A function is convex and concave, show that it has the form $f(x)=ax+b$ A function is convex and concave, it is called affine function. That is the function:
$$f(tx+(1-t)y)=tf(x)+(1-t)f(y),\, \, t\in (0,1) $$  
Force $y=0$(suppose $0$ is in the domain of $f(x)$), we obtain:
$$f(tx)-f(0)=tf(x)-tf(0)=t[f(x)-f(0)]$$
So 
$$F(x):=f(x)-f(0)$$ is linear, but we have a constraint here: $t\in(0,1)$. 
How can we expand this to arbitrary $t$, and assert it is really linear?
Thanks.

I find the answer myself! Using the first order property of convex(concave) function:
$$f(y)-f(x)\ge f'(x)(y-x)$$ for convexity,
$$f(y)-f(x)\le f'(x)(y-x)$$ for concavity.
combine them,
$$f'(x)=\frac{f(y)-f(x)}{y-x}$$
exchange the role of $x,y$
$$f'(y)=\frac{f(y)-f(x)}{y-x}$$
$$f'(x)=f'(y)$$
for $\forall x\neq y$ and $x,y\in (\cal{D-\partial D})$
that is
$$f'(x)$$ is constant for all $x\in int\cal{D}$
integrate it, $f(x)=ax+b$
