0
$\begingroup$

Find all functions $f: \mathbb R\rightarrow \mathbb R$, at the same time satisfying the following two conditions:

a) $f (x + yf (x)) = f (x) f (y)$

b) the function $f$ can be represented in the form $f (x) = (\varphi (x)) ^ 2, x \in \mathbb R,$ where the function $f$ has a finite derivative at $x = 0.$ (not infinite)

I have no clue how to start. Any kind of help will be appreciated.

$\endgroup$
  • $\begingroup$ What do you mean by a 'finite derivative'? Non zero, not infinite? $\endgroup$ – copper.hat Apr 22 '16 at 16:13
  • $\begingroup$ @ copper.hat: not infinite $\endgroup$ – Roman83 Apr 22 '16 at 16:15
  • $\begingroup$ $f(0)=f(0)^2$ $f(0)=0$ $\endgroup$ – kmitov Apr 22 '16 at 16:16
  • $\begingroup$ Is $f(x)=0$ and $f(x)=1$ a solution? Doesn't solve the problem, though. $\endgroup$ – S.C.B. Apr 22 '16 at 16:18
  • $\begingroup$ @kmitov also $f(0)=1$ is possible $\endgroup$ – gt6989b Apr 22 '16 at 16:21
1
$\begingroup$

Here is a possible approach.

Plug in $y=0$ to get $$ f(x) = f(x) f(0), $$ so either $f(x) \equiv 0$ or $f(0)=1$.

Now note that $$ f(x+yf(x)) = f(y+xf(y)) $$ and assuming $f$ is 1-to-1, we have $$ x + yf(x) = y + xf(y)\\ x(1-f(y)) = y(1-f(x))\\ \frac{x}{1-f(x)} = \frac{y}{1-f(y)} $$ for arbitrary $x,y$, and that means both LHS and RHS and constant, say $c$.Then you have $$ c = \frac{x}{1-f(x)} \\ f(x) = 1 - x/c $$

UPDATE If $f$ is not 1-1, $f(x) \equiv 1$ is a solution, but not sure if there are others...

$\endgroup$
  • $\begingroup$ Why $$f(x+yf(x)) = f(y+xf(y)) \Rightarrow x + yf(x) = y + xf(y)$$? $\endgroup$ – Roman83 Apr 22 '16 at 16:38
  • $\begingroup$ @Roman83 if you assume $f$ is one-to-one, then $f(a) = f(b) \implies a = b$. Notice the result satisfies this property as well as that $f(0)=1$... $\endgroup$ – gt6989b Apr 22 '16 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.