find all possible functions : $f(a)f(b)-6ab=\frac{3}{2}f(a+b)$

Question

I'm trying to find all possible functions that satisfy this functional equation:

$f(a)f(b)-6ab=\frac{3}{2}f(a+b),$

$f\in \mathbb{R}.$

My attempt :

$a=b=0$ then $f(0)=0$ or $1.$

But I don't have any ideas to complete .

Please give me hints.

Answer 1

As preferred_anon showed in his comment, if $f(0)=0,$ then $f(b)=0$ for all $b,$ which contradicts what must happen when $a\not=0$ and $b\not=0.$ Therefore, $f(0)=3/2.$ Now let's try $a=0, b=1:$ $$f(0) f(1)-0=\frac32 f(1)\implies \frac32 f(1)=\frac32 f(1).$$ This doesn't narrow down anything, unfortunately, since it's a tautology. The same thing will happen for any $b$ nonzero if $a$ is zero, and, by symmetry, when $b=0$ and $a$ is an unspecified nonzero number.

Let's try to generalize with $a=b$ unspecified, but nonzero: $$(f(a))^2-6a^2=f(2a).$$ Unfortunately, $f$ is not evaluated at the same place on both sides of this equation, and this is a rather intractable recursion relation. Let's try $a=-b$. Then we get $$f(a) f(-a)+6a^2=\frac94=\frac32 f(0). $$ This is a tractable recursion relation. Wolfram Dev Platform yields \begin{align*} f(a)= e^{-a c_1+\frac{1}{2} i \pi (a-1) a} \left(\frac{15}{4}\right)^{a \left(\left\lceil \frac{\pi -\arg (a)}{2 \pi }\right\rceil +\left\lfloor \frac{\arg (a)}{2 \pi }\right\rfloor -1\right)}. \end{align*} This formula doesn't work at $a=0,$ however. It gives $1,$ which we know is incorrect. Therefore, we must still specify that $f(0)=3/2,$ and the above formula works for $a\not=0.$ In sum: $$f(a)=\begin{cases}\frac32,\; &a=0 \\ e^{-a c_1+\frac{1}{2} i \pi (a-1) a} \left(\frac{15}{4}\right)^{a \left(\left\lceil \frac{\pi -\arg (a)}{2 \pi }\right\rceil +\left\lfloor \frac{\arg (a)}{2 \pi }\right\rfloor -1\right)}, &a\not=0 \end{cases}.$$ You can simplify this quite a bit if $a\in\mathbb{R},$ since then $$\arg(a)=\begin{cases}0,\; &a\ge 0\\ \pi, &a<0 \end{cases}. $$ You get $$f(a)=\begin{cases}\frac32,\; &a=0 \\ e^{-a c_1+\frac{1}{2} i \pi (a-1) a} , &a>0 \\ e^{-a c_1+\frac{1}{2} i \pi (a-1) a} (15/4)^{-a}, &a<0\end{cases}.$$ You specified that $f\in\mathbb{R},$ so let's investigate when this can happen. Note that $e^{ix}=\cos(x)+i\sin(x),$ so that $$e^{-a c_1+\frac{1}{2} i \pi (a-1) a}=e^{-ac_1}[\cos(\pi a(a-1)/2)+i\sin(\pi a(a-1)/2)]. $$ We must have the imaginary part zero. The $\sin$ function is zero when its argument is an integer multiple of $\pi$. That is, we must have $$\frac{\pi a(a-1)}{2}=k\pi,$$ for some $k\in\mathbb{Z},$ or $$\frac{ a(a-1)}{2}=k,$$ or $a(a-1)=2k.$ We can actually solve this for $a$ as follows: \begin{align*} a^2-a-2k&=0 \\ a&=\frac{1\pm\sqrt{1-4(-2k)}}{2}\\ &=\frac{1\pm\sqrt{1+8k}}{2}. \end{align*} So every $k\in\mathbb{Z}$ generates these two possible values of $a$. And when that happens, our expression simplifies a bit. When $a(a-1)=2k,$ then $$\cos(\pi a(a-1)/2)=\cos(\pi k)=(-1)^k.$$ It follows that our final expression is given by $$f(a)=\begin{cases}\frac32,\; &a=0 \\ e^{-a c_1}(-1)^{a(a-1)/2} , &a>0, \; a(a-1)=2k, \; k\in\mathbb{Z} \\ e^{-a c_1}(-1)^{a(a-1)/2} (15/4)^{-a}, &a<0, \; a(a-1)=2k, \; k\in\mathbb{Z}\end{cases}.$$

So there you have it!

Answer 2

Making $f(x) = a x + b$ and substituting we have

$$ f(x)f(y)-6xy-\frac 32 f(x+y)=(a^2-6)xy+\left(ab-\frac{3a}{2}\right)(x+y)+b^2-\frac{3b}{2}=0 $$

giving

$$ f(x) = \sqrt 6 x+\frac 32 $$

which is also a solution.