Functional equation from USAMO 2010 preparation session:
Find all functions $f:\mathbb R^+\to\mathbb R^+$ such that
$\big(1 + yf(x)\big)\big(1 – yf(x + y)\big) = 1$
for all $x, y \in \mathbb R^+$, where $\mathbb R^+$ is a set of all positive real numbers.
Well I don't really see what we can do. I mean I could have plugged in some numbers but the best one ($0$) which could have caused simplifications is out of reach and so are the negative numbers. Maybe if we could take two cases:
- Both factors are equal to $\pm 1$.
- The factors are reciprocals of each other.
In the first case $+1$ in fact is only possible when $f(x)=0$ $ \forall x$, which isn't possible as $0$ isnt in the codomain. For $-1$ we would get $f(x+y)=-f(x)$, contradiction again.
So we must have the two factors being reciprocals. Well now what? I'm stuck here. Clearly my approach is not just unprofessional, it's bad too. How can this thing be solved?
EDIT:
After considering @Yesit'sme's comment, I retried the problem and would like to present a solution. PLEASE DO TELL ME WHETHER IT IS CORRECT OR NOT. Here we go.
Since $x,y \in \mathbb R^+$, we may without restriction assume $x,y \neq 0$.
Now from given,
$\begin{align}
\big(1 + yf(x)\big)\big(1 − yf(x + y)\big) &= 1 \\
1 − yf(x + y)&= \frac{1}{1 + yf(x)} \\
1-\frac{1}{1 + yf(x)} &= yf(x + y) \\
\frac{1+yf(x)-1}{1 + yf(x)} &= yf(x + y) \\
\frac{f(x)}{1+yf(x)} &= f(x+y)= \frac{f(y)}{1+xf(y)} \tag 1 \label 1
\end{align}$
$\forall x,y \in \mathbb R^+$.
The last equation follow from symmetry (or plugging in $y+x$ into $f$).
Now by plugging in $y=1$ we see that,
$ f(x+1) =\frac{f(x)}{1+f(x)}<f(x), \forall x \in \mathbb R^+$.
The last inequality follows from the fact that,
$\begin{align} f(x)+1 &>1 \\
1 &>\frac{1}{f(x)+1} \\
f(x)&>\frac{f(x)}{f(x)+1}=f(x+1) \end{align}$
As $f(x) \in \mathbb R^+$.
Thus $f(x)$ is in fact decreasing. We now define a new function $Q$ such that,
$f(x)=\frac{1}{Q(x)}$
where $Q$ is strictly increasing $\forall x$.
Plugging this into \eqref{1} we get,
$\begin{align} \frac{\frac{1}{Q(x)}}{1+\frac{y}{Q(x)}} &= \frac{\frac{1}{Q(y)}}{1+\frac{x}{Q(y)}} \\
\frac{1}{Q(x)+y} &= \frac{1}{Q(y)+x} \\
Q(x)+y &=Q(y)+x \\
Q(y)-y &=Q(x)-x=k \\ \end{align} $
Where $k \in \mathbb R$ is a constant. This gives,
$\begin{align}\frac{1}{f(x)} &= x+k \\
\therefore f(x) &= \frac{1}{x+k} \blacksquare. \\ \end{align} $
Plugging this into the original equation we see that the equation is satisfied and hence the solution is complete.
(I didn't show the checking part as typing out this much already took a boatload of time. Hope you understand. It does satisfy though, I have checked it by hand.)
Best Answer
Aside from two simple observations, your answer after the EDIT makes perfect sense.
One is that the constant $ k $ can't be an arbitrary real number, and must be nonnegative (which you may have implicitly taken into account, but haven't asserted explicitly). That's because if $ k < 0 $, then $ - k \in \mathbb R ^ + $, and thus we should have $ Q ( - k ) - ( - k ) = k $, or equivalently $ Q ( - k ) = 0 $, which can't happen as by definition, codomain of $ Q $ is $ \mathbb R ^ + $. Knowing $ k \ge 0 $, your claim that $ f ( x ) = \frac 1 { x + k } $ is a solution, works perfectly well.
The other one is the comment by @Dylan, which adresses your claim about $ f $ being decreasing. As the comment states, you haven't used the fact that $ f $ is decreasing, but I want to stress that you could prove that fact by changing your argument just a little bit. At that point, you know that $ f ( x + y ) = \frac { f ( x ) } { f ( x ) + y } = \frac 1 { \frac 1 { f ( x ) } + y } $. As $ y > 0 $, you have $ \frac 1 { f ( x ) } + y > \frac 1 { f ( x ) } $, and then $ f ( x + y ) = \frac 1 { \frac 1 { f ( x ) } + y } < \frac 1 { \frac 1 { f ( x ) } } = f ( x ) $, which proves what is desired.
I'd like to add another way of thinking, which is essentially your own argument, but may be useful in my opinion. You could observe from the beginning that $ 1 - y f ( x + y ) = \frac 1 { 1 + y f ( x ) } > 0 $, which shows that $ f ( x + y ) < \frac 1 y $. This means that for every $ x , y \in \mathbb R ^ + $ with $ x < y $ we must have $ f ( y ) < \frac 1 x $, or equivalently for every $ y \in \mathbb R ^ + $, $ f ( y ) \le \frac 1 y $, which in turn shows that letting $ k _ y = \frac 1 { f ( y ) } - y $, we must have $ k _ y \ge 0 $. Then you could rewrite $ f ( x + y ) = \frac 1 { \frac 1 { f ( x ) } + y } $ as $ f ( x + y ) = \frac 1 { \left( \frac 1 { f ( x ) } - x \right) + ( x + y ) } $, which simply means that for every $ x , y \in \mathbb R ^ + $ with $ x < y $, we have $ f ( y ) = \frac 1 { y + k _ x } $. This means that for every $ x , y \in \mathbb R ^ + $, if we choose $ z $ so that $ z > \max ( x , y ) $, we must have $ \frac 1 { z + k _ x } = f ( z ) = \frac 1 { z + k _ y } $, which proves that $ k _ x = k _ y $, so we could simply use a single nonnegative constant $ k $, and we're done.