Problem A. 725. (May 2018)

A. 725. Let \(\displaystyle \mathbb{R}^+\) denote the set of positive real numbers. Find all functions \(\displaystyle f\colon \mathbb{R}^+\to \mathbb{R}^+\) satisfying the following equation for all \(\displaystyle x,y\in\mathbb{R}^+\):

\(\displaystyle f\big(xy+f(y)^2\big) = f(x)f(y)+yf(y). \)

Proposed by: Ashwin Sah, Cambridge, Massachusetts, USA

(5 pont)

Deadline expired on June 11, 2018.

Solution (by the proposer). We claim that the only solution is \(\displaystyle f(x) = x\,\forall x\in\mathbb{R}^+\). This clearly works.

Lemma. If \(\displaystyle g: \mathbb{R}^+\rightarrow\mathbb{R}^+\) such that for all \(\displaystyle x\in\mathbb{R}^+\),

\(\displaystyle g(x + a_1) + b_1 = g(x + c_1) + d_1,\)

\(\displaystyle g(x + a_2) + b_2 = g(x + c_2) + d_2\)

then the ratios \(\displaystyle (d_1 - b_1):(a_1 - c_1)\) and \(\displaystyle (d_2 - b_2):(a_2 - c_2)\) are the same. That is, there is a constant \(\displaystyle \lambda\) such that \(\displaystyle d_i - b_i = \lambda(a_i - c_i)\) for \(\displaystyle i\in\{1, 2\}\). We can further take \(\displaystyle \lambda\ge 0\).

Proof. First, if \(\displaystyle a_1 = c_1\) then \(\displaystyle b_1 = d_1\), and \(\displaystyle a_2 = c_2\) implies \(\displaystyle b_2 = d_2\). Thus either of these cases occurring ends up being trivial, as we only need to solve one or no equation with our choice of \(\displaystyle \lambda\). Now suppose \(\displaystyle a_i\not= c_i, i\in\{1, 2\}\). Without loss of generality let \(\displaystyle a_i > c_i, i\in\{1, 2\}\). We first claim that \(\displaystyle d_i\ge b_i, i\in\{1, 2\}\). Indeed, if say \(\displaystyle d_1 < b_1\) then

\(\displaystyle g(x + N(a_1 - c_1) + a_1) = g(x + c_1) + N(d_1 - b_1)\)

for each integer \(\displaystyle N\ge 0\) by iterating the first relation. Notice \(\displaystyle a_i > c_i\) allows us to continue iterating. We can fix some value of \(\displaystyle x\) and let \(\displaystyle N\rightarrow +\infty\), which eventually yields the range of \(\displaystyle g\) containing negative numbers, a contradiction! Thus \(\displaystyle \lambda\), if it exists, can be taken to be nonnegative. (This also applies to the case where one of the pairs \(\displaystyle a_i=c_i\) but not the other; the case where both pairs equal allow any \(\displaystyle \lambda\), hence we can take \(\displaystyle \lambda\ge 0\).)

Notice what the argument above did. It used the fact that we can increment or decrement the argument of \(\displaystyle g\) by multiples of \(\displaystyle a_i - c_i\), and correspondingly increment or decrement the value of \(\displaystyle g\) by multiples of \(\displaystyle d_i - b_i\). At least, we can do so when the argument of \(\displaystyle g\) is strictly larger than \(\displaystyle \max(a_1, a_2)\). Now we can use both equalities at once. If \(\displaystyle \lambda\) does not exist, then \(\displaystyle a_i\neq c_i,i\in\{1,2\}\), and without loss of generality

\(\displaystyle \frac{d_1 - b_1}{a_1 - c_1} > \frac{d_2 - b_2}{a_2 - c_2}\ge 0,\)

since then the ratios are unequal. If \(\displaystyle d_2 = b_2\) then \(\displaystyle g(x + a_2) = g(x + c_2)\), so \(\displaystyle g\) is eventually periodic with period \(\displaystyle a_2 - c_2\). But then we have

\(\displaystyle g(x) = g(x + N(a_2 - c_2)) = g(x + N(a_2 - c_2) - M(a_1 - c_1)) + M(d_1 - b_1) > M(d_1 - b_1)\)

when \(\displaystyle M, N\) are positive integers and \(\displaystyle x+N(a_2-c_2)-M(a_1-c_1)>0\). For all \(\displaystyle M>0\) we can find an \(\displaystyle N\) such that \(\displaystyle x+N(a_2-c_2)-M(a_1-c_1)>0\), so we find that \(\displaystyle g(x)\) is larger than any real number, a contradiction! Now we can suppose that \(\displaystyle d_2 > b_2\), so that

\(\displaystyle \frac{d_1 - b_1}{d_2 - b_2} > \frac{a_1 - c_1}{a_2 - c_2} > 0.\)

Now let \(\displaystyle m, n\in\mathbb{N}\) such that \(\displaystyle \frac{m}{n}\) is between those two numbers. It is possible since the rationals are dense in the reals. This yields, for sufficiently large \(\displaystyle x\), that

\(\displaystyle g(x) = g(x + m(a_2 - c_2)) - m(d_2 - b_2) = g(x + m(a_2 - c_2) - n(a_1 - c_1)) + n(d_1 - b_1) - m(d_2 - b_2).\)

But \(\displaystyle U = m(a_2 - c_2) - n(a_1 - c_1) > 0\) and \(\displaystyle V = n(d_1 - b_1) - m(d_2 - b_2) > 0\), so this holds for all \(\displaystyle x\in\mathbb{R}^+\), and iterating \(\displaystyle g(x)=g(x+U)+V\), we deduce

