"Uniformly continuous" environment sum of a bijection $\varphi:\mathbb{Z}\times \mathbb{Z} \to \mathbb{Z}$ Given any function $f: \mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}$ we define the environment sum of $(x,y)\in\mathbb{Z}\times \mathbb{Z}$ with respect to $f$ by
$$\text{es}_f(x,y) = \sum\{f(x', y'): |(x',y') - (x,y)|=1\}.$$
Question. Is there a bijection $\varphi:\mathbb{Z}\times \mathbb{Z} \to \mathbb{Z}$ such that there is an integer $K\geq 0$ such that for all $(x,y)\in \mathbb{Z}\times\mathbb{Z}$ the following holds?

whenever $(x',y')\in \mathbb{Z} \times \mathbb{Z}$ with $|(x',y') - (x,y)|=1$ then $|\text{es}_\varphi(x,y) - \text{es}_\varphi(x',y')|<K$.

Bonus question in case of Yes above. (Need not be answered for acceptance) If yes, what is the smallest possible value for the constant $K$?
 A: There exist bijections $f:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ with $\text{es}_f(x,y)$ vanishing identically, which directly answer the bonus question.  
Note that the infinite set of equations that such a bijection must satisfy, 
$$\begin{align} 
         {f(x,y+1)+f(x,y-1)+f(x+1,y)+f(x-1,y)=0,\\ \forall (x,y)\in \mathbb{Z}\times\mathbb{Z},}
         \end{align}\tag{1}$$
splits into two disjoint subsets of equations, according to the parity of  $x+y$. Therefore we may express this $f$  gluing   two maps:
$$\begin{equation}f(x,y)=\cases{g\big({x+y\over2},{x-y\over2}\big)&if $x+y=0\mod 2$\\ \\
h\big({x+y+1\over2},{x-y+1\over2}\big)&if $x+y=1\mod 2$
}  
         \tag{2}\end{equation}$$
where $g:\mathbb{Z}\times\mathbb{Z}\to G$ and $h:\mathbb{Z}\times\mathbb{Z}\to H$ are bijections onto complementary subset $G$, resp., $H$, of $\mathbb{Z}$, and 
satisfy, for all $(u,v)\in\mathbb{Z}\times\mathbb{Z}$ the equations
$$\begin{equation}g(u,v)+g(u+1,v)+g(u,v+1)+g(u+1,v+1)=0 \tag{3}\end{equation}$$
$$\begin{equation}h(u,v)+h(u+1,v)+h(u,v+1)+h(u+1,v+1)=0 \tag{3'}\end{equation}$$
which translate $(1)$. We may choose $G:=3\mathbb{Z}$ and $H:=\mathbb{Z}\setminus3\mathbb{Z}$, and proceed to construct bijections $g:\mathbb{Z}\times\mathbb{Z}\to 3\mathbb{Z}$ and $h:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}\setminus3\mathbb{Z}$ satisfying $(3)$ and $(3')$, thus producing via $(2)$ a bijection $f:\mathbb{Z}\times\mathbb{Z}\to \mathbb{Z}$ satisfying $(1)$ as promised.
Let $A$ the set of nonnegative integer numbers of the form $a=\sum_{k=0}^p a_i 9^{i}$ with $a_i\in\{0,1,2\}$, that is, those $a\in\mathbb{ Z}_+$ whose base $3$ expansion is supported on the even-order places. So $B:=3A$ is the set of the nonnegative numbers $b$ whose base $3$ expansion is supported on the odd-order places. 
Recall: 
Elementary fact: Every $n\in\mathbb{Z}$ can be expressed uniquely as $a-b$.  In other words, every integer $n$ has a unique base $-3$ expansion.  We can summarize it with the (slightly abusing?) notation
$$\begin{equation}\mathbb{Z}=A\oplus(-3A)\tag{4}\end{equation}$$
and with the same meaning, always looking at the digits in base $-3$, 
$$\begin{equation}\mathbb{Z}\setminus3\mathbb{Z}= (A\setminus9A)\oplus(-3A)\tag{4'}\end{equation}.$$
Note the bijection  $\alpha:\mathbb{Z}\to A$ defined by $\alpha(u):=\sum_{i=0}^p u_i9^{i}\in A $ for all $u:=\sum_{i=0}^p u_i(-3)^i\in \mathbb{Z}$. So  $\mathbb{Z}\ni v\mapsto3\alpha(v)\in B$ is a bijection too. Also note $u=\alpha(u)=3\alpha(u)\mod 2$.
Then define:
$$g(u,v):=(-1)^{u+v}\big(3\alpha(u)-9\alpha(v)\big),$$
which immediately satisfies equations $(3)$. To check that $g$ is a bijection  $\mathbb{Z}\times\mathbb{Z}$ onto $3\mathbb{Z}$ as stated, note that the   equation $g(u,v)=3n$ is equivalent to $(-1)^nn= \alpha(u)-3\alpha(v) $ (recall that $u=\alpha(u)=3\alpha(u)\mod 2$ for all $u$). The bijectivity then follows from the above mentioned elementary fact $(4)$, in addition to the maps 
$$\begin{align}  \mathbb{Z}\ni n&\mapsto (-1)^nn\in \mathbb{Z},\\ 
 \mathbb{Z}\ni u&\mapsto \alpha(u)\in A, \\
 \mathbb{Z}\ni v&\mapsto 3\alpha(v)\in 3A, \end{align}$$
all being bijective.
In order to define the map $h:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}\setminus3\mathbb{Z}$ we need a further bijection $\phi:\mathbb{Z} \to \mathbb{Z} \setminus 3 \mathbb{Z}  $ such that $u=\phi(u)\mod 2$. Clearly there exist such  maps because $\mathbb{Z} \setminus 3 \mathbb{Z} $ contains infinitely many odd and infinitely many even numbers; one can easily check that an example is $$\displaystyle\phi(u):={3u+\omega(u)\over 2},$$  where $\omega$ is $4$-periodic, $\omega(0)=-4$, $\omega(1)=-1$, $\omega(2)=-2$, $\omega(3)=1$). Moreover, since $\alpha(\mathbb{Z} \setminus 3 \mathbb{Z} )=A\setminus9A$,  we obtain by composition a bijection $\alpha\circ\phi:\mathbb{Z} \to  A\setminus9A$. Again, $\alpha(\phi(u))=\phi(u)=u\mod 2$ holds for all $u\in\mathbb{Z}$.
Finally define:
$$h(u,v):=(-1)^{u+v}\big(\alpha(\phi(u))-3\alpha(v) \big),$$
which again immediately satisfies $(3')$. To check that $h$ is a bijection of  $\mathbb{Z}\times\mathbb{Z}$ onto $\mathbb{Z}\setminus 3\mathbb{Z}$, note that the   equation $h(u,v)=m\in \mathbb{Z}\setminus 3\mathbb{Z}$ is equivalent to $(-1)^mm= \alpha(\phi(u))-3\alpha(v)$, because $\alpha(\phi(u))-3\alpha(v)=u+v\mod 2$. The bijectivity then follows by the elementary fact $(4')$, together with the bijectivity of the maps 
$$\begin{align}  \mathbb{Z}\ni m&\mapsto (-1)^mm\in \mathbb{Z},\\ 
 \mathbb{Z}\ni u&\mapsto \alpha(\phi(u))\in  A\setminus9A, \\
 \mathbb{Z}\ni v&\mapsto 3\alpha(v)\in 3A, \end{align}$$
concluding the proof.
