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Let $\Omega=\mathbf{R}^{n-j}\times\omega$, where $\omega\subset\mathbf{R}^j$ is a smooth bounded domain. Consider a function $u:\overline\Omega\rightarrow\mathbf{R}$ that satisfies $u(x,y)+k\leq C^{m+1}(u(0,y_0)+k),\quad\mbox{for}\ (x,y)\in\{|x-me_1|\leq1\}\times\overline\omega,$ for constants $k$ and $C$ and for each $m=0,1,...$. We can obtain this inequality for each direction in $\mathbf{R}^{n-j}$. This inequality yields at most exponential growth in the direction $e_1$. In other words, exists positive constants $\alpha$ and $A$ such that $u(x,y)\leq Ae^{\alpha|x|},\quad\mbox{in}\ \Omega.$ Why?

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    What is the set $\{|x-me_1|\leq1\}$? Are you saying that $(x_1-m)^2+x_2^2+\cdots+x_{n-j}^2\leq1$? Also, what is $y_0$, and how does it relate to $y$ or the direction in $\mathbb R^{n-j}$?2012-12-12

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We have: $u(x,y)+k\leq C^{m+1}(u(0,y_0)+k),\quad\mbox{for}\ (x,y)\in\{|x-me_1|\leq1\}\times\overline\omega,$ but since we have a version of this inequality for every direction in $\mathbf R^{n-j}$, we can choose it in the direction of $x$ so that this equation becomes $u(x,y)+k\leq C^{m+1}(u(0,y_0)+k)\quad\mbox{for}\ \ m-1\leq |x|\leq m+1.$

Thus, assuming $C\geq 1$ and $k\geq -u(0,y_0)$, $u(x,y)+k\leq C^{m+1}(u(0,y_0)+k)\leq C^{|x|+2}(u(0,y_0)+k)$ so that (if $k\geq 0$), choosing $A=C^2(u(0,y_0)+k)$ and $\alpha=\log C$, $u(x,y)+k\leq Ae^{\alpha|x|}\Rightarrow u(x,y)\leq Ae^{\alpha|x|}.$

(If $k<0$, the argument is a bit more involved, but the result is still true with judicious choice of $A$. If $C<1$, then $u$ is absolutely bounded, so pick $\alpha=0$.)

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    I understand that your argument is correct and I understood the steps. Then this argument implies the result in all directions, am I right? Because in the first inequality you can replace $e_1$ by $e_n$ for each $n$. And then you can repeat your argument!2012-12-15