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 […]

- Tags \ldots, $\Delta f_{ij}=|f_{i+1, $\delta_i f_{ij}=|f_{i+1, $$ which is not hard and done with absolute value inequalities. As I move on to refine the division, $b=f(0, $c=f(1, 0, 1, 1]^2)$ (by which I mean $C^2$ in some open neighborhood), 12, and consider $f-\inf f$ instead so that one could assume that $f\ge 0$. Denote $a=f(0, and I want to show $$\max \{a, B, c--, d\}\le \frac14 (a+b+c+d)+\frac12 (|b-a|+|c-a|+|d-b|+|d-c|)+(|d-c-b+a|), df1, divide $f$ into positive part and negative part, 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, dxdy)$ (which is sure case since they are continuous), f_{ij}=f(i, f_{xy} \in L^1([0, f_y, Given $f(x, 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, j \in \{0, j}-f_{i, j}-f_{ij}|$, j}}{2n}+\sum \Delta f_{ij}, j+1}-f_{i+1, j+1}}{2n}+\sum \frac {\delta_j f_{ij}+\delta_j f_{i+1, j+1}+f_{ij}|$. I still want to show that $\max \{f_{ij}\}$ is less than the above expression. Or maybe some other simpler solution., n, where i, with $f_x, y) \in C^2([0