Equation of function s.t., $f(x^2)+2f(x)=0$

Question

Are there function $f:\mathbb{R} \to \mathbb{R}$ s.t., $f\not =0$ and $f(x^2)+2f(x)=0$?

I tried polynomial. But it is impossible because of its degree. I tried logarithm, but I couldn't.

Answer 1

First, $f(0)=0$ after we plug in $0$ for $x$. Then $f(x)=f(-x)$, so it is enough to find a function $f:J=(0,\infty)\to\Bbb R$ that satisfies the given equation. Any $x$ in the interval $J$ can be written uniquely as $x=e^t$, $t\in\Bbb R$. So it is enough to search for a function $g:\Bbb R\to\Bbb R$ so that $f(e^t)=g(t)$. The functional equation for $f$ transposes now in an equation for $g$, which is $g(2t)=-2g(t)$.

Of course, we look at the action of $\Bbb Z$ on $\Bbb R$ given by letting $1$ act as $t\to 2t$. Then $0$ is fixed, and any other element in $R$ can be brought by this action to an element in either $[1,2)$, or $[-2,-1)$. We can immediately forget about this action, but this is the reason for the following.

We have immediately $g(0)=0$ as a condition.

Fix now $g$ arbitrarily on $[1,2)\sqcup [-2,1)$, and extend (uniquely) it so that the above $g(2y)=-2g(y)$ is satisfied.

(So there are infinitely many solutions.)

Answer 2

Observations:

1. $f(0) = f(1) = 0$: substitute $x=0$ and $x=1$ respectively.

2. The function must be even: $f(x) = -\frac12f(x^2) = -\frac12f((-x)^2) = f(-x)$, so we focus on $x \ge 0$.

3. Assume $x>0$, $x \ne 1$. If we know $f(x)$, then we know $f(y)$ for all $y = x^{2^n}$ where $n \in \Bbb Z$. Under the right group action, $\{ x^{2^n} \mid n \in \Bbb Z \}$ can be interpreted as the orbit of $x$. Specifically, $f(x^{2^n}) = (-2)^n f(x)$. Moreover, these values are the only values that $f(x)$ can control.

Therefore, we partition $(0,1) \cup (1, \infty)$ by using $\{ x^{2^n} \mid n \in \Bbb Z \}$, which gives us (under ZFC) continuumly many equivalence classes. Each class can have continummly many possible values, and we conclude that there are continuum-powerset ($2^\mathfrak c$) many possible such functions.

With more work, one can actually prove that there is a unique representative of $\{ x^{2^n} \mid n \in \Bbb Z \}$ in $[2,4)$, and then it follows that such functions are in bijection to functions $f_0 : [2,4) \to \Bbb R$, i.e. any such function $f_0 : [2,4) \to \Bbb R$ can be extended to a unique function $f : \Bbb R \to \Bbb R$ satisfying your equation, and (more trivially) any function $f : \Bbb R \to \Bbb R$ can be restricted to a function $f_0 : [2,4) \to \Bbb R$.

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$[2,4)$ can be generalized to any $[y, y^2)$ with $y > 1$ or $(y^2, y]$ with $0 < y < 1$.