Online Algebra Olympiad 2016 Q5

Find all f: \mathbb R \to \mathbb R such that f(xf(x)-yf(y))=f(x-y)(f(x)+y),\forall x,y \in \mathbb R.

All right, looks old-fashioned. For old-fashioned functional equation whose solutions are probably only the obvious ones, we aim to characterise its properties instead of its specific values. Sleeves up, let’s get to work. 

Answer: f(x)=0, \forall x \in \mathbb R and f(x)=x, \forall x \in \mathbb R.

(Checking is omitted for it is trivial.)

Let  S_2 = \{x\in \mathbb R | f(f(x))=x \}, \Omega = \{x \in \mathbb R | f(x)=0\}, and P = \{x \in \mathbb R | f(t)=f(t+x), \forall t \in \mathbb R\}.

That is, S_2, \Omega, P are sets of second-order fixed points, zeroes, and periods of the function, respectively. We will see why we mentioned these sets very soon.

Let the statement of the question be P(x,y). The condition looks rather symmetric, but not quite. This gives us space for manipulation.

P(0,0) \implies f(0)=f(0)^2 \implies f(0)= 0 or 1.

Having conjectured that only the two aforementioned solutions to the functional equation exist, we aim to rule out the second case.

P(x,x) \implies f(0)=f(0)(f(x)+x) \implies f(x)=1-x. Substitution shows that this is NOT a solution. Thus f(0)=0.

Keep going, we are on track.

P(0,y) \implies f(-yf(y)) = yf(-y), P(0,-y) \implies f(yf(-y))=-yf(y).

Look, look, these two switch between each other. Better take this information into account. \forall y \in \mathbb R, -yf(y), yf(-y) \in \mathbb S_2.

But how are we going to exploit the second order fixed points? We have to substitute it into somewhere and compare it with the original value, so we’d better do it with something where x and f(x) are both present, as in such occasion substituting s \in S_2 will make this expression comparable with the original one. Such expression exists in the LHS.

P(s,f(s)), s \in S_2 \\  \implies f(sf(s)-f(f(s))f(s))=f(s-f(s))(f(s)+f(s))\\  \implies f(s-f(s))f(s)=0,

as f(f(s))=s and f(0)=0. If s \ne 0, f(s) \ne 0 as f(f(s))=s \ne 0 and f(0) =0 . Thus f(s-f(s)) =0. If s=0, f(s-f(s))= f(0-0)=0. Thus in whatever case, s-f(s) \in \Omega.

Look at that. That looks like quite some information isn’t it, especially after we substitute s= yf(-y) in…

 yf(-y)-f(yf(-y))=yf(-y)+yf(y) = y(f(y)+f(-y)) \in \Omega, \forall y \in \mathbb R.

That wasn’t a big surprise, since the y and -y are supposed to either cancel or both give vanishing function value anyway. But this expression has an advantage….

P(x, -x) =\implies f(xf(x)+xf(-x))=f(2x)(f(x)-x), and the LHS of this expression is equal to 0. Thus,

\forall x \in \mathbb R, f(2x)=0 or f(x)=x \ \ \ \ \ \ \ \ \ \ \ \ \ (*) .

Once we are here normally there are many ways to finish the problem off.

Claim: \Omega = P.

If p \in P, f(p)=f(0+p)=f(0)=0.

If \omega \in \Omega. P(\omega, -y)\implies f(yf(-y))=f(\omega +y)(0+y)=yf(\omega +y). P(0,-y) \implies f(yf(-y)) = f(y)y. If y \ne 0, we can divide both sides by y and conclude f(\omega + y) =\frac {f(yf(-y))}{y} = f(y). If y =0, f(\omega + y) = f(\omega) =0 =f(0)=f(y). Thus f(\omega + y)=y, \forall y\in \mathbb R.

If f only vanishes on x=0, we recall the statement (*), in which f(2x) \ne 0, \forall x \ne 0, thus f(x)=x \forall x \in \mathbb R \ \{0\}. But f(0)=0 anyway, thus f(x)=x.

If \exists \omega \ne 0, f(\omega)=0, then \omega is a period of the function. P(x, \omega) \implies f(xf(x))=f(x-\omega)(f(x)+\omega)=f(x)(f(x)+\omega)=f(x)^2 +\omega f(x).

But now, P(x, 0) \implies f(xf(x))=f(x)^2. Thus \omega f(x)=0, \forall x \in \mathbb R. Which, since \omega \ne 0, forces us to have f(x)=0, \forall x \in \mathbb R.

Comment: Boring problem, just a few substitutions to crack it open. But the substitutions are not that easy to find, especially the middle part where one needs to play around with the second order fixed point set. The part proving that all zeroes are periods makes use of the ubiquity of f(x) on both sides (since it is ubiquitous we let it vanish to reduce the equation down.)


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