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A function $f$ is concave on $[a,b]$ if $$f((1-\lambda)a+\lambda b) \ge (1-\lambda)f(a) + \lambda f(b),\quad \forall \lambda\in[0,1]$$ and similarly $f$ is convex if $$f((1-\lambda)a+\lambda b) \le (1-\lambda)f(a) + \lambda f(b),\quad \forall \lambda\in[0,1]$$

Then it's clear that $f$ is concave and convex if $$f((1-\lambda)a+\lambda b) = (1-\lambda)f(a) + \lambda f(b),\quad \forall \lambda\in[0,1]$$

But then I'm told that this is equivalent to $f$ being affine. That is that $f(x) = \alpha + \beta x$ for some constants $\alpha, \beta$. How can I prove this from the above?

All I'm getting by setting $(1-\lambda)a+\lambda b=x$ in the equation above is $$f(x) = (1-\lambda)f(\frac{x-\lambda b}{1-\lambda})+\lambda f(\frac{x-(1-\lambda a)}{\lambda})$$ which I don't seem to be able to get into the form $\alpha +\beta x$.

Bobbie D
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    To start, take $a=0$ and $b=1$ to see that $f(\lambda) =\alpha+\beta\lambda$ for $0\le\lambda\le 1$, where $\alpha:=f(0)$ and $\beta:=f(1)-f(0)$. – John Dawkins Oct 15 '16 at 16:24
  • Well if I set $g(\lambda) = \lambda (b-a)+a$, I get $f\circ g(\lambda) = \lambda ( f\circ g(1)-f\circ g(0))-f\circ g(0) = \alpha +\beta \lambda$. But then how do I get rid of that $g$ function? Oh. I got it. Set $x=g(\lambda)$ and I get it. Thanks, @JohnDawkins – Bobbie D Oct 15 '16 at 16:33
  • http://math.stackexchange.com/questions/104810/how-to-prove-convexconcave-affine. – StubbornAtom Jan 29 '17 at 13:16

1 Answers1

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You are on the right track. Set $x=\lambda a+(1-\lambda )b$. Note that $x\in [a,b]$. Furthermore, note that $\lambda=\frac{x-b}{a-b}$.

Now, we have $f(\lambda a+(1-\lambda)b)=f(x)$ and

$$\lambda f(a)+(1-\lambda)f(b)=\frac{bf(a)-af(b)}{b-a}+\left(\frac{f(b)-f(a)}{b-a}\right)\,x$$

which reveals that $f$ is indeed affine on $[a,b]$ as was to be shown!

Mark Viola
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