Is there any functional equation $f(ab+cd)= f(a)+f(b)+f(c)+f(d)$?

I am looking for a real, continuous function that satisfies the functional equation $$f(ab+cd)= f(a)+f(b)+f(c)+f(d)$$ where $$a,b,c,d$$ are real. This is equivalent to a function satisfying these two functional equations: $$f(x+y)=f(x)+f(y)$$ $$f(xy)=f(x)+f(y)$$ or Cauchy's functional equation and the logarithmic functional equation, respectively. I have the suspicion that the only the function $$f(x) = 0$$ satisfies this, but I am very confused as to how a solution may be found analytically. Thank you in advance.

EDIT: I saw in the comments that if f(x) is defined at zero the answer is trivial. If the domain were restricted to not include zero or even to not include all nonpositive numbers, would the answer change? If so, how? What if the function were discontinuous? Thank you again.

• If $f$ is defined at $0$: Taking $y=0$ we see that $f(0)=f(x)+f(0)\implies f(x)=0$. – lulu May 3 at 19:47
• @lulu if the domain were to be restricted to not include zero, would this answer change? – trytryagain May 3 at 19:50
• Not sure. I doubt it, these equations are very strong. But I don't see a proof off the top of my head. The second equation (plus continuity) pretty much tells you the function is a log, and logs don't satisfy the first equation. – lulu May 3 at 19:51
• If $f$ is not defined at $0$, then it is not obvious to me why the given equation is equivalent to the two other equations. In fact, it's not clear to me even when $f$ is defined at $0$. – Quang Hoang May 3 at 20:07

If $$f$$ has the positive numbers as its domain, it is zero regardless of continuity.

Note first that $$f$$ satisfies the relation $$f(x+y)=f(xy)$$ for all $$x,y$$ in its domain. Now, fix any two positive numbers $$u$$ and $$v$$. If $$u^2> 4v$$, then $$u=x+y$$ and $$v=xy$$ for $$x,y=\frac{u \pm \sqrt{u^2-4v}}{2}$$, and these $$x,y$$ are positive. So $$f(u)=f(v)$$.

Similarly, if $$v^2 > 4u$$, then $$f(u)=f(v)$$.

But if $$u^2 \le 4v$$ and $$v^2\le4u$$, then $$u^2\le4(2\sqrt{u})$$, which implies that $$u\le 4$$. So $$4^2 \ge 4u$$, which means that $$f(u)=f(4)$$. Since $$u$$ and $$v$$ are symmetric, we also have $$f(v)=f(4)$$. So $$f(u)=f(v)$$.

Since $$u$$ and $$v$$ are arbitrary, $$f$$ is constant, which means that it is zero.

Let $$a=b=1, c=x, d=0$$. Then $$f(1)=f(ab+cd)=f(1)+f(1)+f(x)+f(0)$$. Thus $$f(x)=C:=-f(1)-f(0)$$ for all $$x$$,i.e., $$f=C$$ is constant. Then the original equation gives $$C=4C$$, i.e., $$C=0$$. So $$f=0$$ is the only solution, even without assuming continuity.

• That was clever! +1 – Dr. Mathva May 11 at 9:13

Observe that all continuos solutions to Cauchy's functional equation $$f:\Bbb R\to \Bbb R$$ $$f(x+y)=f(x)+f(y)$$ are of the form $$f(x):=ax\quad \text{for some a\in\Bbb Z}$$

Can you end it now?

Plugging this into the second functional equation $$f(xy)=axy=f(x)+f(y)=a(x+y)\iff axy=a(x+y)$$ This holds $$\forall x,y\in\Bbb R$$ if and only if $$a=0$$ and so $$f(x)=0$$ is the only continous solution.