# Spring 2017, problem 38

### Comments

The proof is a little bit longer . Seemingly you took x = 1 + y which works for x greater than 1 .

I may be wrong but after the second implication you find that the function has necssarily

some form for every x grater than one. Then you verify that this form works for every x. I think that this does not imply that there are no other solutions which are identical to the one you found only for every x greater than one. The fact that there are no other solutions follows by also assuming y to be one for every x and use f(1+x) with y in place of x in your solution

I am not enterely satisfied either with Hubert's proof and I would offer the folowing:

First let us assume that the function is continuous, then the original functional equation is equivalent to: f(x+y) - f(x) = -y f(x) f(x+y) This can be written as: (f(x+y) -f(x))/y = -f(x)f(x+y)

Now when y tends to 0 the left hand side is simply df/dx (the derivative and the right hand side is -f(x)^2 We now have a non linear differential equation whose general solution is 1/( a+x) So we have at least captured all continuous functions which satisfy thie original functional equation.

Hello Michel:

I too obtained the hyperbola $f(x) = \frac{1}{x + c}$ as the set of functions that satisfy the original equation. However, we need to be careful since we are concerned with mapping $\mathbb{R^{+}} \rightarrow \mathbb{R^{+}}$ only in the problem statement. This can be realized when $c \ge 0$ (i.e. a negative-valued vertical asymptote or the $y$-axis). If $c$ is negative (i.e. a positive-valued vertical asymptote), then we get a discontinuity at $x = c$ and $f(x)$ is negative for $x \in (0,c)$.

The best function that does this mapping for $x, y \in \mathbb{R^{+}}$ is $f(x) = \frac{1}{x}$ since it encompasses ALL positive reals in its domain and range. Otherwise, $f(x) = \frac{1}{x + c}$ has $y \in (0, \frac{1}{c})$, $x > 0$.

Let me know if you or anyone else here has questions/concerns.

Thanks, Tom Engelsman

The original functional equation is equivalent to: f(x+y) - f(x) = -y f(x) f(x+y) This can be written as: (f(x+y) -f(x))/y = -f(x)f(x+y). |

$x=1, \; a=f(1)\Rightarrow~f(1+y)=\frac a{1+ay}\Rightarrow~~f(x)=\frac a{1-a+ax}$ $f(x)\ge0\Rightarrow~~a\in(0,1] \Rightarrow~~$

$f(x)=\frac 1{\alpha+x}, \; \alpha\ge 0$ which works.