Functional Equation $f(x+y)=f(x)+f(y)+f(x)f(y)$

Question

I need to find all the continuous functions from $\mathbb R\rightarrow \mathbb R$ such that $f(x+y)=f(x)+f(y)+f(x)f(y)$. I know, what I assume to be, the general way to attempt these problems, but I got stuck and need a bit of help. Here is what I have so far:

Try out some cases:

Let $y=0$: $$ \begin{align} f(x)&=f(x)+f(0)+f(x)f(0) \\ 0&=f(0)+f(x)f(0) \\0 & = f(0)[1+f(x)] \end{align}$$ Observe that either $f(0)=0$ or $f(x)=-1$. So this gives me one solution, but I am having trouble finding the other solution(s). Somebody suggested to me that $f(x)=0$ is also a solution but I can't find a way to prove what they said is true. Can anyone please, without giving away the answer, give me a teeny hint? I really want to figure this out as much as I can. I've tried the case when $y=-x$ and $x=y$ but I don't feel like those cases help me towards the solution.

Thanks in advance

Answer 1

We have $$f(x+y) + 1 = (f(x)+1)(f(y)+1)$$ If we let $f(z) + 1$ as $g(z)$, we then have $$g(x+y) = g(x) g(y)$$ Now this is the good old Cauchy functional equation, which you should be able to solve.

Below are some relevant links:

Is there a name for function with the exponential property $f(x+y)=f(x) \cdot f(y)$?

Classifying Functions of the form $f(x+y)=f(x)f(y)$

If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t

If $f\colon \mathbb{R} \to \mathbb{R}$ is such that $f (x + y) = f (x) f (y)$ and continuous at $0$, then continuous everywhere

continuous functions on $\mathbb R$ such that $g(x+y)=g(x)g(y)$

What can we say about functions satisfying $f(a + b) = f(a)f(b) $ for all $a,b\in \mathbb{R}$?

https://math.stackexchange.com/questions/171806/

Answer 2

Set $g(x)=f(x)+1$. Then $g(x+y)=g(x)g(y)\Rightarrow g(x)=g^2(\frac{x}{2}).$

Now we will prove that $g(x+y)=g(x)g(y)$ implies that $g(x)=c^x$.

First of all, $g(x)=g^2(\frac{x}{2})$ implies that $g(0)=0$ , or $g(0)=1$.

• If $g(0)=0$ then $f(0)=-1$ and using the given relationship for $y=0, x\in \mathbb R$, we get:

$$f(x+0)=f(x)+f(0)+f(x)f(0)\Rightarrow f(x)=f(x)-1-f(x)\Rightarrow f(x)=-1,\forall x\in \mathbb R$$

• Now, If $g(0)=1$ , we have to follow some steps in order to reach the desired conclusion:
  2. For $n\in\mathbb N$ it is easy to see that $$g(n)=g(1+\dots+1)=g(1)\times\dots\times g(1)=[g(1)]^n$$
  3. Also note that $$g(1+(-1))=g(1)g(-1)\Rightarrow g(0)=g(1)g(-1)\Rightarrow g(-1)=\dfrac{1}{g(1)}=[g(1)]^{-1}$$
  4. Now for a negative integer $-m$ we have : $$g(-m)=g(-1\dots-1)=g(-1)\times\dots\times g(-1)=[g(1)]^{-m}$$
  5. Next, prove it for $\dfrac{1}{n},n\in \mathbb N$. You have already proved that $g(2x)=g^2(x).$ If you extend this, you can easily get $g(nx)=g^n(x)$. So for $x=\frac{1}{n}$ $$g^n\left(\frac{1}{n}\right)=g(1)\Rightarrow g\left(\frac{1}{n}\right)=[g(1)]^{1/n}$$
  6. Now the big step: Prove it for rationals $\dfrac{m}{n}$ $$g\left(\frac{m}{n}\right)=g\left(\frac{1}{n}+\dots+\frac{1}{n}\right)=g\left(\frac{1}{n}\right)\times\dots\times g\left(\frac{1}{n}\right)=g\left(\frac{1}{n}\right)^m=[[g(1)]^{1/n}]^m=[g(1)]^{m/n}$$
  7. Finally, if you take a $x\in\mathbb R$, then there exists a sequence of rationals $\{q_n\}_{n=1}^\infty$ such that $q_n\xrightarrow{n\rightarrow \infty}x$ and: $$g(x)=\lim\limits_{n\rightarrow\infty}g(q_n)=\lim\limits_{n\rightarrow\infty}[g(1)]^{q_n}=[g(1)]^x$$ where we used the continuity of $g$ as well as the continuity of the exponential function.

So we reached the desired result : $$g(x)=[g(1)]^x=c^x$$ $$\Rightarrow f(x)=g(x)-1=c^x-1$$ (or $f(x)=-1$ do not forget!)