A function is convex and concave, show that it has the form $f(x)=ax+b$

Question

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.

Answer 1

You're almost done. The convex/concave property only speaks about $t\in[0,1]$, but the $x$ there is arbitrary, so you can just let that grow.

For example, $y=0, x=1$ shows that the function coincides with an affine* function on $[0,1]$. Then $y=0, x=1000$ shows that the function coincides with an affine* function on $[0,1000]$, which must be the same as before, or the values on $[0,1]$ wouldn't match up. In particular $f(1000)$ must be the value of the expression for the function that works at $[0,1]$, so this expression happens to work everywhere.

For negative numbers, similar reasoning works -- the function is affine* on $[-1000,1]$ and therefore must be the one you can extrapolate from $[0,1]$.

(* Whoops, for me "affine" means "of the form $x\mapsto ax+b$". I've kept that usage in my answer because I have no better short word to use for it).

Answer 2

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, $f(x)=ax+b$

Answer 3

$f(tx+(1-t)y)=tf(x)+(1-t)f(y),\forall t\in(0,1)$

Case 1: $x\in(0,1)\Rightarrow x=(1-x)\cdot0+x\cdot1\Rightarrow f(x)=(1-x)f(0)+xf(1)$.

Case 2: $x>1\Rightarrow0<\frac{1}{x}<1\Rightarrow1=(1-\frac{1}{x})\cdot0+\frac{1}{x}\cdot x$$\Rightarrow f(1)=(1-\frac{1}{x})f(0)+\frac{1}{x}f(x)\Rightarrow f(x)=(1-x)f(0)+xf(1)$

Case 3: $x<0\Rightarrow0<\frac{1}{1-x}<1\Rightarrow0=\frac{1}{1-x}\cdot x+(1-\frac{1}{1-x})\cdot1$$\Rightarrow f(0)=\frac{1}{1-x}f(x)+(1-\frac{1}{1-x})f(1)\Rightarrow f(x)=(1-x)f(0)+xf(1)$

So $f(x)=(1-x)f(0)+xf(1)=[f(1)-f(0)]x+f(0),\forall x\in\mathbb{R}$.

Answer 4

First show it's homogenous. That is, $F(ax) = aF(x)$ for any $a\in\mathbb{R},x\in\mathbb{R}^n$. If $a \in (0,1)$, we have: $$ F(ax) = F(ax + (1-a)0) = aF(x) + (1-a)F(0)=aF(x) $$ If $a\ge 1$, he have: $$ F(x) = F((1/a)ax + (1-1/a)0) = (1/a)F(ax) + (1-1/a)F(0)=F(ax)/a $$ If $a<0$, we use the above cases combined with the fact that $F(-x)=-F(x)$ since $$ 0 = F(0) = F((1/2)x +(1/2)(-x))=(1/2)F(x)+(1/2)F(-x). $$ Lastly, to show linearity, we use homogeneity: $$ \begin{align} F(x+y) &= F((1/2)(2x) +(1/2)(2y))\\ &= (1/2)F(2x) +(1/2)F(2y)\\ &=(1/2)(2F(x)) +(1/2)(2F(y))\\ &=F(x) +F(y) \end{align} $$