Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 .

share|cite|improve this question
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) . – Ester May 19 '12 at 3:39
In the problem statement, does the domain of your functions include 0? – Alex R. May 19 '12 at 3:45
No , that is actually where the problem lies . – Ester May 19 '12 at 3:48
up vote 6 down vote accepted

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.

share|cite|improve this answer
Nice alternative proof. – Ewan Delanoy May 24 '12 at 9:37
Excellent work Mercio – 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.

share|cite|improve this answer
How can you assume that f , g , h are continuous ? – Ester May 19 '12 at 17:20
@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 $(2)$. – Christian Blatter May 19 '12 at 17:38
Can you give an example of such weird functions? – Alex R. May 19 '12 at 17:42
@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. – Dejan Govc May 19 '12 at 17:52
@Sam: You might be interested in the following question. (And the wikipedia article mentioned there.) – 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

share|cite|improve this answer
Since when $0$ defined to be positive? – Asaf Karagila May 19 '12 at 10:13
Oops..I have not seen. Ok i will tell. – Prasad G May 19 '12 at 10:15
It's my understanding that in France zero is considered to be positive. – 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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