Prove that, the function $f$ is injective: $f \big(f(x)f(y)\big) + f(x +y) = f(xy).$

Question

I need to learn, by which method, I can prove that the function $f$ is injective. I would like to ask you to explain this problem using more detailed, more understandable, clearer and simpler English words. There's only one way I can understand. For example;

Let $f: \mathbb{R} \to\mathbb{R}$ and $f(x)=x-10$.

We have, $f(y)=y-10$, then $f(x)=f(y)$, we get $f(x)=f(y) \Rightarrow x-10=y-10 \Rightarrow x=y$. So, $f$ is injective.

And here is my problem:

Why can't we apply this method to this problem?

$f: \mathbb{R} \to\mathbb{R}$ and $f(0)≠0$, such that, for all real numbers $x$ and $y$, $$f \big(f(x)f(y)\big) + f(x +y) = f(xy).$$ Prove that, $f$ is injective.

Thank you very much for teaching.

Answer 1

You cannot apply the simple test straightforwardly, mainly because we do not have explicit information about $f$. You may be able to arrive at a contradiction by assuming $f(a)=f(b)$ for some $a\ne b$ and playing around cleverly with the functional equation.

For example, if $f(a)=f(b)$, we find $$f(a^2)-f(2a)=f(ab)-f(a+b)=f(b^2)-f(2b)=f(f(a)f(b)) $$ This gives you further points of $f$ that are somwehat related. This may help, but it is not straightforward to see how this could shortcut the way to injectivity (as in, as was said in the comments, being able to see that $f$ is injective much easier than determining the complete set of solutions of the functional equation).

Another approach: By letting $x=y=0$, we find $$f(f(0)^2)=0 $$ and from $y_0:=f(0)\ne 0$ can conclude that $f$ has a non-zero root, i.e., there exists $x_0(=f(0)^2)$ such that $f(x_0)=0$. Then $$f(f(x_0)f(y)) +f(x_0+y)=f(x_0y)$$ for all $y$, i.e., $$f(x_0y)=f(x_0+y)+y_0\ne f(x_0+y). $$ Here, we arrive at a contradiction if we exhibit $y$ with $x_0y=x_0+y$. This is possibly (namely, $y=\frac{x_0}{x_0-1}$) as long as $x_0\ne 1$. So we conclude $$ x_0=1$$ before getting any closer to the goal of showing that $f$ is injective (i.e., we seem to only exclude that $|a-b|=1$). We can use this (letting $x=y=1$) to show $$ f(2)=-f(0)$$ and other tricky things (in fact, ultimately that $f(x)=\pm(x-1)$), but I still do not see any way to show injectivity much simpler than solving the functional equation completely.

Answer 2

We can prove the solution is injective and from this fact, solve the equation completely, getting the solution $f(x) = 1-x,\forall x\in\mathbf{R}$ or $f(x)= x-1, \forall x$. I'll start from the scratch.

Let $\gamma := f(0)$. Then, from the FE, we have $f(\gamma^2) =0$. Our first claim is that $$f(y) = 0 \Rightarrow y = 1.$$ Assume to the contrary that $y\neq 1$, but $f(y)=0$. Then by setting $x=\frac{y}{y-1}$, we should have $$0 = f(f(x)f(y)) = f(0) =\gamma$$, contradicting $\gamma \neq 0$.
From the above argument we get that $\gamma^2 = 1$ and for all $y \neq 1$, it holds that $$f(y)f(\frac{y}{y-1})=1 \cdots(*)$$ So far we have $\gamma =\pm1$, and we will see if $\gamma =1$, then $f(x) = 1-x$ and otherwise, $f(x) = x-1$.
Assume that $\gamma = -1$. From the functional equation, we have $$f(x+1) = f(x) + 1,\quad\forall x\in \mathbf{R},$$ $$f(x+n) = f(x) + n,\quad\forall x\in \mathbf{R}, \;n\in \mathbf{Z},$$ $$f(n) = n-1, \quad \forall n\in \mathbf{Z}.$$ Using this fact, after a change of variable $y-1 \mapsto y$, we get $$\left(f(y)+1\right)\left(f(\frac{1}{y})+1\right)=1,\quad \forall y\neq 0 \cdots(**)$$ Notice this implies $f^{-1}(-1) = \{0\}$. Our next claim is that $$f(\alpha) =f(\beta) \Rightarrow f(q\alpha) = f(q\beta), f(\alpha+q) = f(\beta+q),\quad\forall q\in \mathbf{Q}.$$ We may assume $\alpha \neq 0, \beta \neq 0$. To prove this, note that $f(\alpha) =f(\beta)$ implies $f(n\alpha)=f(n\beta)$ for all $n\in\mathbf{Z}$ by the FE. Then, by $(**)$, for non-zero $n$, we have $f(\frac{1}{n\alpha})=f(\frac{1}{n\beta}).$ Hence, it holds $f(\frac{m}{n\alpha})=f(\frac{m}{n\beta})$ and again by $(**)$, $$f(\frac{n}{m}\alpha)=f(\frac{n}{m}\alpha), \quad\forall n\in\mathbf{Z}, m\in \mathbf{Z} \setminus \{0\}.$$ This prove the fisrt half. For the second half, note that by the FE, $$f(\alpha x) -f(\beta x) = f(\alpha +x )-f(\beta +x) \quad\forall x\in \mathbf{R}.$$ So, in particular, $f(\alpha +q ) -f(\beta +q) =0,\forall q\in \mathbf{Q}$.
Our (almost) final claim is that $f$ is injective. Assume to the contrary that $f(\alpha) = f(\beta) $ for $\alpha\neq\beta$. We proceed by finding the solution $(x,y,q)\in \mathbf{R}\times\mathbf{R}\times\mathbf{Q}$ satisfying: $$xy = \alpha +q,\; x+y = \beta +1 +q.$$ Once this is done, then by the FE, $$f(f(x)f(y)) = f(xy) -f(x+y) = f(\alpha +q)-f(\beta +1 +q) = f(\alpha +q)-f(\beta +q)-1 = -1, $$ getting $f(x)f(y) = 0$. Without loss of generality, $f(x) = 0$ and hence $x=1$. So we end up with $y -q=\alpha = \beta ,$ contradiction!
We notice that the real solution $(x,y)$ exists if and only if its quadratic discriminant $$(x-y)^2 = (x+y)^2 -4xy = (\beta+q+1)^2 -4(\alpha+q)= q^2 +2(\beta-1)q +(\beta+1)^2 -4\alpha \geq 0.$$ Of course we can choose $q_0\in\mathbf{Q}$ such that $$q_0^2 +2(\beta-1)q_0 +(\beta+1)^2 -4\alpha > 0.$$
Hence, the injectivity of $f$ is established. Finally, by letting $y=0$ in the FE, it holds $$f(-f(x)) =-f(x) -1,$$ which is saying that $f(t) = t-1$ if $(-t)$ belongs to the range of $f$. Let $f(u) = u'$. By the above FE, $-(1+u')$ also belongs to the range of $f$. Thus, $f(1+u') = u'$. By injectivity, we get $u= 1+u'$, hence $f(u) = u-1$ for all $u\in \mathbf{R}$, as desired.

Note: exactly the same argument shows that $f(u) = 1-u$ in the case $\gamma =1.$