Convex Point exists? Let $f$ be differentiable on $(a,b)$. Suppose $f$ is not a linear function, show that $f$ has at least one convex or concave point in $(a,b)$.
Here, a point $c\in (a,b)$ is called a convex (concave) point of $f$ povided that there exists a neighborhood $U$ of $c$ such that $x\in U\Rightarrow f(x)>\ (<)f(c)+f'(c)(x-c).$
 A: It is not a complete answer, since I supposed some conditions on $f'$. (In particular, notice that there exist continuous functions not locally monotonic at any point.)


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*If there is an interval $I \subset (a,b)$ on which $f'$ is strictly monotonic:


Without loss of generality, suppose $I$ open and $f'$ increasing on $I$. Let $c,x \in I$. 
If $x<c$, then there exists $d_x \in (x,c)$ such that $\displaystyle \frac{f(x)-f(c)}{x-c}=f'(d_x)<f'(c)$, hence $f(x)>f(c)+f'(c)(x-c)$; if $x>c$, then there exists $d_x \in (c,x)$ such that $\displaystyle \frac{f(x)-f(c)}{x-c}=f'(d_x)>f'(c)$, hence $f(x)>f(c)+f'(c)(x-c)$.


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*If $f$ is $C^2$:


If $c \in (a,b)$ is neither a convex point nor a concave point, then every neighborhood $U$ of $c$ contains a point $x$ such that $f(x)=f(c)+f'(c)(x-c)$; differentiating twice, $f''(x)=0$.
Therefore, if there is no convex or concave point on $(a,b)$, $\{x \mid f''(x) =0 \}$ is dense in $(a,b)$, and closed since $f''$ is continuous. So $f''=0$ and $f$ is linear.
