Taha1381 wrote: Let $a,b,c$ be non-zero distinct real numbers so that there exist functions $f,g:\mathbb{R}^{+} \to \mathbb{R}$ so that: $af(xy)+bf(\frac{x}{y})=cf(x)+g(y)$ Prove that there exists a function $h:\mathbb{R} \to \mathbb{R}^{+}$ so that: $f(xy)+f(\frac{x}{y})=2f(x)+h(y)$ Wrong. Choose as counter-example $f\equiv g\equiv 0$

@above Seems like the typo is fixed.

Taha1381 wrote: Let $a,b,c$ be non-zero distinct real numbers so that there exist functions $f,g:\mathbb{R}^{+} \to \mathbb{R}$ so that: $af(xy)+bf(\frac{x}{y})=cf(x)+g(y)$ Prove that there exists a function $h:\mathbb{R}^{+} \to \mathbb{R}$ so that: $f(xy)+f(\frac{x}{y})=2f(x)+h(y)$ With this modification : Let $P(x,y)$ be the assertion $af(xy)+bf(\frac xy)=cf(x)+g(y)$ 1) If $a+b-c\ne 0$ $P(x,1)$ $\implies$ $f(x)=\frac{g(1)}{a+b-c}$ constant and so $f(xy)+f(\frac xy)=2f(x)$ and so $\boxed{h(y)=0\quad\forall y}$ fits Q.E.D. 2) If $a+b-c=0$ $P(x,y)$ may be written as new assertion : $Q(x,y)$ : $a(f(xy)-f(x))+b(f(\frac xy)-f(x))=g(y)$ Subtracting $Q(1,y)$ from $Q(x,y)$, this becomes (1) : $a(f(xy)-f(x)-f(y)+f(1))+b(f(\frac xy)-f(x)-f(\frac 1y)+f(1))=0$ Moving in (1) $y\to\frac 1y$, we get : (2) : $b(f(xy)-f(x)-f(y)+f(1))+a(f(\frac xy)-f(x)-f(\frac 1y)+f(1))=0$ We obviously have determinant $a^2-b^2\ne 0$ (since $a\ne b$ and $a+b=c\ne 0$) and so $f(xy)=f(x)+f(y)-f(1)$ $\forall x,y>0$ $f(\frac xy)=f(x)+f(\frac 1y)-f(1)$ $\forall x,y>0$ Adding, this becomes $f(xy)+f(\frac xy)=2f(x)+f(y)+f(\frac 1y)-2f(1)$ And so $\boxed{h(y)=f(y)+f(\frac 1y)-2f(1)\quad\forall y}$ fits Q.E.D.

Literally nothing different to above. This is for people who want to represent $h$ in terms of $g$. 0. $a+b \ne c$, put $y=1$ we get $f$ is constant and $h(y)=0$ fits. 1. $a+b=c$ and if $a=b$, then $h(y)=g(y)/a$ fits. 2. $a+b=c$ and if $a\ne b$ Write the given equation as $a(f(xy)-f(x))-b(f(x)-f(x/y))=g(y)$. Put $1/y$ in $y$ and we get $b(f(xy)-f(x))-a(f(x)-f(x/y))=g(1/y)$ Subtracting, we get $(a-b)(f(xy)+f(x/y)-2f(xy))=g(y)-g(1/y)$. Therefore, $h(y)=(g(y)-g(1/y))/(a-b)$ fits. By the way, is it really problem 3? Seems too easy for Iranian TST #3

bumjoooon wrote: Literally nothing different to above. This is for people who want to represent $h$ in terms of $g$. 0. $a+b \ne c$, put $y=1$ we get $f$ is constant and $h(y)=0$ fits. 1. $a+b=c$ and if $a=b$, then $h(y)=g(y)/a$ fits. 2. $a+b=c$ and if $a\ne b$ Write the given equation as $a(f(xy)-f(x))-b(f(x)-f(x/y))=g(y)$. Put $1/y$ in $y$ and we get $b(f(xy)-f(x))-a(f(x)-f(x/y))=g(1/y)$ Subtracting, we get $(a-b)(f(xy)+f(x/y)-2f(xy))=g(y)-g(1/y)$. Therefore, $h(y)=(g(y)-g(1/y))/(a-b)$ fits. By the way, is it really problem 3? Seems too easy for Iranian TST #3 This isn’t TST. As far as I know there is another condition for both equations. Parameter $y$ have to be sufficiently large. So we can’t put $y=1$ inside the equation.

Dadgarnia wrote: bumjoooon wrote: Literally nothing different to above. This is for people who want to represent $h$ in terms of $g$. 0. $a+b \ne c$, put $y=1$ we get $f$ is constant and $h(y)=0$ fits. 1. $a+b=c$ and if $a=b$, then $h(y)=g(y)/a$ fits. 2. $a+b=c$ and if $a\ne b$ Write the given equation as $a(f(xy)-f(x))-b(f(x)-f(x/y))=g(y)$. Put $1/y$ in $y$ and we get $b(f(xy)-f(x))-a(f(x)-f(x/y))=g(1/y)$ Subtracting, we get $(a-b)(f(xy)+f(x/y)-2f(xy))=g(y)-g(1/y)$. Therefore, $h(y)=(g(y)-g(1/y))/(a-b)$ fits. By the way, is it really problem 3? Seems too easy for Iranian TST #3 This isn’t TST. As far as I know there is another condition for both equations. Parameter $y$ have to be sufficiently large. So we can’t put $y=1$ inside the equation. Oops.. Sorry, yes this is the 3rd round. But still I think this was too easy for 3rd round's #3. Good to hear that there are more conditions.! Thanks.

