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I am trying to prove the following: If a function $f$ is concave and convex, then it is affine.

Since $f$ is convex we have $f(\lambda x +(1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y))$ and $f(\lambda x +(1-\lambda)y) \geq \lambda f(x) + (1-\lambda)f(y)) $. Thus we know $f(\lambda x +(1-\lambda)y) = \lambda f(x) + (1-\lambda)f(y))$. This shows that $f$ is linear and hence it is affine.

Is this proof right? Could anyone tell me if there is another way to prove this? Thanks!

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    Two things: (i) You can't conclude that $f$ is linear. In general, it is not. (ii) You need to show that the last equation holds for all $\lambda$ not just $\lambda \in [0,1]$. – 2017-01-26
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    @copper.hat: I agree, although it's possible OP is using linear to mean "has no terms greater than linear". – 2017-01-26
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    Thanks for the comments! I did mean that f was a linear function not just that it doesn't have terms greater than linear. Why can't we conclude that $f$ is linear? – 2017-01-26
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    Linear implies $f(tx) = t f(x)$, the function $f(t) =1 +t $ is affine but not linear. – 2017-01-26
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    $f(x)=a+bx$ is affine but not linear. $f(x)=bx$ is linear (and affine). You cannot conclude that $f$ is affine from your last equation since the statements hold $\forall \lambda\in[0,1]$ (which you are missing). Hence you can only conclude that $f$ is 'affine' between $x$ and $y$. – 2017-01-26
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    But since we have $f(\lambda x + (1-\lambda)y) = \lambda f(x) + (1-\lambda)f(y))$ doesn't that mean that f is linear? What would be the difference if we only had $ f(\lambda x ) = \lambda f(x) $, would that show $f$ is linear? – 2017-01-26
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    I gave an example above of a function that is affine but not linear. – 2017-01-26
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    Thanks I saw it as soon as I posted my question. Thanks! – 2017-01-26

1 Answers 1

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You need to show that $f(\lambda x + (1-\lambda ) y) = \lambda f(x) + (1-\lambda ) f(y)$ for all $\lambda$.

You have shown that this is true for $\lambda \in [0,1]$.

Assume $x \ne y$ (otherwise there is nothing to prove).

Suppose $\lambda >1$ and let $p=\lambda x + (1-\lambda ) y$.

Write $x = t p +(1-t) y$ and show that $\lambda = {1 \over t}$. In particular, $t \in [0,1]$ and so $f(x) = t f(p) + (1-t) f(y)$. Hence $f(x) = {1 \over \lambda} f(p) + (1-{1 \over \lambda}) f(y)$ which gives $f(\lambda x + (1-\lambda ) y) = \lambda f(x) + (1-\lambda ) f(y)$.

The same sort of analysis applies for $\lambda <0$ (except we start with $y = sx + (1-s) p$).