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$\mathbf {The \ Problem \ is}:$ Let, $f,g,h$ be three functions defined from $(0,\infty)$ to $(0,\infty)$ satisfying the given relation $f(x)g(y) = h\big(\sqrt{x^2+y^2}\big)$ for all $x,y \in (0,\infty)$, then show that $\frac{f(x)}{g(x)}$ and $\frac{g(x)}{h(x)}$ are constant.

$\mathbf {My \ approach} :$ Actually, by putting $x$ in place of $y$ and vice-versa, we can show that $\frac{f(x)}{g(x)}$ is a constant, let it be $c .$ Then, I tried that $g(x_i)g(y_i)=g(x_j)g(y_j)$ whenever $(x_i,y_i)$, $(x_j,y_j)$ satisfies $x^2+y^2 =k^2$ for every $k \in (0,\infty)$ . But, I can't approach further.

Any help would be greatly appreciated .

Mohsen Shahriari
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Rabi Kumar Chakraborty
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2 Answers2

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Yes you're right at the beginning: $$P(x,y) \implies f(x)g(y)=h\left(\sqrt{x^2+y^2}\right)$$ $$P(y,x) \implies f(x)g(y)=g(x)f(y)\implies f(x)/g(x)=f(y)/g(y)=c$$ $$\implies f(x)=cg(x) \implies cg(x)g(y)=h\left(\sqrt{x^2+y^2}\right)$$ We can now substitute $y$ with $k$, where $k$ is an infinitesimally small value close to zero from the positive side, getting $g(x)/h\left(\sqrt{x^2+k^2}\right)=c_2$ for $c_2=1/cg(k)$, and since $x^2 \ge 0$ we have $\sqrt{x^2+k^2} \ge k >0$, if we perceive it as $k=0$ we get $g(x)/h(x) = c_2$ and we're done, otherwise we can believe that $\sqrt{x^2+k^2}$ is infinitesimally close to $x$.

