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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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Can you clarify the meaning of $\overline\Omega$? – Mario Carneiro Dec 12 '12 at 6:26
This is not exactly an answer, since I'm not quite clear on all the notation in the hypothesis, but it seems to me that it is saying (translated to 1D) that $u(x)+k\leq C^2u(0)$ for $|x|<1$, $u(x)+k\leq C^3u(0)$ for $|x|<2$, and so on, so that for large $x$ $u(x)\lesssim C^xu(0)$. – Mario Carneiro Dec 12 '12 at 6:37
Mario, $\overline\Omega=\Omega\cup\partial\Omega$. What I can't prove is that the inequalite implies that for some positive constants $\alpha$ and $C$, we have $$u(x,y)\leq Ce^{\alpha|x|}, \ \ \ \mbox{in} \ \ \Omega.$$ – José Carlos Dec 12 '12 at 8:12
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 $\mathbb R^{n-j}$? – Mario Carneiro Dec 12 '12 at 8:56
up vote 0 down vote accepted

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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Hi mario, I think that your answer is correct! I forgot to say that $k\geq0$. I just didn't understand the part that you says: "...we can choose it in the direction of $x$ so that this equation becomes..." This argument that you constructed can be used for one direction. I don't know why do I need the first inequality in all directions? Now, I'm thinking that I need the first inequality in just one direction. Did you understand my questioning? – José Carlos Dec 13 '12 at 0:55
Your expected result was $u(x, y)\leq Ce^{\alpha|x|}$, so it is supposed to be true in every direction. Otherwise, you just get $u(x, y)\leq Ce^{\alpha x_1}$, and only for points with $x_1\geq0$ relatively close to the $x_1$-axis. What exactly do you want the answer to look like? – Mario Carneiro Dec 13 '12 at 7:24
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! – José Carlos Dec 15 '12 at 20:42

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