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Let $f$ be a real function defined on some interval $I$.

Assuming that $f$ both convex and concave on $I$, i.e, for any $x,y\in I$ one has $$f(\lambda x+(1-\lambda)y)=\lambda f(x)+(1-\lambda)f(y),\, \, \lambda\in (0,1) .$$
I would like to show that $f$ is of the form $f=ax+b$ for some $a,b$.

I was able to prove it when $f$ is differentiable, using the relation $$f'(x)=f'(y).$$

Anyway, I was not able to provide a general proof (without assuming that $f$ is differentiable, and without assuming that $0\in I$).

Any answer will be will be appreciated.

Edit: It is little bit different from tte other question How to prove convex+concave=affine?. Here $f$ is defined on some interval, so $o$ not necessary in the domain. Please remove the duplicate message if this possible

Community
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2 Answers2

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Let $x<y$ be in $I$. Then for $z=\lambda x + (1-\lambda)y\in [x,y]$, $$f(z)=f(\lambda x + (1-\lambda)y)=\lambda f(x)+(1-\lambda)f(y)=\frac{f(y)-f(x)}{y-x}(z-x)+f(x)$$ Hence, $f(z)=az+b$ for some $a,b$ on every closed interval in $I$. Every interval can be expressed as the limit of a non-decreasing sequence of closed intervals $(I_n)_{n\in\mathbb{N}}$, and since $f\rvert I_n$ must coincide with $f\rvert I_{n+1}$ on $I_n$, we find that $f$ extends to a linear function $az+b$ on all of $I$.

Jacob
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Consider a fixed point $c \in I$. We have $$\dfrac{f(c+\lambda(x-c))-f(c)}{\lambda(x-c)} = \dfrac{\lambda(f(x)-f(c))}{\lambda(x-c)}$$ Hence, $$f'(c)=\lim_{\lambda \to 0}\dfrac{f(c+\lambda(x-c))-f(c)}{\lambda(x-c)} = \lim_{\lambda \to 0} \dfrac{\lambda(f(x)-f(c))}{\lambda(x-c)} = \dfrac{f(x)-f(c)}{x-c}$$ Hence, for all $x \in I$, we have $$f(x) = f(c) + (x-c)f'(c)$$

Adhvaitha
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  • Are you assuming that $f$ is differentiable? thanks. –  Apr 15 '15 at 20:30
  • @JohnJohn No. We infact prove that $f$ is differentiable, if $f(\lambda x+(1-\lambda )y) = \lambda f(x)+(1-\lambda)f(y)$ – Adhvaitha Apr 15 '15 at 20:31
  • I will note that such functions are only provably differentiable Lebesgue-a.e. If memory serves, the set of non-differentiable points may be countable (but no larger) – FShrike Jan 25 '22 at 22:49
  • I don’t see anything guaranteeing the existence of a fixed point (in fact, there is none in the case of $f(x)=x+b$ with $b\neq 0$) – Jacob Jan 26 '22 at 08:18
  • @Jacob I guess he only means fixed choice of a point instead of fixed point. You can replace it by a sentence like "Let $c \in I$ be arbitrary". – Jochen Jul 01 '22 at 17:54