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I need help solving this functional equations problem.

Find all $\ f : \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y\ \epsilon \ \mathbb{R}$, the two following equations hold:

$f(3x)=f\left(\frac{x+y}{(x+y)^2+1}\right) + f\left(\frac{x-y}{(x-y)^2+1}\right)$

$f(x^2-y^2)=(x+y)f(x-y)+(x-y)f(x+y)$

What I did was using the second equation to get that $f(0)=0$ by setting $x=y=0$.

then setting $x=0, \ y$ arbitrary in the first equation to get:

$0=f(0)=f\left(\frac{y}{y^2+1}\right) + f\left(\frac{-y}{y^2+1}\right)$ $\implies f\left(\frac{y}{y^2+1}\right)=-f\left(\frac{-y}{y^2+1}\right) \implies f$ is odd.

But then I got stuck here, I tried many ways but always end up with a variation

of $f(-y^2)=2yf(-y)=-2yf(y)$

Any suggestions how to go on from here?

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I'm going to disregard the first equation altogether, not because it is ugly (though it is), but because one equation is just enough for a problem. Another one does not feel like it belongs here; for all I know, it could have been pasted by mistake.

Now to the point. $$f(x^2-y^2)=(x+y)f(x-y)+(x-y)f(x+y)\tag1$$

Let $a=x+y,\;b=x-y$ $$f(ab)=af(b)+bf(a)\tag2$$

Plugging $b=a$, we get $f(a^2)=2af(a)$. Now plugging $b=a^2$, we get $f(a^3)=3a^2f(a)$, and continuing in this manner, we end up with $f(a^n)=na^{n-1}f(a)$.

Now let's apply that in reverse: $f(a)=na^{n-1\over n}f(a^{1/n})$, hence $f(a^{1/n})={1\over n}f(a)/a^{n-1\over n}$.

Now let's put the two together: $$f(a^{p\over q}) = {1\over q}f(a^p)/a^{p(q-1)\over q}={p\over q}f(a)\cdot a^{p-1-{p(q-1)\over q}}={p\over q}f(a)\cdot a^{{p\over q}-1}\tag3$$ or, in other words, for any rational $r$ we have $$f(a^r)=r\cdot f(a)\cdot a^{r-1}\tag4$$

Well, rationals are dense in $\mathbb R$, but what's between them? The discontinuous solutions are plenty. If you know what the Hamel basis is, then you know where to find them, and if you don't know, then trust me, you don't want to know. But if we stick to $f$ being continuous everywhere (or anywhere, for that matter), the only way to achieve that is for (4) to hold at any real $r$.

Say we fix $a$ and looks at some $x$ (not the same $x$ as in the beginning):

$$x=a^{\log_ax}\implies f(x)=\log_ax\cdot f(a)\cdot a^{\log_ax-1}=C\cdot x\cdot\ln x\tag5$$

So this is the only family of continuous solution to your second equation. If you are still interested, you may go and check how well does this sit with the first equation (hint: not well, except the trivial case of $C=0$).

That's it.

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  • $\begingroup$ I know this is late but at that time I thought this was above my level. But now I understand everything except the last equation (5). where does x.lnx come from? $\endgroup$ Mar 1 '20 at 15:59
  • $\begingroup$ Why, it comes from what is right before it. $a^{\log_ax}$ is just $x$, and $\log_ax$ is basically $\ln x$ times stuff. $\endgroup$ Mar 1 '20 at 22:11

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