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Suppose $\ f:\mathbb{R}\to\mathbb{R}\ $ has the property:$\ f(x_1)f(x_2) = f\left( \frac{x_1+x_2}{2} \right)^2\ $ for all $\ x_1,\ x_2\in\mathbb{R}$.

I made some educated guesses and stumbled upon the fact that if $\ A,\alpha\in\mathbb{R},\ $ then $\ f(x) = A e^{\alpha x}\ $ satisfies this property.

I also realise that $\ f\ $ must be convex if $\ f>0\ $ and concave if $\ f<0$.

So now I'm wondering if any other functions satisfy the property, and if not, how to prove uniqueness of $\ f(x) = A e^{\alpha x}\ $ in satisfying the property. Edit: I want something stronger: to classify all the solutions to this functional equation.

Mohsen Shahriari
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Adam Rubinson
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  • Since $f(x_1)f(x_2=f(x_2)f(x_1)$ and $x_1+x_2=x_2+x_1$, the condition $x_1<x_2$ doesn't seem to mean much. Is it just there to exclude $x_1=x_2$? But all functions satisfy this if $x_1=x_2$. – lulu Jan 16 '22 at 19:24
  • And did you mean to assume continuity? I would think there are discontinuous examples. – lulu Jan 16 '22 at 19:25
  • https://math.stackexchange.com/q/2370719/42969 (applied to $\log(f)$) shows that there are discontinuous examples. – Martin R Jan 16 '22 at 19:25
  • See https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation#Existence_of_nonlinear_solutions_over_the_real_numbers for nonintuitive results. – kimchi lover Jan 16 '22 at 19:26
  • @lulu - I acknowledge your first point. @ Martin R I'm not sure what you mean. Please elaborate. – Adam Rubinson Jan 16 '22 at 19:49
  • @kimchilover I am not familiar with Cauchy's functional equation. However, I don't see how this is related to it. – Adam Rubinson Jan 16 '22 at 19:57
  • Suppose $g$ is a function that satisfies $g(x_1) + g(x_2) = 2g((x_1+x_2)/2)$. Then $f = A\exp(Bg)$ satisfies your equation for any constants $A,B$. If there was a non-continuous function that may serve as $g$, this would imply that ${A\exp(Bx)}$ is not the only class of functions that solves your equation. The references above tell you where to read about non-continuous solutions to the equation in $g$ - either in the answer Marin linked, or in stuff on the Cauchy equation, because any function that satisfies the latter can serve as $g$. – stochasticboy321 Jan 16 '22 at 20:13
  • Note that $,\frac{f(a)f(b)}{f(a+b)}/\frac{f(b)f(a-b)}{f(a)}=\frac{f(a)^2}{(f(a-b)f(a+b)}.$ – Somos Jan 16 '22 at 20:21

2 Answers2

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As stochasticboy commented while I was drafting this:

Let $\lambda(x)$ be any discontinuous solution of the additive form of Cauchy's functional equation. One checks trivially that $\lambda(2u)=2\lambda(u)$ for all $u$, equivalently $\lambda(u/2)=\lambda(u)/2$. Now let $f(x)=\exp(\lambda(x))$, so $$ f(x)f(y)=e^{\lambda(x)+\lambda(y)}=e^{\lambda(x+y)} =e^{2\lambda((x+y)/2)}=\left(f\left(\frac{x+y}2\right)\right)^2,$$ taking $u=x+y$.

Adam Rubinson
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kimchi lover
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  • @AdamRubinson Thanks; I wasn't. – kimchi lover Jan 16 '22 at 21:09
  • Oh. I get this now. So the natural follow-up question: Are there any functions other than $exp$ of solution's to Cauchy's functional equation, that satisfy my inequality? The answer is yes, like $\ f(x) = Ae^{\alpha \lambda(x)},\ A\neq 1;\ \alpha\neq 1.\ $ I guess I want all the solutions to my original functional equation. – Adam Rubinson Jan 16 '22 at 21:58
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It can be shown that $ f : \mathbb R \to \mathbb R $ satisfies $$ f ( x ) f ( y ) = f \left ( \frac { x + y } 2 \right ) ^ 2 \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $ iff there is a constant $ c \in \mathbb R $ and an additive $ A : \mathbb R \to \mathbb R $ such that $ f ( x ) = c \exp \bigl ( A ( x ) \bigr ) $ for all $ x \in \mathbb R $. The fact that those $ f $ that have this form are solutions is already discussed in the comments under the original post and the answer by kimchi lover. To show that those are the only solutions, first note that if there is $ z \in \mathbb R $ with $ f ( z ) \ne 0 $, then letting $ y = 2 z - x $ in \eqref{0} we get $ f ( x ) f ( 2 z - x ) = f ( z ) ^ 2 > 0 $, and therefore $ f ( x ) \ne 0 $ for all $ x \in \mathbb R $. This then shows that the right-hand side of \eqref{0} is always positive, which by looking at the left-hand side shows that $ f $ must be always positive or always negative. Let $ \epsilon = \operatorname {sgn} \bigl ( f ( 0 ) \bigr ) $, and note that $ \epsilon f ( x ) > 0 $ for all $ x \in \mathbb R $, which lets us define $ g : \mathbb R \to \mathbb R $ with $ g ( x ) = \log \bigl ( \epsilon f ( x ) \bigr ) $ for all $ x \in \mathbb R $. Taking logarithms from both sides of \eqref{0}, we get $$ g \left ( \frac { x + y } 2 \right ) = \frac { g ( x ) + g ( y ) } 2 \tag 1 \label 1 $$ for all $ x , y \in \mathbb R $. \eqref{1} is known as Jensen's functional equation, and its solutions are exactly functions of the form $ g ( x ) = A ( x ) + b $ for some constant $ b \in \mathbb R $ and some $ A : \mathbb R \to \mathbb R $ satisfying Cauchy's functional equation. See "Function that is both midpoint convex and concave: $f\left(\frac{x+y}{2}\right) = \frac{f(x)+f(y)}{2}$" for a proof. Then we will have $$ f ( x ) = \epsilon ^ 2 f ( x ) = \epsilon \exp \bigl ( g ( x ) \bigr ) = \epsilon \exp \bigl ( A ( x ) + b \bigr ) = c \exp \bigl ( A ( x ) \bigr ) $$ for all $ x \in \mathbb R $, where $ c = \epsilon \exp b $. Note that we did all this with the assumption that there is some $ z \in \mathbb R $ with $ f ( z ) \ne 0 $, and ended up with some $ c \ne 0 $. In case there is no such $ z $, $ f $ must be the constant zero function, which is of the same form as before, only with $ c = 0 $.

Note that the axiom of choice implies the existence of nonlinear additive functions, which have wild behaviors. In case you have regularity conditions on $ f $ such as monotonicity, being bounded above/below on some interval, continuity and measurability, $ g $ and $ A $ will satisfy the same regularity condition, and we know that the only regular additive functions are linear ones. See "Overview of basic facts about Cauchy functional equation" for more information.

Mohsen Shahriari
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