Let $f$ , $g$ , $h$ be three functions from the set of positive real numbers to itself satisfying $$f(x)g(y) = h\left((x^2+y^2)^{\frac{1}{2}}\right)$$ for all positive real numbers $x$ , $y$ . Show that $\dfrac{f(x)}{g(x)}$ , $\dfrac{g(x)}{h(x)}$ and $\dfrac{h(x)}{f(x)}$ are all constant functions .

I have proved that $\dfrac{f(x)}{g(x)}$ is constant and can see that proving either of the last two will prove the final one , but I am not able to prove any of the last two .

Thanks for any help .

  • $\begingroup$ Yes , I have tried that . I can take x as xcosa and y as xsina . Alternatively I can also take x and y as x/2^(1/2) . $\endgroup$ – Ester May 19 '12 at 3:39
  • $\begingroup$ In the problem statement, does the domain of your functions include 0? $\endgroup$ – Alex R. May 19 '12 at 3:45
  • $\begingroup$ No , that is actually where the problem lies . $\endgroup$ – Ester May 19 '12 at 3:48

forall $x,y,z>0$, $h(\sqrt{x^2+y^2})f(z) = f(x) g(y) f(z) = f(x) h(\sqrt{y^2+z^2})$,
thus $\frac {h(\sqrt{x^2+y^2})}{h(\sqrt{z^2+y^2})} = \frac{f(x)}{f(z)} $.

Therefore, forall $x,y,z,t > 0$ : $\frac {h(\sqrt{x^2+z^2})}{h(\sqrt{y^2+z^2})} = \frac {f(x)}{f(y)} = \frac {h(\sqrt{x^2+z^2+t^2})}{h(\sqrt{y^2+z^2+t^2})} = \frac {f(\sqrt{x^2+z^2})}{f(\sqrt{y^2+z^2})}$, thus $\frac {h(\sqrt{x^2+z^2})}{f(\sqrt{x^2+z^2})} = \frac{h(\sqrt{y^2+z^2})}{f(\sqrt{y^2+z^2})} $, which proves that $h/f$ is a constant function.

  • $\begingroup$ Nice alternative proof. $\endgroup$ – Ewan Delanoy May 24 '12 at 9:37
  • $\begingroup$ Excellent work Mercio $\endgroup$ – Ester May 24 '12 at 16:41

We may assume $f(1)=g(1)=1$. It follows that $f(x)=h\bigl(\sqrt{x^2+1}\bigr)=g(x)$ for all $x>0$. Put $$H(t):=h\bigl(\sqrt{t}\bigr)\qquad(t>0)\ ,$$ then $$f(x)\ f(y)=H(x^2+y^2)\qquad(x>0, \ y>0)\ .$$ Taking logarithms we obtain $\log f(x)+\log f(y)=\log H(x^2+y^2)$ or $$\log H(x^2+1)+\log H(y^2+1)=\log H(x^2+y^2)\ .\qquad(1)$$ We now write $x^2:=1+u$, $\ y^2:= 1+v$ with $u$ and $v$ near $0$ and introduce the new function $\phi(t):=\log H(2+t)$. Then $(1)$ becomes the familiar functional equation $$\phi(u)+\phi(v)=\phi(u+v)\ .\qquad(2)$$ If we insist that $f$, $g$, $h$ are continuous then the only solutions to $(2)$ are the functions $\phi(t)=C\,t$, and going all the way backwards the claim about $f$, $g$, $h$ follows.

  • $\begingroup$ How can you assume that f , g , h are continuous ? $\endgroup$ – Ester May 19 '12 at 17:20
  • $\begingroup$ @Sopu: If we assume $f$, $g$, $h$ continuous then the stated claims hold. When $f$, $g$, $h$ are allowed to be arbitrary then there are very weird functions $\phi$ satisfying $(2)$. $\endgroup$ – Christian Blatter May 19 '12 at 17:38
  • $\begingroup$ Can you give an example of such weird functions? $\endgroup$ – Alex R. May 19 '12 at 17:42
  • $\begingroup$ @ChristianBlatter: If my calculations are correct, the discontinuous solutions of $(2)$ also satisfy the conclusion in the original post. (Since then $h(t)=e^{\phi(t^2-2)}$ and $f(t)=e^{\phi(t^2-1)}$, then $\frac{h(t)}{f(t)}=e^{\phi(t^2-2)-\phi(t^2-1)}=e^{-1}$.) Very elegant solution, +1. $\endgroup$ – Dejan Govc May 19 '12 at 17:52
  • $\begingroup$ @Sam: You might be interested in the following question. (And the wikipedia article mentioned there.) $\endgroup$ – Dejan Govc May 19 '12 at 17:57

$f , g , h$ are three functions from the set of positive real numbers to itself

Hence $f(0)$ and $g(0)$ are constants.

$$f(x)g(0) = h\left((x^2+0^2)^{\frac{1}{2}}\right)$$ $$f(x)g(0) = h(x)$$ $$ \dfrac{h(x)}{f(x)} = g(0)$$ There fore, $\dfrac{h(x)}{f(x)}$ is a constant function

$$f(0)g(x) = h\left((0^2+x^2)^{\frac{1}{2}}\right)$$ $$f(0 )g(x) = h(x)$$ $$ \dfrac{h(x)}{g(x)} = f(0)$$ There fore, $\dfrac{h(x)}{g(x)}$ is a constant function $$f(x)g(y) = h\left((x^2+y^2)^{\frac{1}{2}}\right)= f(y)g(x)$$ Take $y=0$, $$f(x)g(0) = f(0)g(x)$$ $$\dfrac{f(x)}{g(x)} = \dfrac{f(0)}{g(0)}$$ There fore, $\dfrac{f(x)}{g(x)}$ is a constant function

  • $\begingroup$ Since when $0$ defined to be positive? $\endgroup$ – Asaf Karagila May 19 '12 at 10:13
  • $\begingroup$ Oops..I have not seen. Ok i will tell. $\endgroup$ – Prasad G May 19 '12 at 10:15
  • $\begingroup$ It's my understanding that in France zero is considered to be positive. $\endgroup$ – Gerry Myerson May 19 '12 at 12:40

We suppose $f,g$, and $h$ are all continuous. I claim that $\lim_{x \rightarrow 0}f(x)$, $\lim_{x \rightarrow 0}g(x)$, and $\lim_{x \rightarrow 0}h(x)$ all exist and are positive. To see this, note that $f(x) = {h(\sqrt{x^2 + 1}) \over g(1)}$ and take limits as $x \rightarrow 0$ on the right. One does the symmetrical argument to show $\lim_{x \rightarrow 0} g(x)$ exists and is positive, and then since $h(x) = f({x \over \sqrt{2}})g({x \over \sqrt{2}})$ we have that $\lim_{x \rightarrow 0} h(x)$ is a positive number as well.

Thus we can extend the domain of definition of all three functions and assume that $f,g,$ and $h$ are continuous positive functions on $[0,\infty)$. By continuity (taking limits as $x$ and/or $y$ go to zero) the functional equation will hold for any $x$ and $y$ in $[0,\infty)$. Now we may plug $y = 0$ into the functional equation and get that $f(x)g(0) = h(x)$, and plugging $x = 0$ into the functional equation we get that $f(0)g(y) = h(y)$ and we are done.


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