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I'm having difficulties in following construction used in proof.

If $u$ is continuous function on open set $\Omega \subset R^n$ and $ p \in R^n$ satisfies $$\displaystyle \limsup_{y\to x} \frac{u(y)-u(x)-p \cdot (y-x)}{|y-x|} \leqslant 0 $$ then show that there exists a positive $\delta$ and continuous, increasing function $\sigma $ defined on $[0,\delta]$ such that $\sigma(0)=0$ and $$u(y) \leq u(x)+p \cdot (y-x) + \sigma(|y-x|) \cdot |y-x|\mbox{ if }|y-x|<\delta.$$

I thought $p$ can be understood as a gradient of a tangent line which lies above $u$, but I can't proceed it to construct such $\sigma$..

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In your lim sup formula, are $x$ and $y$ both allowed to move ? Or $x$ is fixed and only $y$ moves ? –  Ewan Delanoy Dec 6 '12 at 10:19
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There is something wrong in your last line: When $\sigma$ is real valued then $\sigma(|y-x|)\cdot(y-x)$ is a vector. –  Christian Blatter Dec 6 '12 at 14:12
    
I just corrected mistake, thanks. –  Detectives Dec 6 '12 at 15:30
    
If $\Omega$ is not connected, $\sigma$ may not exist. For example, let $n=1$, $\Omega=\mathbb{R}\setminus\{0\}$, $u(x)=\frac{x}{|x|}$ and $p=0$. Then $u$ and $p$ satisfy the inequality but $u(y)-u(x)=2$ for any $x<0<y$. –  23rd Dec 7 '12 at 18:36

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When $\Omega$ is connected, actually $u(x)-p\cdot x$ is a constant function. Consequently, the choice of $\sigma$ is arbitrary.

To show $u(x)-p\cdot x$ is constant, it suffices to show that it is constant along any line segment contained in $\Omega$. To this end, given $x_0\in\Omega$, $v\in\mathbb{R}^n$ with $|v|=1$ and $r>0$ such that $x_0+tv\in\Omega$ when $|t|\le r$, let us show that $$f(t)=u(x_0+tv)-p\cdot(x_0+tv),\quad t\in[-r,r]$$ is a constant function of $t$.

By definition, $\limsup_{h\to 0}\frac{f(t+h)-f(t)}{|h|}\le 0$ on $[-r,r]$. For every $\epsilon>0$, define $f^\pm_\epsilon(t)=f(t)\mp \epsilon t$. It follows that, for every $t\in[-r,r]$,

$$\limsup_{h\to 0^+}\frac{f^\pm_\epsilon(t\pm h)-f^\pm_\epsilon(t)}{h}=\limsup_{h\to 0^+}\frac{f(t\pm h)-f(t)-\epsilon h}{h}\le -\epsilon.$$ On the one hand, the inequality above implies that $f^+_\epsilon$ is decreasing, i.e. $f(s)-f(t)\le \epsilon(s-t)$ when $s\le t$. Letting $\epsilon\to 0$, we can conclude that $f$ is decreasing. On the one hand, the inequality above implies that $f^+_\epsilon$ is increasing, so similarly we can conclude that $f$ is increasing. Therefore, $f$ must be constant.

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You are assuming that the assumption holds for all $x$, with $p$ independent of $x$? –  Lukas Geyer Dec 7 '12 at 17:40
    
@LukasGeyer: Yes, I think the question assumes that $p$ is a constant vector. –  23rd Dec 7 '12 at 17:44
    
I understood $x$ to be fixed in the question as well. –  Lukas Geyer Dec 7 '12 at 18:48
    
Yeah, maybe the poster should clarify. –  Lukas Geyer Dec 7 '12 at 19:09
    
@LukasGeyer: Now I am a little confused. If $x$ is fixed, the question is more or less trivial; if $x$ is variable but $p$ is fixed, the question seems wierd and the conclusion could be false when $\Omega$ is not connected; if $x$ is variable and $p$ is dependent on $x$, the conclusion is false in general. –  23rd Dec 7 '12 at 19:12

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