Given $f(x,y) \in C^2([0,1]^2)$ (by which I mean $C^2$ in some open neighborhood), with $f_x, f_y, f_{xy} \in L^1([0,1]^2, dxdy)$ (which is sure case since they are continuous), does the following hold? $$ \sup |f| \le \iint |f|+|f_x|+|f_y|+|f_{xy}|\, dxdy $$
I think it is safe to
- discretise this function under the assumptions,
- divide $f$ into positive part and negative part, and
- consider $f-\inf f$ instead so that one could assume that $f\ge 0$.
Denote $a=f(0,0)$, $b=f(0,1)$, $c=f(1,0)$, $d=f(1,1)$, and I want to show $$\max \{a, b, c, d\}\le \frac14 (a+b+c+d)+\frac12 (|b-a|+|c-a|+|d-b|+|d-c|)+(|d-c-b+a|),$$ which is not hard and done with absolute value inequalities.
As I move on to refine the division, I find difficulties with those coefficients. RHS reads to me as $$\frac1{(n+1)^2} \sum f_{ij}+\sum \frac {\delta_i f_{ij}+\delta_i f_{i,j+1}}{2n}+\sum \frac {\delta_j f_{ij}+\delta_j f_{i+1,j}}{2n}+\sum \Delta f_{ij},$$ where $i,j \in \{0, 1, \ldots, n\}, f_{ij}=f(i,j)$, $\delta_i f_{ij}=|f_{i+1,j}-f_{ij}|$,$\Delta f_{ij}=|f_{i+1, j+1}-f_{i+1,j}-f_{i,j+1}+f_{ij}|$.
I still want to show that $\max \{f_{ij}\}$ is less than the above expression. Or maybe some other simpler solution.
I guess the claim is wrong considering an indicator function of a tiny square $[0,r]^2$ and some smooth approximations of it.