Find all continuous functions with the property that f(x)f(y)f(z) = 3, for every x , y , z ∈ R, with x + y + z = 0.I am quite new to new to functional equations and I have been stuck on this problem for quite a while.I tried to somehow get and transform into a Cauchy equation but failed during the process.

Question

Find all continuous functions with the property that f(x)f(y)f(z) = 3, for every   x  ,  y  ,  z  ∈  R, with x + y + z = 0.I am quite new to new to functional equations and I have been stuck on this problem for quite a while.I tried to somehow get and transform into a Cauchy equation but failed during the process.

Coraducci0d · Accepted Answer

First of all, note that   f  (  0      )    3    =  3, by plugging   x  =  y  =  z  =  0. This way, we have that   f  (  0  )  =      3    3  , so let us denote   g  (  x  )  =  f  (  x  )      /        3    x  . The original condition tells us that   g  (  x  )  g  (  y  )  g  (  z  )  =  1 for every   x  +  y  +  z  =  0, and   g  (  0  )  =  1 by construction.Now, by plugging   y  =  −  x  ,  z  =  0, we have that   g  (  x  )  g  (  −  x  )  g  (  0  )  =  1; or   g  (  −  x  )  =  g  (  x      )          −      1      . In particular, this tells us that   g  (  x  )  ≠  0 for all   x.To conclude, note that for every   x and   y, we can take   z  =  x  +  y, so  g  (  x  )  g  (  y  )  g  (  −  x  −  y  )  =  1    ⟹    g  (  x  )  g  (  y  )  =  g  (  x  +  y  )  .This is now a Cauchy functional equation.

acsalagi3l · Accepted Answer

Nice

Open question

Answer & Explanation

New Questions in Differential Calculus