Solving $f(3x)=f\Big(\frac{x+y}{(x+y)^2+1}\Big)+f\Big(\frac{x-y}{(x-y)^2+1}\Big)$ and $f\big(x^2-y^2\big)=(x+y)f(x-y)+(x-y)f(x+y)$

Question

I need help solving this functional equations problem.

Find all $ f : \mathbb R \to \mathbb R $ such that for all $ x , y \in \mathbb R $, the two following equations hold: $$f(3x)=f\left(\frac{x+y}{(x+y)^2+1}\right) + f\left(\frac{x-y}{(x-y)^2+1}\right)\text;$$ $$f\left(x^2-y^2\right)=(x+y)f(x-y)+(x-y)f(x+y)\text.$$

What I did was using the second equation to get that $f(0)=0$ by setting $x=y=0$.

Then setting $x=0$ and $y$ arbitrary in the first equation to get: $$0=f(0)=f\left(\frac{y}{y^2+1}\right) + f\left(\frac{-y}{y^2+1}\right)$$ $$\implies f\left(\frac{y}{y^2+1}\right)=-f\left(\frac{-y}{y^2+1}\right) \implies f\text{ is odd.}$$

But then I got stuck here. I tried many ways but always end up with a variation of $$f\left(-y^2\right)=2yf(-y)=-2yf(y)\text.$$ Any suggestions how to go on from here?

Answer 1

I'm going to disregard the first equation altogether, not because it is ugly (though it is), but because one equation is just enough for a problem. Another one does not feel like it belongs here; for all I know, it could have been pasted by mistake.

Now to the point. $$f(x^2-y^2)=(x+y)f(x-y)+(x-y)f(x+y)\tag1$$

Let $a=x+y,\;b=x-y$ $$f(ab)=af(b)+bf(a)\tag2$$

Plugging $b=a$, we get $f(a^2)=2af(a)$. Now plugging $b=a^2$, we get $f(a^3)=3a^2f(a)$, and continuing in this manner, we end up with $f(a^n)=na^{n-1}f(a)$.

Now let's apply that in reverse: $f(a)=na^{n-1\over n}f(a^{1/n})$, hence $f(a^{1/n})={1\over n}f(a)/a^{n-1\over n}$.

Now let's put the two together: $$f(a^{p\over q}) = {1\over q}f(a^p)/a^{p(q-1)\over q}={p\over q}f(a)\cdot a^{p-1-{p(q-1)\over q}}={p\over q}f(a)\cdot a^{{p\over q}-1}\tag3$$ or, in other words, for any rational $r$ we have $$f(a^r)=r\cdot f(a)\cdot a^{r-1}\tag4$$

Well, rationals are dense in $\mathbb R$, but what's between them? The discontinuous solutions are plenty. If you know what the Hamel basis is, then you know where to find them, and if you don't know, then trust me, you don't want to know. But if we stick to $f$ being continuous everywhere (or anywhere, for that matter), the only way to achieve that is for (4) to hold at any real $r$.

Say we fix $a$ and looks at some $x$ (not the same $x$ as in the beginning):

$$x=a^{\log_ax}\implies f(x)=\log_ax\cdot f(a)\cdot a^{\log_ax-1}=C\cdot x\cdot\ln x\tag5$$

So this is the only family of continuous solution to your second equation. If you are still interested, you may go and check how well does this sit with the first equation (hint: not well, except the trivial case of $C=0$).

That's it.

Answer 2

I will completely disregard the second equation, not because putting the expression $ C x \log | x | $ (derived by Ivan Neretin) in the first equation isn't conclusive (of course it is!) or that it involves formally scary calculations (well, it does!), but because the first equation alone implies that only the constant zero function is a solution, and it is a rather simple task to verify this.

