \infty} v(y)$ makes sense if $v$ is continuous, and hence also if $v \in H^1(\R)$. With the above theorem, you can safely define $$ E= \left\{ u\in H^1(\R^2): \lim_{y\to\infty} T_x u(y)=0 \quad \text{for a.e. } x \in \R \right\}. $$ Good luck!

Answer 2

What does a limit mean when the functions are not really functions, but distributions? Regardless, let's assume that you also restrict $E$ to continuous functions. Even then, it seems tough, since limits do not work too well with $L^p$ convergence.

Hint for a counterexample. Take a continuous function $u$ in $H^1 \backslash E$. Then construct a sequence $(u_n)$ of $E$ that converges to $u$. For instance, slightly modify $u$ away from the origin, in order to ensure that you have a vanishing limit.