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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.

jackson
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    Is there any reason why you don't allow 0 in the domain? Also, do you require any conditions like continuity? – Alan Chung Aug 09 '23 at 16:36
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    The constant $f(x)=-\frac 12$ also works, and more things like sgn are possible (of course, nothing like that is surjective). – lulu Aug 09 '23 at 16:41
  • @AlanChung When I originally came across the problem, I was interested in defining a homomorphism from $\mathbb R^\times$ to a certain subgroup of $GL(2,\mathbb R)$. More relevant to the broader context I think the question properly lives in, I am interested in solutions (like $\mathrm{sgn}(x)$) that are not necessarily continuous at $0$. – jackson Aug 09 '23 at 16:44
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    Mathematica outputs $f(x)=\cos(c \ln(x))$ as a solution. Can be seen that it works, after taking the exponent out of the log you get $\cos(2 c \ln(x)) = 2(\cos(c \ln(x)))^2-1$ which is true by the cosine double angle identity. Note that this is not surjective though as cosine is constrained to $[-1, 1]$. – dgeyfman Aug 09 '23 at 16:48
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    Note that it's sufficient to consider only $x>0$. Once those are nailed down, the values for $x < 0$ are only constrained by $f(-x)^2 = f(x)^2$. Once you assume $x > 0$, you can let $f(x) = g(\ln x)$ to get $g(2x) = 2 g(x)^2-1$. – eyeballfrog Aug 09 '23 at 16:50
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    @DanielGeyfman Note that $\cosh$ also works (set $c$ to be imaginary). Of course, $\cosh$ isn't surjective either, but it is at least unbounded. – eyeballfrog Aug 09 '23 at 17:00
  • I'm not 100% sure if my reasoning is correct, but because $2f(x)^2-1 > f(x)$ if $f(x) > 1$, so provided $x > 1$ and $f(x) > 1$ we have that $f(x)$ is strictly increasing. – dgeyfman Aug 09 '23 at 17:12
  • @DanielGeyfman that seems right—possibly even $f$ is monotonic on (each component of) $f^{-1}(1,\infty)$? – jackson Aug 09 '23 at 17:14
  • Seems like you might be able to use: If $f(x)=\cosh z=\frac12\left(e^{z}+e^{-z}\right),$ then $f(x^2)=\frac12\left(e^{2z}+e^{-2z}\right)$ and $$f(x^{2^n})=\frac12\left(e^{2^nz}+e^{-2^nz}\right).$$ – Thomas Andrews Aug 09 '23 at 17:17
  • Anyway if you glue together $g=\mathrm{sgn}(x)\cos{(x)}$ on $\pm(0,1]$ and $g=\mathrm{sgn}(x)\mathrm{cosh}(x)$ on $\pm[1,\infty)$, you get a continuous, surjective solution—but not an injective one – jackson Aug 09 '23 at 17:17
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    Okay, I've found an infinite class of such functions, and updated my answer. – Thomas Andrews Aug 09 '23 at 18:58

1 Answers1

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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 $f(x)\in(-1,1).$

Thomas Andrews
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  • I love this answer—to tie it in with a comment from @eyeballfrog, if $f(x) = g(\ln x)$, then taking $w(x) = x$ is equivalent to $g(x) = \cosh x$, as $\cosh{(\ln x)} = \frac{x + 1/x}{2}$ – jackson Aug 10 '23 at 15:57