# Find all functions for which $\ x \cdot f(xy)+f(-y)=x \cdot f(x)$

Does anyone have idea for solution (for all non-zero numbers)?

$$\ x ≠ 0,y ≠ 0$$

$$\ f: R \setminus \{0\} → R$$

$$\ x \cdot f(xy)+f(-y)=x \cdot f(x)$$

Thanks!

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Is that equation supposed to be true for all $x$ and $y$? For some pair $x$ and $y$? For some function $y$ of $x$? For some function $x$ of $y$? –  Chris Eagle Dec 2 '12 at 15:56
for all non-zero numbers –  Miloš Havlíček Dec 2 '12 at 15:58
Right, now put that in the question. Important facts should not lurk in the comments. –  Chris Eagle Dec 2 '12 at 15:59
The nil function works! Do I get a candy for finding a solution? –  gniourf_gniourf Dec 2 '12 at 16:41
If you will say me how to solve it, I will be incredibly thankful :-) –  Miloš Havlíček Dec 2 '12 at 16:48

I can solve this completely given the additional assumption that $f$ is continuously differentiable on its domain. I suspect that it's possible to relax this to merely "continuous", but you'll probably need some sort of niceness assumption to get anywhere.

If we set $x=1$, the functional equation becomes $$f(y)+f(-y)=f(1) \, .$$ So the equation is equivalent to $$x \, f(xy)-f(y)+f(1)=x \, f(x) \, ,$$ which I will suggestively rewrite as $$x[f(xy)-f(x)]=f(y)-f(1) \, .$$

Now, restrict to the case where $y$ is positive and set $y=1+\epsilon$. Since $f$ is differentiable, we can apply the mean value theorem to find $c_\epsilon \in [x, xy]$, $d_\epsilon \in [1, y]$ such that $$f(xy)-f(x)=x \epsilon \, f'(c_\epsilon)\\ f(y)-f(1)= \epsilon f'(d_\epsilon)$$ (where we take the convention that interval endpoints are commutative: that is, if $a>b$, $[a,b]=[b,a]$).

So our equation becomes $$x^2\epsilon f'(c_\epsilon)=\epsilon f'(d_\epsilon) \, .$$ But $f'$ is continuous by assumption, and we have $\lim_{\epsilon \to 0} c_\epsilon=x$, $\lim_{\epsilon \to 0} d_\epsilon=1$. So, passing to the limit, $$x^2 f'(x)=f'(1) \, ;$$ that is, $f'(x)=\frac{C}{x^2}$ for all $x$ and some constant $C$, and so $f(x)=\frac{C}{x}+D$ for all $x$ and some constants $C,D$.

Now, we can plug this $f$ back into the original functional equation to find which values of $C,D$ actually work. After a little bit of algebra, we see that the left-hand side of the equation reduces to $D+Dx$, while the right-hand side reduces to $C+Dx$. So we must have $C=D$.

That is, the continuously differentiable $f$ which satisfy the equation are precisely those of the form $f(x)=C(1+\frac{1}{x})$ for some constant $C$.

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Partial Solution

$$x \cdot f(xy)+f(-y)=x \cdot f(x)$$

Multiplying by $y$ we get

$$xy \cdot f(xy)+yf(-y)=xy \cdot f(x) \,.$$

We define a new function $g(t):= tf(t)$.

Then, the functional equation becomes

$$g(xy)-g(-y)=yg(x) \,.$$

Thus

$$g(xy)=yg(x)+g(-y)$$

In particular, we get

$$\frac{g(xy)-g(-y)}{xy+y}=\frac{g(x)}{x+1} \,.$$

It is easy to see that $g(-1)=0$. Thus, with the extra assumption that $g(t)$ is differentiable at $-1$, by setting $x \to -1$, we get that $g$ is differentiable at $-y$ and $g'(-y)=g'(-1)$. Let $g'(-1)=C$, then $g(t)= Ct+D$ for some constants $C,D$, and the functional equation Yields $c=d$.

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I guess there doesn't lies any solution to this one . Replacing all y's by x. We get x∗f(x^2)+f(−x)=x∗f(x)....(1)in this equation if we put x=1,we get f(-1)=0 .So let us consider there lies a function Q(x) such that , f(x)=(x+1)Q(x). Replacing the f(x) in equation (1) we get one more equation, if we consider the lading coefficient of Q(x) be a1, and it be of degree n.....we see that n=0. so there doesn't lie a function as Q(x). we get f(x)=(x+1);which actually does't satisfy equation 1. so we conclude that f(x) is a constant function and is identically equal to zero.

ALTERNATIVELY considering the given functional equation if we differentiate both sides with respect to x we get

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The first solution works only for polynomials... The second one only for differentiable functions.... –  N. S. Dec 2 '12 at 17:17
I am sure that there is at least one possible solution. –  Miloš Havlíček Dec 2 '12 at 17:20