Determine all $f:\Bbb R\to \Bbb R$ such that $f\big(a-3f(b)\big)=f\left(a+f(b)+b^3\right)+f\left(4f(b)+b^3\right)+1$ for every $a,b\in\Bbb R$.

Question

Im struggling with this functional equation:

Determine all $f: \Bbb R \to \Bbb R$ such that $$f\big(a-3f(b)\big)=f\left(a+f(b)+b^3\right)+f\left(4f(b)+b^3\right)+1$$ for all $a,b\in\Bbb R$.

Clearly the constant function $f(x)=-1$ for all $x\in\mathbb{R}$ is a solution. If $k=f(0)$, then $k\neq 0$. Otherwise, if $a,b=0$, we have $$f\big(-3f(0)\big)=f\big(f(0)\big)+f\big(4f(0)\big)+1\,,$$ which would give $0=0+0+1$ if $k=0$. By taking $b=0$, we have $$f(a-3k)=f(a+k)+f(4k)+1,$$ or $$f(a+4k)=f(a)+r,$$ where $r=-f(4k)-1$. This proves that $$f(a+4kn)=f(a)+nr$$ for all integers $n$. What to do next?

Thanks in advance for your time.

Answer 1

Let $t\in\mathbb{R}$. Suppose that $f:\Bbb R\to\Bbb R$ satisfies the functional equation $$f\big(a-3f(b)\big)=f\big(a+f(b)+b^3\big)+f\big(4f(b)+b^3\big)+t\tag{1}$$ for all $a,b\in\Bbb{R}$. Substituting $a+3f(b)$ for $a$ yields $$f(a)=f\big(a+g(b)\big)+f\big(g(b)\big)+t,\tag{2}$$ where $g(b)=4f(b)+b^3$. From $(2)$, when $a=0$, we have $$f\big(g(b)\big)=\frac{f(0)-t}{2}.\tag{3}$$ This shows that $$f\big(a+g(b)\big)=f\big(a+g(c)\big),$$ or equivalently $$f\Big(a+\big(g(b)-g(c)\big)\Big)=f(a)$$ for all $a,b,c\in\Bbb R$.

Let $S=\big\{g(b):b\in\Bbb{R}\big\}$, $T=S-S=\big\{x-y:x,y\in S\big\}$, and $U=\operatorname{span}_{\Bbb Z}T$. It follows that $$f(a+u)=f(a)\tag{4}$$ for every $a\in \Bbb R$ and $u\in U$. Define $V=\operatorname{span}_{\Bbb Q}T$. Then $V$ has a Hamel basis $\mathcal{H}$, and every element of $U\subseteq V$ is a linear combination of elements of $\mathcal{H}$. In other words, $$g(b)-g(0)=4\sum_{x\in\mathcal{H}}h_x(b)x\in U$$ for some functions $h_x:\Bbb R\to\Bbb Q$ such that for each $b\in\Bbb{R}$, $h_x(b)= 0$ for all but finitely many $x\in\mathcal{H}$. Therefore, $$g(b)=4f(0)+4\sum_{x\in\mathcal{H}}h_x(b)x$$ and $$f(b)=-\frac{b^3}{4}+f(0)+\sum_{x\in\mathcal{H}}h_x(b)x.\tag{5}$$ Note that $$\sum_{x\in\mathcal{H}}h_x(b)x\in\frac{1}{4}U=\left\{\frac{u}{4}:u\in U\right\}.$$

Using $(2)$ and $(3)$, we have $$-\frac{a^3}{4}+f(0)+\sum_{x\in\mathcal{H}}h_x(a)x=-\frac{\left(a+4f(0)+4\sum_{x\in\mathcal{H}}h_x(b)x\right)^3}{4}+f(0)+\sum_{y\in\mathcal{H}}h_y\left(a+4f(0)+4\sum_{x\in\mathcal{H}}h_x(b)x\right)y+\frac{f(0)+t}{2}.$$ Therefore $$\frac{\left(a+4f(0)+4\sum_{x\in\mathcal{H}}h_x(b)x\right)^3-a^3}{4}-\frac{f(0)+t}{2}=\sum_{y\in\mathcal{H}}h_y\left(a+4f(0)+4\sum_{x\in\mathcal{H}}h_x(b)x\right)y\in \frac{1}{4}U.$$ The left hand side is a continuous function in $a$ whose range is a connected subset of $\mathbb{R}$ (i.e., an interval). Therefore, unless $$f(0)+\sum_{x\in\mathcal{H}}h_x(b)x=0$$ for every $b\in\mathbb{R}$, we must have $U=\mathbb{R}$.

If $f(0)+\sum_{x\in\mathcal{H}}h_x(b)x=0$ for all $b\in\mathbb{R}$, then by $(5) $ we have $$f(b)=-\frac{b^3}{4}$$ for all $b\in\Bbb R$. This can happen if and only if $t=0$. We now assume that $U=\mathbb{R}$.

If $U=\mathbb{R}$, then by $(4)$, $f$ is constant. Thus, from $(2)$, $$f(a)=-t$$ for every $a\in\mathbb{R}$.

In general, let $p\neq 0$ and $\gamma\neq -1$ be real numbers, and $q:\Bbb R\to\Bbb R$ a continuous function such that for every real number $\tau\neq 0$, the function $\Delta_\tau q:\mathbb{R}\to\mathbb{R}$ defined by $$\Delta_\tau q(z)=q(z+\tau)-q(z)$$ is not a constant function. For example, $q$ may be a polynomial of degree at least $2$. Then all functions $f:\mathbb{R}\to\mathbb{R}$ such that $$f(a)=f\big(a+pf(b)+q(b)\big)+\gamma f\big(pf(b)+q(b)\big)+t$$ are

• the constant function $f(a)=-\frac{t}{\gamma}$ if $\gamma\neq 0$;
• the function $f(a)=-\frac{1}{p}q(a)$ if $t=\frac{\gamma}{p}q(0)$.

In the case $\gamma=0$ and $t\neq 0$, there are no solutions.