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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!

  • 0
    The set of solutions is a linear space... this generalizes the remark of gniourf.2012-12-02

3 Answers 3

3

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 $C$.

2

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

  • 0
    I am sure that there is at least one possible solution.2012-12-02