While studying functional equation, I came across the following problem.
Problem. find all functions $ f : \mathbb R \to \mathbb R $ such that for some $ a , b \in \mathbb R $, $$ f ( x ) f ( y ) = x ^ a f \left( \frac y 2 \right) + y ^ b f \left( \frac x 2 \right) $$ holds for all reals $ x $ and $ y $.
The solution presented by the author is as follows.
Solution. If $ a = b $, we can denote $$ \frac { f ( x ) } { x ^ a } = g ( x ) \text , $$ so that $$ g ( x ) g ( y ) = g ( x + y ) \text , \quad \forall x , y \in \mathbb R \text , $$ and here use the new function $ \require {color} \color {red} h ( x ) = g ( e ^ x ) $ to get $ h ( x + y ) = h ( x ) + h ( y ) $. This is the classic Cauchy equation with the solution $ h ( x ) = k x $ for all real $ x $ and any real $ k $.
I am not able to understand that how we can obtain $$ h ( x + y ) = h ( x ) + h ( y ) $$ by using the new function $ h ( x ) = g ( e ^ x ) $.
My try:
We have $ h ( x ) = g ( e ^ x ) $ which is same as saying $ h ( \ln x ) = g ( x ) $. On substituting $ h ( \ln x ) = g ( x ) $ in $$ g ( x ) g ( y ) = g ( x ) + g ( y ) $$ we get $$ h ( \ln x ) h ( \ln y ) = h ( \ln x ) + h ( \ln y ) \text . $$ But I am really not able proceed any further.
A detailed explanation of what's actually going on and how to proceed further would be highly helpful. As it is mentioned in the problem "... for some $ a , b \in \mathbb R $", could you please explain that why and how $ a = b $ in the solution?
Also another elegant solution for the very problem would be highly appreciated.
As I've explained in a comment, the question is not well-posed. To make this go away, one can add the requirement that $ a $ and $ b $ are positive integers. Alternatively, on can consider $ f : \mathbb R ^ + \to \mathbb R $ and require that the functional equation $$ f ( x ) f ( y ) = x ^ a f \left( \frac y 2 \right) + y ^ b f \left( \frac x 2 \right) \tag 0 \label 0 $$ holds for all positive reals $ x $ and $ y $. Some (seemingly more complicated) alternatives are possible, too.
Assuming we've taken care of the ambiguity of the problem, one can see that in case $ a \ne b $, the only solution is the constant zero function. To see this, use \eqref{0} and see that $$ x ^ a f \left( \frac y 2 \right) + y ^ b f \left( \frac x 2 \right) = f ( x ) f ( y ) = f ( y ) f ( x ) = y ^ a f \left( \frac x 2 \right) + x ^ b f \left( \frac y 2 \right) \text , $$ which in particular shows that $$ f ( x ) = \frac { f ( 1 ) } { 2 ^ a - 2 ^ b } \left( ( 2 x ) ^ a - ( 2 x ) ^ b \right) \text . \tag 1 \label 1 $$ \eqref{1} shows that $ f \left( \frac 1 2 \right) = 0 $, and then letting $ x = y = 1 $ in \eqref{0} we get $ f ( 1 ) = 0 $, and thus by \eqref{1}, $ f $ is the constant zero function. It's trivial to check that this indeed gives a solution.
For the case $ a \ne b $, you can follow the given solution. But you need to note that as $ g ( x ) $ is defined to be equal to $ \frac { f ( x ) } { x ^ a } $, this is only possible when $ x ^ a \ne 0 $, and thus $ 0 $ cannot be assumed to be in the domain of $ g $. Also, as $ h ( x ) $ is defined to be equal to $ g ( e ^ x ) $, determining $ h $ only determines the restriction of $ g $ to the set of positive reals (since $ e ^ x > 0 $). If we've chosen the alternative in which we have $ f : \mathbb R ^ + \to \mathbb R $, we're indeed done at this point. But if we've chosen the alternative in which $ f : \mathbb R \to \mathbb R $ but $ a $ and $ b $ couldn't be just any real numbers, we need to continue further, and find out what $ f ( x ) $ can be for negative $ x $ and also for $ x = 0 $. That shouldn't be much difficult, assuming you've made your peace with the given solution in the original post.