I am interested in the functional equation
$$
f(x^2) = 2f(x)^2 – 1,
$$
where the domain of $f$ is nonzero real numbers. Are there any nontrivial continuous solutions? I know $f(x) = 1$ and $f(x) = \mathrm{sgn}(x)$ are solutions, but I want to know if there are any that are surjective on nonzero real numbers.
Does anyone know how to approach this problem? It seems simple but I'm a bit at sea here.
Best Answer
There is a simple example:
$$f(x)=\frac12\left(x+x^{-1}\right),$$ or more generally:
$$f(x)=\frac12\left(x^k+x^{-k}\right).$$
We can define lots of such functions. Contrary to my prior argument, these functions are continuous at $\pm 1.$
We will construct a class of continuous functions $w:\mathbb R^+\to\mathbb R^+$ such that $w(x^2)=w^2(x).$
Then we can define $f(x)=\frac{1}{2}\left(w(x)+\frac1{w(x)}\right)$ which has the property needed.
Define $w$ as any continuous function on $[2,4]$ with the property that $w(4)=w^2(2)$ and $w(x)\geq1$ for all $x\in[2,4].$ The case $w(x)=x^k$ will give the general example above.
Then we extend $w$ to $x>1$ with:
$$w(x)=w^{2^n}\left(x^{2^{-n}}\right), x\in [2^{2^n},2^{2^{n+1}}),n\in\mathbb Z,\\ w(1)=1$$
So we've partitioned the reals greater than $1$ into the intervals:
$$\dots,[2^{1/8} ,2^{1/4}),[2^{1/4},2^{1/2}),[2^{1/2},2),[2,2^2),[2^2,2^4),\dots$$
This gives a $w$ such that $w(x^2)=w^2(x).$
(The case $w(x)=x^k$ on $[2,4]$ gives $w(x)=x^k$ on all $x>1.$)
The condition that $w(4)=w^2(2),$ ensures we get that $w$ is continuous on $x>1.$
We define it for $0<x<1$ similarly by first defining on an interval on $[1/4,1/2].$ Then we solve for the negative $x$ values, and define it for $x=1$ as $f(1)=1.$
But as $x\to 1^+,$ $w(x)\to 1,$ since it corresponds to $n\to-\infty$ and $w^{2^{-n}}(x)<\sup_{y\in[2,4]} w(y)$ and thus $w(x)\to 1.$
Similarly, for $x\to 1^-.$
So $w$ is continuous at $x=1.$
We can, in fact, have $w(x^2)=w^{-2}(x)$ as another option.
Trying to do this to get $f(1)=-1/2$ will not work with continuous functions. In general, given $f(1)=-1/2,$ $w(1)$ must be a primitive cube root of $1,$ and it is, I think, impossible to define a continuous complex-valued function $w(x^2)=w^2(x)$ or $w(x^2)=w^{-2}(x)$ which converges to a primitive cube root of $1$ as $x\to 1,$ other than the constant $w.$ Maybe I'm wrong?
You can define more generally $w(x)$ with range $$S=\{x\in\mathbb R\mid |x|\geq 1\}\cup \{z\in\mathbb Z\mid |z|=1\}.$$
But you'd need a continuous square root defined on the range to let you extend $w(x).$ Or at least you need that $w(x)$ doesn't "go around" the circle - that $w$ on $[2,4]$ is homotopic to some map which doesn't hit all points of the circle, and likewise for $w$ on $[1/4,1/2].$
For example, if $w(x)=e^{x\pi i}$ for all $x\in[2,4],$ we couldn't find a continuous value for $w^{1/2}(x).$
But if $w(x)=e^{2x\pi i}, x\in[2,3]$ and $w(x)=e^{-2x\pi i}$ for $x\in[3,4],$ then $w$ loops around once counter-clockwise and once clockwise, and we can define $w^{1/2}(x)$ for any $x$ and get a still-continuous function.
This allows you to get examples with $f(x)\in(-1,1).$