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Solve the following functional equation: $ k \in \mathbb R $, $ f : \mathbb R \to \mathbb R $ and $$ f ( x + y ) = f ( x ) f ( y ) + k x y ( x + y ) \text , \forall x , y \in \mathbb R \text . $$

I have only got $ f ( 0 ) = 1 $, by letting $ f ( 0 ) = a $ and solving for $ a $.

If I use this, everything cancels out and leaves me with $ f ( x ) = f ( x ) $ which is fruitless.

Mohsen Shahriari
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1 Answers1

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Sketch: Assume $k \ne 0$.

Note that \begin{align} f(x)f(-x)= f(0) = 1 \end{align} then it follows $f$ is never zero and \begin{align} f(x) = \frac{1}{f(-x)}. \end{align} Also, we have that \begin{align} f(2x) = f(x)^2+2kx^3 \end{align} which means \begin{align} 1=&\ f(2x)f(-2x) \\ =&\ (f(x)^2+2kx^3)(f(-x)^2-2kx^3) \\ =&\ (f(x)^2+2kx^3)(\frac{1}{f(x)^2}-2kx^3)\\ =&\ 1+\frac{2kx^3}{f(x)^2}-2kx^3f(x)^2-4k^2x^6 \end{align}

Hence \begin{align} &0=2kx^3f(x)^4+4k^2x^6f(x)^2-2kx^3\\ &\implies f(x)^4+2kx^3f(x)^2-1=0 \end{align}

So, we have that \begin{align} f(x)^2=\frac{-2kx^3\pm \sqrt{4k^2x^6+4}}{2}= -kx^3\pm \sqrt{k^2x^6+1} \end{align} which means \begin{align} f(x) = \sqrt{\sqrt{k^2x^6+1}-kx^3}. \end{align}

However, by plugging $f$ back into the functional equation, we immediately arrive at a contradiction. Hence $f(x)^4+2kx^3f(x)^2-1\ne 0$, which means $k = 0$. This reduces the functional equation to $f(x+y) = f(x)f(y)$, which means $f(x) = e^{cx}$.

Jacky Chong
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  • Hey @JackyChong, when I submit this my friend says to think about the problem more because there is a mistake. He says the problem is somewhere in $\sqrt{\sqrt{k^2 x^6+1}+kx^3}$ –  Oct 28 '20 at 12:43
  • The formula you've found is derived by the assumption $k\ne0$. You can put it in the functional equation and do the messy calculations, only to get a contradiction. That means it cannot be true, thus we must have $k=0$, and hence $f$ must satisfy Cauchy's exponential functional equation. – Mohsen Shahriari Oct 28 '20 at 18:04
  • @MohsenShahriari Yes. That's correct. I got to check the validity of the solution when I toss out the $2kx^3$ factor. – Jacky Chong Oct 29 '20 at 18:04