The question says

Find all functions that satisfy $$f(xf(x)) = 2f(x)$$.

By comparing the degree of the two sides of the equation, one can easily see that $f$ can not be a non-zero polynomial or a power function. The function that satisfy the functional equation I found so far is the trivial $f(x) =0$. Can anyone show a way to find the rest or prove the one I found is only one ? Thanks in advance.

  • $\begingroup$ What is the domain and codomain? Real numbers? Integers? Natural numbers (with or without zero)? $\endgroup$ Oct 18, 2020 at 21:23
  • $\begingroup$ In the question, it doesn't explicitly state the domain. I think it is all real numbers though. $\endgroup$ Oct 18, 2020 at 21:25
  • 2
    $\begingroup$ That's unfortunate. The answer could depend quite delicately on which domain you pick. $\endgroup$ Oct 18, 2020 at 21:26
  • $\begingroup$ @QiaochuYuan It seems to me that this query touches on a special area in math : functional equations. Do you know of any online resource (e.g. pdf) that is appropriate for someone conversant in Real Analysis but new to the area of functional equations? $\endgroup$ Oct 18, 2020 at 22:08

1 Answer 1


Let us look for all continuous (a very string restriction!) solutions $\Bbb R\to\Bbb R$ (a domain allowing many tools!).

With $x=0$, we find $f(0)=0$.

If $f(x_0)=1$, then $$2=2f(x_0)=f(x_0f(x_0))=f(x_0)=1,$$ contradiction. We conclude $f(x)\ne1 $ for all $x$.

If $y=f(x)$ is in the image of $f$, then so is $2y=f(xf(x))$. In particular, if $f$ attains any positive value, then $f$ is unbounded from above. But then the Intermediate Value Theorem implies that $1$ is in the image of $f$. We conclude that $f(x)\le 0$ for all $x$.

And if $f$ attains any negative value, then $f$ is unbounded from below. Then $f(x_1)=-\frac12$ for some $x_1$. Then $$ f(-\tfrac12x_1)=f(x_1f(x_1))=2f(x_1)=-1.$$ By the IVT again, there exists $x_2$ between $0$ and $-\frac12x_1$ with $f(x_2)=-\frac12$. By repeating this process, we obtain a sequence $\{x_n\}$ such that $f(x_n)=-\frac 12$ and $|x_{n+1}|<\frac12|x_n|$, i.e., $x_n\to 0$. As $f(0)=0$, this contradicts continuity.

We conclude that the only continuous solution $f\colon \Bbb R\to\Bbb R$ is the trivial solution.


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