Vanishing at the infinity of a function in the Sobolev space.

Question

If $f \in H^s (\Bbb R^n)$ for $s > 1 + \frac{n}{2} $ then the Sobolev inequality implies that $f$ and $\nabla^\alpha f$ ($|\alpha| =1$) vanishes at the infinity. ($\alpha$ : multi-index). But in this case can I conclude that $\nabla^\alpha f$ vanishes at the infinity also for all $|\alpha| \le s$ ? ($H^s$ : general Sobolev space, $s=0,1,2,\cdots$.)

Answer 1

Yes. But this has nothing to do with Sobolev inequalities. Let $A_j = B(0,2^{j}) \setminus B(0,2^{j-1})$ where $B(x,R)$ is the ball of radius $R$ centered at $x$. So $A_j$ are precisely the dyadic annuli. Then we have that for any $k > 0$

$$ \|f\|_{L^2(\mathbb{R}^n)}^2 \geq \|f\|_{L^2(B(0,1)}^2 + \sum_{j = 1}^k \|f\|_{L^2(A_k)}^2 $$

Hence the series $\sum_{j = 1}^\infty \|f\|_{L^2(A_k)}^2$ converges absolutely which implies that for every $\epsilon$ there exists some $K$ such that

$$ \|f\|_{L^2(\mathbb{R}^n \setminus B(0,2^k))}^2 \leq \epsilon \qquad \forall k > K $$

In other words, $f$ decays (in the $L^2$ sense) at infinity. Now, if you insist on asking about point-wise behaviour, let me point out to you that Sobolev "functions" are actually equivalent classes for functions which may differ on sets of measure zero. Given any function $g$ I can get another representative of the equivalent class by setting

$$\tilde{g}(x) = \begin{cases} \ell & x = (\ell,0,0,\ldots,0) \quad \ell \in \mathbb{Z} \\ g(x) & \text{otherwise} \end{cases} $$

and only one of $g(x)$ and $\tilde{g}(x)$ can be pointwise bounded at infinity.

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What you may have gotten confused is with regards to the statement of Sobolev inequality. Sobolev inequality (in the case of Morrey which gives embedding of $H^{n/2 + \epsilon} \to C^0$) does not guarantee that a function in a given Sobolev class is continuous. What Sobolev inequality guarantees is that you can find a (uniformly) continuous representative in the that equivalent class of functions, which when coupled with the $L^2$ decay indicated above would imply a pointwise decay at infinity.

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In terms of PDEs (since you tagged it), how Sobolev inequality guarantees regularity properties is in that we use it as an a priori control for the solutions. That is to say, suppose we have a sequence of continuous approximate solutions $f_k \in C^0 \cap H^s$ with $s > n/2$ such that $f_k \to f$ in $H^s$, we can use Sobolev's inequality to conclude that $f_k \to $ a continuous representative of $f$ uniformly in $C^0$, and thus we can a fortiori assume that $f$ is $C^0\cap H^s$ and so on and so forth.