\(\displaystyle g(x) = g(x+NU) + NV>NV\)

for all \(\displaystyle x\). Fix some \(\displaystyle x\) and send \(\displaystyle N\to\infty\), then again \(\displaystyle g(x)\) is larger than any real number, a contradiction! \(\displaystyle \blacksquare\)

Denote by \(\displaystyle P(x, y)\) the equality

\(\displaystyle f(xy + f(y)^2) = f(x)f(y) + yf(y).\)

Let's put the left-hand side into a bracket on the right-hand side: \(\displaystyle P(xz + f(z)^2, y), P(x, z)\) implies

\(\displaystyle f((xz + f(z)^2)y + f(y)^2) = (f(x)f(z) + zf(z))f(y) + yf(y),\)

\(\displaystyle f((yz)x + (f(y)^2 + yf(z)^2)) = (f(y)f(z))f(x) + (yf(y) + zf(z)f(y)),\)

\(\displaystyle f((yz)x + (f(z)^2 + zf(y)^2)) = (f(y)f(z))f(x) + (zf(z) + yf(y)f(z)),\)

where the second is a cleaning of the first and the third is the second with \(\displaystyle y, z\) switched. We can subtract to get

\(\displaystyle f\big((yz)x + \underbrace{(f(y)^2 + yf(z)^2)}_{a}\big) + \big(\underbrace{zf(z) + yf(y)f(z)}_{b}\big) = f\big((yz)x + \underbrace{(f(z)^2 + zf(y)^2)}_{c}\big) + \big(\underbrace{yf(y) + zf(z)f(y)}_{d}\big),\)

and plugging \(\displaystyle \frac{x}{yz}\) in for \(\displaystyle x\) demonstrates that we are in the regime of the Lemma. Suppose that \(\displaystyle a\neq c\) for some pair \(\displaystyle y_0,z_0\). Then, comparing \(\displaystyle y_0,z_0\) with \(\displaystyle y,z\), the Lemma shows that for some \(\displaystyle \lambda\ge 0\) depending on \(\displaystyle y,z\), we have

where we must always have \(\displaystyle \lambda=\frac{d(y_0,z_0)-b(y_0,z_0)}{a(y_0,z_0)-c(y_0,z_0)}\), so \(\displaystyle \lambda\) is the same for all pairs \(\displaystyle y,z\). This is also true if \(\displaystyle a=c\) for all pairs \(\displaystyle y,z\), as that implies \(\displaystyle b=d\), so \(\displaystyle \lambda=0\) works.

Letting \(\displaystyle z = 1\) in \(\displaystyle (*)\) gives

\(\displaystyle f(y)(y +f(1)- yf(1)) - f(1) = \lambda(yf(1)^2 - f(1)^2)\)

so that

\(\displaystyle f(y)((1 - f(1))y + f(1)) = (\lambda f(1)^2)y + (f(1) - \lambda f(1)^2).\)

Thus for all but possibly one positive real \(\displaystyle y\), we must have

\(\displaystyle f(y) = \frac{(\lambda f(1)^2)y + (f(1) - \lambda f(1)^2)}{(1 - f(1))y + f(1)}.\)

At this point we essentially are able to plug in, but we will use some tricks to simplify the argument. Suppose \(\displaystyle f(1)\not= 1, \lambda\not= 0\). Then as \(\displaystyle y\rightarrow +\infty\), \(\displaystyle f(y)\) limits towards \(\displaystyle L = \frac{\lambda f(1)^2}{1 - f(1)}\not= 0\). In the original equation, \(\displaystyle f(xy + f(y)^2) = f(x)f(y) + yf(y)\), sending \(\displaystyle x = y\rightarrow +\infty\) will hence send the left to \(\displaystyle L\) and the right to \(\displaystyle \pm\infty\) since \(\displaystyle L\not= 0\). This is a contradiction. If \(\displaystyle \lambda = 0, f(1)\not= 1\) then \(\displaystyle L=0\) and \(\displaystyle f(y) = \frac{f(1)}{(1 - f(1))y + f(1)}\) for all but possibly one positive real \(\displaystyle y\). Then when \(\displaystyle x = y\rightarrow +\infty\) we have the left tending towards \(\displaystyle L = 0\) and the right towards \(\displaystyle L^2 + \lim_{y\rightarrow +\infty}\frac{f(1)y}{(1 - f(1))y + f(1)} = \frac{f(1)}{1 - f(1)}\not= 0\). This is a contradiction. Finally, we deduce that \(\displaystyle f(1) = 1\) and hence

\(\displaystyle f(y) = \lambda y + (1 - \lambda)\)

for all positive real \(\displaystyle y\). Then \(\displaystyle P(1,1)\) shows \(\displaystyle f(1+1)=1+1\), whence \(\displaystyle \lambda=1\). Therefore, \(\displaystyle f(x) = x\forall x\in\mathbb{R}^+\), which works. This is our only solution and we are done. \(\displaystyle \Box\)

Author's note. It is not much harder to solve the equation for \(\displaystyle f: \mathbb{R}^+\rightarrow\mathbb{R}\). If there exists \(\displaystyle x\) with \(\displaystyle f(x) < 0\) then taking \(\displaystyle y=-f(x)\), we find a value \(\displaystyle z\) with \(\displaystyle f(z) = 0\). Therefore if the range is not contained in \(\displaystyle \mathbb{R}^+\), we will have a positive real \(\displaystyle z\) with \(\displaystyle f(z) = 0\). Then taking \(\displaystyle y=z\) yields \(\displaystyle f(xz) = 0\), so that \(\displaystyle f\) is identically zero. This allows us to add the solution \(\displaystyle f(x) = 0\forall x\in\mathbb{R}^+\), but otherwise does not affect the solution.

Statistics:

4 students sent a solution.
5 points:	Schrettner Jakab.
0 point:	3 students.

Problems in Mathematics of KöMaL, May 2018