Let: $$K=[0,5]\times[0,5]\subset \mathbb R^2,$$ $$L \text{ - the edge of the square } K,$$ $$f: K \to \mathbb R \text{ which is continuous at square } K, \text{ differentiable in the interior } K \text{ and } \forall _{x,y \in (0,5)} \frac{\partial f}{\partial x}\left(x,y\right)+\frac{\partial f}{\partial y}\left(x,y\right)\le 0$$ Prove that $\sup _K f=\sup _L f$ and $ \inf _K f=\inf _L f$
My try:
Let $g(t)=f(tx,ty)$ where $tx,ty \in (0,1)$. $f$ is differentiable in the interior $K$ so: $$g'(t)=f'(tx,ty)=x\frac{\partial f}{\partial x}\left(x,y\right)+y\frac{\partial f}{\partial y}\left(x,y\right)$$ Since $\forall _{x,y \in (0,5)} \frac{\partial f}{\partial x}\left(x,y\right)+\frac{\partial f}{\partial y}\left(x,y\right)\le 0$ we have also that: $$\begin{cases} x\frac{\partial f}{\partial x}\left(x,y\right)+x\frac{\partial f}{\partial y}\left(x,y\right)\le 0 \\ y \frac{\partial f}{\partial x}\left(x,y\right)+y \frac{\partial f}{\partial y}\left(x,y\right)\le 0 \end{cases} \Rightarrow g'(t)+ x\frac{\partial f}{\partial y}\left(x,y\right)+y\frac{\partial f}{\partial x}\left(x,y\right)\le 0$$ Then I think that one can somehow conclude that $f$ is monotonic, and then come to the thesis, but I don't know exactly what to do next.
Take a point $(x,y)$ in the interior of $K$ and connect it to the boundary with a line of slope $(1,1)$, i.e. consider the function $$ g(t) = f(x-t,y-t). $$ Because $$ g'(t) = \partial_x f (-1) + \partial_y f (-1) \geq 0, $$ $g$ is increasing so $g(0)=f(x,y)\leq g(T) = f(x-T,y-T)$, where $(x-T,y-T)$ belongs to $[0,5]\times \{0\} \cup \{0\}\times [0,5] \subset L$. To find the exact point study separately the cases $x\geq y$ and $x<y$.
To prove $\min_K f=\min_L f$, repeat the same but this time take $$ h(t) = f(x+t,y+t)$$ and show that $h(0)\geq h(T)$, where $(x+T,y+T)\in [0,5]\times \{5\}\cup \{5\}\times [0,5]$.