I have a differential inequality of the kind: $$ \Delta f \ge f_x (1 + f_x^2 + f_y^2) $$ where $f \colon \Omega \subset \mathbb{R}^2 \to \mathbb{R}$.
Assume to know that $f_x > 0$ on $\Omega$ and $f$ is convex, so in particular $f_{xx} \ge 0$ and $f_{yy} \ge 0$.
From this I would like to show that $\Omega$ can not be the whole $\mathbb{R}^2$. I tried to show one of the partial derivatives must blow up in a bounded region, but all the ideas I tried didn't work.
Any hint would be really appreciated! Thanks a lot!
I will prove the $\Omega\subset\mathbb R^1$ case, and give you a hint about the $\Omega\subset\mathbb R^2$ case: apply the scaling $y\mapsto y/k$ for $k\gg 1$ to make the $f_{yy}$ derivative small. If you would like more details, I can provide a complete solution.
Let us consider the simpler case of $f:\mathbb R\to\mathbb R$ solving $f_{xx}\ge f_x(1+f_x^2)$, assuming similar conditions on $f$. We have $\frac{1}{f_x(1+f_x^2)}=\frac{1}{f_x}-\frac{f_x}{1+f_x^2}$, so after rearranging, the inequality becomes $$ \frac{d}{dx}\ln\frac{f_x}{\sqrt{1+f_x^2}}\ge 1. $$ Let $g(x)=e^{-x}\frac{f_x}{\sqrt{1+f_x^2}}$. Then integrating the inequality implies $g(x)\ge g(0)$. But $g(x)\le e^{-x}$, so $g(0)=0$, a contradiction to $f_x>0$.
EDIT: The above hint was wrong! (The rescaling is supposed to be in all variables, not just $y$) I have included (what I believe is) the correct solution below, but if this also has errors, please unaccept my answer.
First, let $f$ be such a solution of the inequality. For each $k>1$, the rescaled function $f_k(x,y)=k f(k^{-1}x,k^{-1}x)$ is also a solution, and it has the same positivity properties. We will derive an inequality valid for all solutions $g$, then substitute $g=f_k$ and re-express in terms of $f$.
For simplicity, let us remove the $g_y^2$ term: $$ g_{xx}+g_{yy}\ge g_x(1+g_x^2), $$ such that $$ \partial_x\ln\frac{g_x}{\sqrt{1+g_x^2}}\ge 1-\frac{g_{yy}}{g_x(1+g_x^2)}. $$ Hence, $$ e^{-x}F[g](x,y)\ge F[g](0,y)G[g](x,y), $$ where \begin{align} F[g](x,y)&:=\frac{g_x(x,y)}{\sqrt{1+g_x^2(x,y)}},\\ G[g](x,y)&:=\exp-\int_0^x\frac{g_{yy}(t,y)}{g_t(t,y)(1+g_t^2(t,y)}dt. \end{align} Let us now put $g=f_k$. Observe that \begin{align} F[f_k](x,y)&=F[f](k^{-1}x,k^{-1}y),\\ G[f_k](x,y)&=G[f](k^{-1}x,k^{-1}y). \end{align} This means $$ e^{-kx}F[f](x,y)\ge F[f](0,y)G[f](x,y). $$ Fixing $x>0$ and letting $k\to \infty$ gives the result.