Take a look at the function $ h : \mathbb R \to \mathbb R $ defined with $ h ( x ) = \frac x { x ^ 2 + 1 } $ for all $ x \in \mathbb R $. Don't let the appearance of $ h ( x \pm y ) $ as arguments for the function $ f $ trick you in thinking that the equation is ugly! As we will see in a moment, all we need to know is that the function $ g : \mathbb R \to \mathbb R $ mapping each $ x \in \mathbb R $ to $ \frac 3 2 x - h ( x ) $ is surjective. I mean, it's not even important what $ h ( x ) $ actually looks like, as long as we know that it has such a property. As a matter of fact, in case we know $ h ( z ) = \frac 3 2 z $ for some $ z \in \mathbb R $ (for example, $ h ( 0 ) = 0 $, which is true in our case), it's even sufficient that the range of $ g $ contains a Hamel basis. If you don't want to bother taking Hamel bases into account (I won't blame you if you don't), just know that any non-degenerate interval contains a Hamel basis, and thus it's sufficient that the mentioned range contains an open interval.

Back to the main task, let $ f : \mathbb R \to \mathbb R $ be a function satisfying $$ f ( 3 x ) = f \bigl ( h ( x + y ) \bigr ) + f \bigl ( h ( x - y ) \bigr ) \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $. Putting $ y = 0 $ in \eqref{0} gives $$ f \bigl ( h ( x ) \bigr ) = \frac { f ( 3 x ) } 2 \tag 1 \label 1 $$ for all $ x \in \mathbb R $. Substitute $ \frac { x + y } 6 $ for $ x $ and $ \frac { x - y } 6 $ for $ y $ in \eqref{0} and use \eqref{1} to get $$ f \left ( \frac { x + y } 2 \right ) = \frac { f ( x ) + f ( y ) } 2 \tag 2 \label 2 $$ for all $ x , y \in \mathbb R $. \eqref{2} is known as Jensen's functional equation, and its solutions are exactly functions of the form $ f ( x ) = A ( x ) + c $ for some constant $ c \in \mathbb R $ and some $ A : \mathbb R \to \mathbb R $ satisfying Cauchy's functional equation. See "Function that is both midpoint convex and concave: $f\left(\frac{x+y}{2}\right) = \frac{f(x)+f(y)}{2}$" for the simple proof. Now, use the fact that an additive function is $ \mathbb Q $-linear (see "Overview of basic facts about Cauchy functional equation" for more information), and rewrite \eqref{1} in terms of $ A $ and $ c $ to get $$ A \bigl ( g ( x ) \bigr ) = \frac c 2 \tag 3 \label 3 $$ for all $ x \in \mathbb R $. Now, knowing that $ g $ is surjective, \eqref{3} shows that $ f $ takes the value $ \frac { 3 c } 2 $ constantly, which then by \eqref{0} implies $ c = 0 $. As the constant zero function satisfies both \eqref{0} and the other given functional equation, we've found the unique solution.

Some Notes:

1. The fact that $ x \mapsto \frac 3 2 x - \frac x { x ^ 2 + 1 } $ is surjective can be proven by solving the cubic algebraic equation $ 3 x ^ 3 + y x ^ 2 + x + y = 0 $ for $ x \in \mathbb R $ given any $ y \in \mathbb R $: just let $ x = - \frac 1 9 \left ( y + z + \frac { y ^ 2 - 9 } z \right ) $, where $ z = \sqrt [ 3 ] { y ^ 3 - 9 \sqrt 3 \sqrt { y ^ 4 + 47 y ^ 2 + 3 } + 108 y } $. This is given by Cardano's formula; see "Cubic equation" for more information. Alternatively, one can use the intermediate value theorem for a proof of surjectivity: as the value of the expression $ \frac x { x ^ 2 + 1 } $ is bounded between $ - \frac 1 2 $ and $ \frac 1 2 $, $ x \mapsto \frac 3 2 x - \frac x { x ^ 2 + 1 } $ is a continuous function with the limit $ - \infty $ at $ - \infty $ and the limit $ + \infty $ at $ + \infty $.
2. There are other ways of concluding $ c = 0 $ after we've got \eqref{3}, depending on what $ h $ is. for example, if you know that the range of $ g $ contains two elements $ y $ and $ z $ such that $ y = q z $ for some $ q \in \mathbb Q \setminus \{ 1 \} $, then \eqref{3} and $ \mathbb Q $-linearity of $ A $ directly imply $ c = 0 $, since then $ c $ must be equal to $ q c $.