What are the continuous solutions to the functional equation $f(x) = \tfrac{1}{2}f(x^2)+\tfrac{1}{2}f(2x-x^2)$?

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.)

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 .