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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, $s=0,1,2,\cdots$.)

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up vote 2 down vote accepted

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.

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.

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.

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A similar concept is that for the trace theorems in Sobolev classes. We can think about this slightly differently as "given some fact (an estimate) which is true on the dense subset $H^s \cap C^\infty$, we try to take the completion of this fact to $H^s$ by using continuity". – Willie Wong Nov 28 '12 at 8:53

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