# $f(xf(x)) = 2f(x)$

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.

• What is the domain and codomain? Real numbers? Integers? Natural numbers (with or without zero)? Oct 18, 2020 at 21:23
• In the question, it doesn't explicitly state the domain. I think it is all real numbers though. Oct 18, 2020 at 21:25
• That's unfortunate. The answer could depend quite delicately on which domain you pick. Oct 18, 2020 at 21:26
• @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? Oct 18, 2020 at 22:08

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.