Anas A. Ibrahim
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  • Sir,can we give a more concrete logic rather than perceiving $k =0$ or infinitesimally close to $0 ?$ – Rabi Kumar Chakraborty Jun 04 '20 at 15:45
  • Well I'm afraid whenever you substitute with $y=a$ you get $\sqrt{x^2+a^2} \ge a$, so the argument of $h$ can never have a value of less than $a$ – Anas A. Ibrahim Jun 05 '20 at 00:53
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    But when you do it with $k$, we know now that the argument of $h$ will never get to $0$, but can get infinitesimally close, just like the function is defined! Moreover, it is very difficult to get $g(x)/h(x)$ with $g$ and $h$ having the same argument, except if you can define $h(x)=h(\sqrt{x^2+a^2})/c_3$ for some constants $c_3$ and $a$ – Anas A. Ibrahim Jun 05 '20 at 00:58
  • As far as I can see, this argument only makes sense when $ \lim _ { k \to 0 ^ + } \frac { g ( x ) } { h \left( \sqrt { x ^ 2 + k ^ 2 } \right) } $ exists and is equal to $ \frac { g ( x ) } { h ( x ) } $; in other words, when $ h $ is continuous from the right at $ x $. This is not an assumption of the problem as stated, and needs proof if you want to use it. – Mohsen Shahriari Jun 05 '20 at 07:53
  • To see the gap in the argument, note that the variable $ c _ 2 $ is in fact a function of $ k $. If you choose another positive $ k $ near $ 0 $, you can end up with another value for $ g ( k ) $, and thus for $ c _ 2 ( k ) = \frac 1 { c g ( k ) } $. When $ k $ approaches $ 0 $ from the right, the value of $ c _ 2 $ may not have a limit; you need to show that it has, if you want to argue as you did. – Mohsen Shahriari Jun 05 '20 at 07:53
  • In fact, while on can validly argue for what the question asks (see my own answer, for example), your argument is not valid; as using the axiom of choice, one can prove the existence of noncontinuous functions $ f $, $ g $ and $ h $ satisfying the desired functional equation. – Mohsen Shahriari Jun 05 '20 at 11:12
  • Take any additive function $ A $ (see https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation or https://math.stackexchange.com/questions/423492/overview-of-basic-facts-about-cauchy-functional-equation/423494#423494), choose any real numbers $ a $ and $ b $, and let $ f ( x ) = \exp \Big( A \big( x ^ 2 \big) + a \Big) $, $ g ( x ) = \exp \Big( A \big( x ^ 2 \big) + b \Big) $ and $ h ( x ) = \exp \Big( A \big( x ^ 2 \big) + a + b \Big) $, and you'll have a solution to the desired functional equation. – Mohsen Shahriari Jun 05 '20 at 11:13
  • In terms of the somehow "inaccurate" language you've used, the "infinitesimal" $ k $ can result in an "infinite" value for $ c _ 2 $, when $ g $ is not supposed to be continuous. You can make sense of the terms "infinitesimal" and "infinite" accurately in the context of nonstandard analysis, for example (see https://en.wikipedia.org/wiki/Nonstandard_analysis). Your argument fails anyway, as there are noncontinuous functions satisfying the equation, even when one can make sense of the nonstandard language. – Mohsen Shahriari Jun 05 '20 at 11:20
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To make the formulas look simpler, define the functions $ \tilde f $, $ \tilde g $ and $ \tilde h $ from $ ( 0 , + \infty ) $ to $ ( - \infty , + \infty ) $ by: $$ \tilde f ( x ) = \log \frac { f \left( \sqrt x \right) } { f ( 1 ) } \qquad \tilde g ( x ) = \log \frac { g \left( \sqrt x \right) } { g ( 1 ) } \qquad \tilde h ( x ) = \log \frac { h \left( \sqrt x \right) } { h \left( \sqrt 2 \right) } $$ Then the functional equation $$ f ( x ) g ( y ) = h \left( \sqrt { x ^ 2 + y ^ 2 } \right) \tag 0 \label 0 $$ transforms to $$ \tilde f ( x ) + \tilde g ( y ) = \tilde h ( x + y ) \text , \tag 1 \label 1 $$ together with $ \tilde f ( 1 ) = \tilde g ( 1 ) = \tilde h ( 2 ) = 0 $. Now, substituting $ y $ for $ x $ and $ x $ for $ y $ in \eqref{1} and comparing the result with \eqref{1}, you get $ \tilde f ( x ) + \tilde g ( y ) = \tilde f ( y ) + \tilde g ( x ) $. Letting $ y = 1 $ in the last equation, you have $$ \tilde f ( x ) = \tilde g ( x ) \text . \tag 2 \label 2 $$Now, use \eqref{1} twice to get $$ \tilde f ( x + 1 ) + \tilde g ( y ) = \tilde h \big( ( x + 1 ) + y \big) = \tilde h \big( x + ( y + 1 ) \big) = \tilde f ( x ) + \tilde g ( y + 1 ) \text. $$ Letting $ y = 1 $ in the above equation you have $ \tilde f ( x + 1 ) = \tilde f ( x ) + \tilde g ( 2 ) $, which using \eqref{1} and \eqref{2} gives $$ \tilde g ( x + y ) = \tilde f ( 1 ) + \tilde g ( x + y ) = \tilde h \big( 1 + ( x + y ) \big) = \tilde h \big( ( x + 1 ) + y \big) \\ = \tilde f ( x + 1 ) + \tilde g ( y ) = \tilde f ( x ) + \tilde g ( 2 ) + \tilde g ( y ) = \tilde g ( x ) + \tilde g ( y ) + \tilde g ( 2 ) \text . $$ Substituting $ \frac x 2 $ for both $ x $ and $ y $ in the above equation you get $ \tilde g ( x ) = 2 \tilde g \left( \frac x 2 \right) + \tilde g ( 2 ) $. Now, using \eqref{1} and \eqref{2} you get $$ \tilde h ( x ) = \tilde h \left( \frac x 2 + \frac x 2 \right) = \tilde f \left( \frac x 2 \right) + \tilde g \left( \frac x 2 \right) = 2 \tilde g \left( \frac x 2 \right) \text . $$ This, together with the previous result shows that $$ \tilde g ( x ) = \tilde h ( x ) + \tilde g ( 2 ) \text . \tag 3 \label 3 $$ You can now rewrite \eqref{2} and \eqref{3} in terms of the original functions and get $$ \frac { f ( x ) } { g ( x ) } = \frac { f ( 1 ) } { g ( 1 ) } $$ and $$ \frac { g ( x ) } { h ( x ) } = \frac { g \left( \sqrt 2 \right) } { h \left( \sqrt 2 \right) } \text , $$ as desired.

Of course, you could avoid defining $ \tilde f $, $ \tilde g $ and $ \tilde h $, and use \eqref{0} and some messier equations corresponding to the above ones, in terms of $ f $, $ g $ and $ h $. But I find this way more elegant, and in fact I think this way you can see the simple idea behind the solution more easily.

Mohsen Shahriari
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