Assume there is a Sobolev function $f\in D^{1,2}(\mathbb{R}^2)\cap D^{2,1}(\mathbb{R}^2)$, in another word $\nabla f\in L^{2}(\mathbb{R}^2)$ and $\nabla^2 f\in L^{1}(\mathbb{R}^2)$. What can we say about $f$ ?
Can we deduce that $f$ has a limit at infinity? i.e. $\lim\limits_{|x|\to\infty}f(x)$ exists.
Generally speaking, under what condition in Sobolev spaces a function would have a limit at infinity?
On
In fact, this question is not difficult. For any smooth function with compact support, it holds that
$$u(x_{1},x_{2})=\int_{-\infty}^{x_{1}}\int_{-\infty}^{x_{2}}\frac{\partial^{2}u}{\partial{x_{1}}\partial{x_{2}}}(y_{1},y_{2})dy_{1}dy_{2}$$
Hence, $||u||_{L^{\infty}(\mathbb{R}^{2})}\leq C||\nabla^{2}u||_{L^{1}(\mathbb{R}^{2})}$. Then by approximation this is also true for $D^{2,1}$ function, and one has obtained the embedding $D^{2,1}\subset C_{0}$. The existence of limit at infinity is an easy corollary.
By Theorem 4.12 in Adams and Fournier's book Sobolev spaces, $W^{2,1}(\mathbb R^2)$ embeds into $C_0(\mathbb R^2)$ (and in general $W^{n,1}(\mathbb R^n)$ embeds into $C_0(\mathbb R^n)$), so the answer is yes.
The proof goes like this. By taking the second-order Taylor expansion with integral remainder and integrating it with respect to the point of expansion (not the point of evaluation), we show the estimate
$$ \| u \|_{L^\infty(\mathbb R^2)} \le C \| u \|_{W^{2,1}(\mathbb R^2)} \quad \text{for } u \in C_c^\infty(\mathbb R^2). $$ Since $W^{2,1}$ is the closure of $C_c^\infty$ in $W^{2,1}$-norm, the above estimate implies that $W^{2,1}$ is contained in the closure of $C_c^\infty$ in $L^\infty$-norm, i.e., in the space $$ C_0(\mathbb R^2) = \{ u \in C(\mathbb R^2) : u(x) \to 0 \text{ when } x \to \infty \}. $$