Constructing function as follows? 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 $\sigma$.. 
 A: 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.