Very nice!

I'm very late here, but it's worth mentioning that this is almost the same as this problem from Thailand MO 2018. All we need to do here is define three new functions $f_1(x)\equiv f(e^x),g_1(x)\equiv g(e^x),h_1(x)\equiv h(e^x)$.

Observe that if $$ a = b = \frac{c}{2} \qquad \qquad (\star) $$then $$ h(y) \equiv \frac{g(y)}{a} $$just works. Otherwise, if $(\star)$ is not true, we will show $$ h(y) \equiv 0 $$works. Suppose that the equation $$ af(xy) + b f \left( \frac{x}{y} \right) = cf(x) + g(y) \qquad \qquad (1)$$is satisfied for all $y \ge \lambda$. We may assume $\lambda \ge 1$ (hence now if $y$ is large, then so are $y^2,y^3,\ldots$). Fix a (large) $y$ and any $t$. We define $$ s_i = t \cdot y^i $$Then from $(1)$ we have the recursion $$ as_{i+1} + bs_{i-1} = cs_i + d_1 \qquad \qquad (2)$$where $d_1 = g(y)$. But the point is that we could have done the same thing for $y^m$ instead of $y$. Hence we have $$ as_{i+m} + bs_{i-m} = cs_i + d_m \qquad \qquad (3)$$for all $m \in \mathbb N$, where $d_m = f(y^m)$. We want to show that $$ s_{i+1} + s_{i-1} - 2s_i = 0 ~~ \text{if } (\star) \text{ is not satisfied} \qquad \qquad (\diamondsuit)$$Define $$ r_i = s_{i+1} - s_i $$Note $(\diamondsuit)$ is equivalent to $$ \text{sequence } \{r_i\} \text{ is constant if } (\star) \text{ is not satisfied} \qquad \qquad (\heartsuit)$$Replacing $i$ by $i+1$ in $(3)$ gives $$ as_{i+m+1} + bs_{i-m+1} = cs_{i+1} + d-m \qquad \qquad (4) $$Subtracting $(3)$ from $(4)$ gives $$ a(s_{i+m+1} - s_{i+m}) + b(s_{i-m+1} - s_{i-m}) = c(s_{i+1} - s_i) $$Hence we obtain $$ ar_{i+m} + br_{i-m} = cr_i \qquad \qquad (5) $$Putting $m=1$ in $(5)$ gives $$ ar_{i+1} + br_{i-1} = cr_i \qquad \qquad (6)$$Putting $m=2$ in $(5)$ and using $(6)$ gives \begin{align*} cr_i = ar_{i+2} + br_{i-2} = (cr_{i+1} - br_i) &+ (cr_{i-1} - ar_i) = c(r_{i+1} + r_{i-1}) - (a+b)r_i \\ \implies (a+b+c)r_i &= c(r_{i+1} + r_{i-1}) \end{align*}Let $$k = \frac{a+b+c}{c}$$Note $k$ well defined as $c \ne 0$. We obtain $$ r_{i+1} + r_{i-1} = kr_i \qquad \qquad (7) $$Combining $(6)$ and $(7)$ we obtain \begin{align*} cr_i = ar{i+1} + br_{i-1} = ar_{i+1} + ar_{i-1} + (b-a)r_{i-1} \\ \implies cr_i = (ak)r_i + (b-a)r_{i-1} \qquad \qquad (8) \end{align*}We have several cases now. We will assume that all $r_i$ are not $0$, as in that case $(\heartsuit)$ is trivially satisfied. Case 1: All $b-a \ne 0$ and $c-ak \ne 0$. We obtain there is a $\alpha \ne 0$ such that $$ r_i = \alpha r_{i-1} ~~ \forall ~ i $$Using $r_i \ne 0$, from $(5)$ we obtain $$ a \alpha^m + \frac{b}{\alpha^m} = c ~~ \forall ~ m \in \mathbb N $$If $\alpha \ne 1,-1$, then for large $m$, the LHS becomes too large, contradiction. Hence $\alpha = \pm 1$. If $\alpha = -1$, then putting $m=1,2$ we get $c = -c$, hence $c=0$, contradiction. Hence $$ \alpha = 1 $$This means all $r_i$ are equal. Case 2: $b-a \ne 0$ while $c - ak = 0$. Then we get all $r_i$ are $0$, which contradicts our assumption. Case 3: $a =b$. As all $r_i$ are not $0$, so from $(8)$ we obtain $$ ak = c $$Hence we have $$ \frac{a+b+c}{c} = \frac{c}{a} \implies c = -a , 2a $$If $c=-a$, then from $(5)$ we obtain $$ r_{i+m} + r_{i-m} + r_i = 0 ~~ \forall ~ i \in \mathbb Z , m \in \mathbb N \qquad \qquad (9)$$This directly gives $$ r_{i+3} = r_i ~~ \forall ~ i $$Then putting $m=3$ in $(9)$ we obtain $$ r_i = 0 ~~ \forall ~ i $$which is a contradiction. Hence we conclude $$ c = 2a $$Thus $$ a = b = \frac{c}{2} $$ So we proven $(\heartsuit)$, which completes the proof. $\blacksquare$