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The solution set is a vector space and includes all functions of the form $f(x)=ax+b$. But apart from these observations I have nothing to say about the problem. If it helps, restrict the domain of $f$ to the unit interval.

The problem came up when I was trying to find the end behavior of the following martingale. Let $\epsilon_i$ be an IID sequence of fair coinflips. Set $X_0$ to $x_0$ with $0\le x_0\le 1$, and then set $X_{i+1}$ to $X_i^2$ if $\epsilon_i$ lands tails but to $2X_i-X_i^2$ if $\epsilon_i$ lands heads. Then $X_i$ is a martingale and, being $\mathrm{L}^1$-bounded, converges almost surely by a theorem of Doob. Clearly, the only numbers the martingale can converge to are $0$ and $1$. The chances of convergence to $1$ as a function of $x_0$ must satisfy the functional equation above.

Ghwosque
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2 Answers2

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Suppose you have such a continuous function $f:[0,1]\to\mathbb{R}$. The equation is linear in $f$, so by subtracting a linear function, we may assume that $f(0)=f(1)=0$. Now let $M$ be the maximum value of $f$, achieved at some point $c\in [0,1]$. The functional equation now implies that $f(c^2)=f(2c-c^2)=M$, since otherwise their average would have to be less than $M$. But now we can repeat the same argument with $c$ replaced by $c^2$ to find $f(c^4)=M$, and similarly $f(c^{2^n})=M$ for all $n$. Taking the limit, we conclude that either $f(0)=M$ or $f(1)=M$, depending on whether $c<1$ or $c=1$. Either way, $M=0$. By a similar argument, the minimum value of $f$ is also $0$, so $f$ is identically $0$.

Thus the only such functions are linear functions.

More generally, a similar argument applies to any continuous function $f:[a,b]\to\mathbb{R}$ satisfying a functional equation $$f(x)=\dfrac{f(g(x))+f(h(x))}{2}$$ where $g,h:[a,b]\to[a,b]$ satisfy $x=\dfrac{g(x)+h(x)}{2}$ and $g(x),h(x)\neq x$ for all $x\in (a,b)$. (Note that the argument may in general require transfinite induction, since iterating the functions $g$ and $h$ may not reach the endpoints of the interval in just $\omega$ steps.)

Eric Wofsey
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2

Take $\alpha$=$x^{2}$ and $\beta$=$2x-x^{2}$ See that given equation can be written in form of $$f(\frac{\alpha + \beta}{2})=\frac{f(\alpha) + f(\beta)}{2}$$ Then use the fact that the equality condition in jensen's inequality holds if $f$ is linear or a constant .

Chinmaya mishra
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