IMO

# Iran TST 2011 Q5

Find all SURJECTIVE functions $f:\mathbb R \to \mathbb R$ such that $\forall x,y \in \mathbb R$,

$f(x+f(x)+2f(y))=f(2x)+f(2y)$.

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

Let the equation be $P(x,y)$.

All right… Surjective… That means we will be able to let $f(y)$ be whatever we want…

$f(x)=x$ obviously works, and $f(x)=kx, k\ne 1$ don’t work. $f(x)=x+c,c\ne 0$ don’t work either. Since the right hand side is symmetric in $x$ and $y$, we’d expect the LHS to be basically symmetric as well (hmm we might be able to twiddle take this, at some point), thus we expect $f(x)=x$ to be the only solution. (Constant solutions work but they are not surjective.

Turns out that $x+f(x)$ term is hard to control, and we don’t find $f(0)$ that easily.

We want to show it is injective. Because if it was injective we can twiddle take the equation and show $f(y)-f(x)=y-x$ by switching $y$ and $x$, from which the solution follows straightaway.

So LHS, as function of $y$ only depends on $f(y)$, and the RHS has only $f(2y)$. Thus we see that $f(2y) = h(f(y))$, i.e. $f(2y)$ only depends on $f(y)$.

Now, let $f(x_1)=f(x_2)$, then $P(x_1,y), P(x_2,y) \implies f(x_1 + f(x_1)+2f(y))=f(x_2 + f(x_1)+2f(y))$ as $f(2x_1)=f(2x_2)$. Let $x_1 - x_2 = t, 2f(y)+f(x_1)+x_2 = a$. We then have $f(a+t)=f(a), \forall a \in \mathbb R$. (This uses surjectivity!!!! It is only because the function is surjective that we can make $a$ range over the entire real number.)

Seems like our injectivity scheme is not going well… Well we will twiddle take the equation anyway.

Define $x\equiv y$ as $f(x)=f(y)$. That is, $x-y$ is a period of $f$. It is easy to check that this is an equivalence relation that is preserved by addition of equal term.

$P(x,y). P(y,x) \implies x+f(x)+2f(y) \equiv y+f(y) + 2f(x)$, or,

$(f(x)-x) \equiv (f(y)-y) , \forall x, y \in \mathbb R \,\,\,\,\,\,\,(*)$.

This is like a pseudo $f(y)-f(x)=y-x$….  Let’s sub in values and see…

Let $f(r)=0$. (by surjectivity it exists.)

In (*) let $y=0$, we see that $f(x)\equiv x+f(0)$.

$P(0,r) \implies f(f(0))=f(0)+f(2r)$. LHS$\equiv f(0)+f(0)\equiv 2f(0)$, RHS$\equiv f(0)+2r+f(0)$. Thus $2r \equiv 0$.

$P(r,y) \implies f(r+f(r)+2f(y))=f(2r)+f(2y)$. $P(0, y) \implies f(0+f(0)+2f(y))=f(0)+f(2y)$. Since $2r \equiv 0$, $r \equiv f(0) \equiv f(f(r)) \equiv r + 2f(0)$. Thus $2f(0)\equiv 0$.

$\displaystyle P(\frac r2 , y) \implies f(\frac r2 +f(\frac r2 )+2f(y))=f(2y) \\ \implies \frac r2 +f(\frac r2 )+2f(y) \equiv 2y \\ \implies 2\times \frac r2 + f(0) + 2y + 2f(0) \equiv 2y \\ \implies r \equiv -f(0) \equiv f(0)\\ \implies 0=f(r) =f(f(0))$.

But since $r \equiv f(0), 2r \equiv r+r\equiv f(0)+f(0)\equiv 2f(0)\equiv 0$, thus $f(f(0))=f(0)+f(2r)=2f(0)$. But $f(f(0))=0 \implies f(0)=0$.

So, we tried to show that $f$ is injective, but ended up showing $f(0)=0$. That is still great though, because now we have $f(x)\equiv x, \forall x \in \mathbb R$. All these substitutions are found by trial and error, which makes the problem quite challenging indeed. The idea behind, though, is that if the “$\equiv$” sign is replaced by equal, we will be able to determine $f(0)=0$ by equating coefficients of LHS and RHS of $P(x,y)$. We are just doing in modulo the set of periods.

Since $f(f(x))=f(x)$, all values in the image of $f$ are fixed points of $f$.

But $f$ is surjective. Thus $f(\mathbb R)=\mathbb R$, thus every real number is a fixed point. Or, $f(x)=x, \forall x \in \mathbb R$.

Comment: the middle part of finding the $f(0)$ is tricky, and it is useful to notice the structure that all difference between pair of number giving off the same value are periods of the function. We used the surjectivity twice, once when showing that they are periods, the other when showing that the image of $f$ is the entire real number. The second time, despite that the equation could be reduced to additive Cauchy function, the zero function can still work, so we need to use the surjectivity to do this.