How prove this $f(a)\le f(b)$ Suppose $f(x)$ is continuous on $[a,b]$, and that for any $x\in (a,b)$, the limit
$$\lim_{h\to 0}\dfrac{f(x+h)-f(x-h)}{h}$$ exists and $$\lim_{h\to 0}\dfrac{f(x+h)-f(x-h)}{h}\ge 0.$$
Show that $f(a)\le f(b)$
this problem is from http://wenku.baidu.com/view/a643e6c26137ee06eff91855.html
 A: The statement in the question is a direct corollary of Theorem 2 in the linked article. The following self-contained argument is essentially borrowed from the proof of Lemma 1 in the same article.
Given $\epsilon>0$, define
$$f_\epsilon(x):=f(x)+\epsilon x,\quad \forall x\in[a,b].$$
By definition, for every $x\in(a,b)$, 
$$\lim_{h\to 0}\frac{f_\epsilon(x+h)-f_\epsilon(x-h)}{h}=2\epsilon+\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{h} \ge 2\epsilon.\tag{1}$$
Since $\epsilon$ is arbitrary, to show $f(a)\le f(b)$, it suffices to prove that $f_\epsilon(a)\le f_\epsilon(b)$. By reduction to absurdity, assume that $f_\epsilon(a)> f_\epsilon(b)$. Fix an arbitrary $y\in (f_\epsilon(b),f_\epsilon(a))$ and define
$$x_y:=\inf\,\{x\in[a,b]: f_\epsilon(x)<y\}.$$
By definition and the continuity of $f_\epsilon$, $x_y\in (a,b)$ and (i) $f_\epsilon(x)\ge y$ when $x\in[a,x_y]$; (ii) for every $x\in(x_y,b]$, there exists $x'\in (x_y,x)$, such that $f_\epsilon(x')<y$. (i) and (ii) implies that
$$\liminf_{h\to 0}\frac{f_\epsilon(x_y+h)-f_\epsilon(x_y-h)}{h}\le 0,$$
which contradicts to $(1)$ and completes the proof. 
