The functional equation $f(x) + f\bigl(x+f(y)\bigr) = y + f\Bigl(f(x) + f\bigl(f(y)\bigr)\Bigr)$

Question

I'm having trouble with this functional equation:

Find all functions $f:\mathbb R \to \mathbb R$ for which the following is valid for all $x,y\in \mathbb R$: $$f(x) + f\bigl(x+f(y)\bigr) = y + f\Bigl(f(x) + f\bigl(f(y)\bigr)\Bigr)$$

I haven't been able to pinpoint any solution, nor to prove surjectivity or injectivity, So no progress. I tried $x=y=0$, $x=0$, $y=0$, $x=-f(y)$, $y=f(x)$, and even with some assumptions on bijectivity, but nothing helped.

I don't know the source, but it seems like a pretty difficult problem, maybe even impossible to solve.

Answer 1

I arrive at a proof that no such funtion exist (assuming the statement is correct).

---

Replace $x$ with $f(x)$ in the original equation: $$f(f(x)) - y = f(f(f(x)) + f(f(y))) - f(f(x) + f(y)).\tag{1}$$

The right hand side is symmetric in $x$ and $y$, so we get $$f(f(x)) - y = f(f(y)) - x,$$ or $f(f(x)) + x$ is constant for all $x$. Write $c$ for this constant. Thus $$f(f(x)) = c - x.\tag{2}$$

(In particular, this shows that $f$ is both injective and surjective.)

---

It follows that $$f(c - x) = f(f(f(x))) = c - f(x),\tag{3}$$ which gives $f(\frac c 2) = \frac c 2$.

Using (2), the equation (1) becomes $$x + y + f(2c - x - y) = c + f(f(x) + f(y)).\tag{4}$$

Put $x = \frac c 2$ and $y = c$: $$c = f(\frac c 2 + f(c)).$$

Applying $f$ again and using (2), we get $f(c) = \frac c 4$, and then $f(0) = \frac {3c}4 $ by (3).

---

In (4), replace $x$ with $0$ and $y$ with $x + y$: $$x + y + f(2c - x - y) = c + f(\frac {3c}4 + f(x + y)).$$

We compare this with (4). From injectivity of $f$, we get

$$f(x + y) + \frac {3c} 4 = f(x) + f(y).\tag{5}$$

---

If we define $g(x) = f(x) - \frac{3c} 4$, then we have $g(x + y) = g(x) + g(y)$. This implies that $g(rc) = rg(c) = -\frac{rc}2$ for any rational number $r$.

Therefore $f(rc) = (\frac 3 4 - \frac r 2)c$ for any rational number $r$.

Applying $f$ again gives $f(f(rc)) = (\frac 3 8 + \frac r 4) c$. Together with (2), we conclude that $c = 0$.

---

Thus (5) becomes $f(x + y) = f(x) + f(y)$ and (2) becomes $f(f(x)) = -x$.

The original equation can then be rewritten as: $$f(x) + f(x) + f(f(y)) = y + f(f(x)) + f(f(f(y))),$$ or: $$2f(x) + f(y) = 2y - x.\tag{6}$$

Apply $f$ again: $$-2x - y = 2f(y) - f(x).\tag{7}$$

We take a linear combination of (6) and (7) and get: $$5f(x) = 5y.$$

This obviously cannot hold for all $x, y \in \Bbb R$.

Answer 2

I do not have a solution, but I want to share with you my ideas so far. To make things every to read, we write $f^k$ for $f\circ \dots\circ f$ $k$-times.

Applying $f$ to the left hand side, we obtain the expression $$ f\big{(} \underbrace{f(x)}_{=:x'} + f(\underbrace{x + f(y)}_{=:y'})\big{)} = f(x' + f(y')) = y' + f\left(f(x') + f^2(y')\right) - f(x'), $$ i.e. $$ f(f(x) + f(x+f(y))) = x + f(y) + f(f^2(x) + f^2(x+f(y))) - f^2(x). $$ But we can also write $$ f\big{(} f(\underbrace{x}_{=:y'}) + \underbrace{f(x + f(y))}_{=:y'}\big{)} = x + f(f^2(x + f(y)) + f^2(x)) - f^2(x+f(y)). $$ Comparing the two quantities gives $$ f(y) = f^2(x) - f^2(x+f(y)). $$ For $y = f(x)$, this implies $$ f^2(x + f^2(x)) = 0 $$ for all $x\in\Bbb R$. Also, if $f$ was surjective, we could find $y$ such that $f(y) = -x$ giving $$ f^2(x) = f^2(0) - x. $$ Combining this with the previous yields $$f^4(0) = 0.$$

(Edit: more precisely, we get $$ f^4(x) = f^2(0)- f^2(x) = x. $$ In particular, subjectivity implies bijectivity and gives the inverse function explicitly.)

Naturally, the last bit completely relies on surjectivity which might not be given.

That are my thoughts so far.