Finding all $f: \mathbb{R} \to \mathbb{R}$ such that $f\bigl(xf(y)+y\bigr)+f\bigl(-f(x)\bigr)=f\bigl(yf(x)-y\bigr)+y $

Question

Find all $f: \mathbb{R} \to \mathbb{R}$ such that $$f\bigl(xf(y)+y\bigr)+f\bigl(-f(x)\bigr)=f\bigl(yf(x)-y\bigr)+y $$ for all $x,y \in \mathbb{R}$.

Help me solving this. My expectation of the answer is $f(x) = x+1$.

My try: $$ P(x, y): f\bigl(xf(y)+y\bigr)+f\bigl(-f(x)\bigr)=f\bigl(yf(x)-y\bigr)+y \text. \\ P(x, 0): f\bigl(xf(0)\bigr)+f\bigl(-f(x)\bigr) = f(0) \text. \\ P(0, y): f(y)+f\bigl(-f(0)\bigr) = f\Bigl(y\bigl(f(0)-1\bigr)\Bigr)+y \text. \\ P(0, 0): f(0)+f\bigl(-f(0)\bigr) = f(0) \implies f\bigl(-f(0)\bigr)=0 \text. \\ \text {Assume) } f(a)=1 \text. \\ P(x, a): f(x+a) + f\bigl(-f(x)\bigr)=f\bigl(af(x)-a\bigr)+a \text. \\ x = 0 \text; \ 1 = f\Bigl(a\bigl(f(0)-1\bigr)\Bigr)+a \text. \\ x = a \text; \ f(2a)+f(-1) = f(0)+a \text. \\ $$

Answer 1

$$f(xf(y)+y)+f(−f(x))=f(yf(x)−y)+y$$ Define $f(0)=c$ for some $c\in \mathbb{R}$.

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• $(x,y)\equiv (0,0)$ $$f(0)+f(-f(0))=f(0)\implies f(-c)=0$$
• $(x,y)\equiv (-c,-c)$ $$f(-c)+c=f(c)-c\implies f(c)=2c$$
• $(x,y)\equiv (0,c)$ $$f(c)+f(-c)=f(c^2-c)+c\implies c=f(c^2-c)$$
• $(x,y)\equiv(0,-c)$ $$f(-c)+f(-c)=f(-c^2+c)-c\implies c=f(-(c^2-c))$$
• $(x,y)\equiv(-c,c^2-c)$ $$f(-c)+c=f(-c^2+c)+c^2-c \implies c^2=c\implies c\in\{0,1\}$$

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If $c=f(0)=0$,

• $(x,y)\equiv(x,0)$ $$f(-f(x))=0\;, \;\;\; \forall x\in \mathbb{R}$$
• $(x,y)\equiv(-x/f(x),x)$ $$-f(xf(-x/f(x))-x)=x \\ f(-f(xf(-x/f(x))-x))=f(x) \\ \;\;\;\therefore\; f(x)=0\;, \;\;\; \forall x\in \mathbb{R}\;\;$$ But $f(x)=0$ for real $x$ is not a solution of the original F.E.

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If $c=f(0)=1$,

• $(x,y)\equiv(0,x)$ $$f(x)=f(cx-x)+x\implies f(x)=x+1.$$ Plugging this back in the original F.E, we verify that $\boxed{f(x)=x+1}\; \forall x\in \mathbb{R}$ is the only solution.

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Answer 2

Throughout my answer, the expression $(n)(x_1, x_2, \ldots, x_n)$ will mean the substitution in the equation with the number $n$.

We want to find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that for every $x, y \in \mathbb{R}$ following holds $$ f(xf(y)+y)+f(−f(x))=f(yf(x)−y)+y \label{eq:main} \tag*{$(1)(x, y)$}$$

Substitutions $(1)(0, 0),\ (1)(0, y),\ (1)(-y, y),\ (1)(y, 0)$ lead us to \begin{gather} f(-f(0)) = 0 \tag*{$(2)$} \\[0.35em] f(y) = f\big(y(f(0) - 1)\big) + y && \tag*{$(3)(y)$} \\[0.35em] f\big(-y(f(y) - 1)\big)+f(-f(-y)) = f\big(y(f(-y)-1)\big) + y \tag*{$(4)(y)$} \\[0.35em] f(f(0)y) + f(-f(y)) = f(0) \tag*{$(5)(y)$} \end{gather}

Adding equalities $(4)(y)$ and $(4)(-y)$, we obtain $$ f(-f(y)) = -f(-f(-y)) \tag*{$(6)(y)$} $$

Case $f(0) = 0$

If we assume that $f(0) = 0$, then $(3)(y)$ and $(5)(y)$ will simplify to \begin{gather} f(y) = f(-y) + y && \tag*{$(7)(y)$} \\[0.35em] f(-f(y)) = 0 \tag*{$(8)(y)$} \end{gather}

Combining $(7)(f(y))$ and $(8)(y)$ we have that \begin{gather} f(f(y)) = f(y) && \tag*{$(9)(y)$} \end{gather}

Then $(1)(-f(x), f(y))$ is equivalent to \begin{gather} f\big(f(y)(1-f(x))\big) = f(y) && \tag*{$(9)(y)$} \end{gather}

Now we use together $(1)(x, f(y)),\ (7)(f(y)(f(x)-1))$ and $(9)(y)$ \begin{align*} f\big(f(y)(1+x)\big) &= f\big(f(y)(f(x)-1)\big) + f(y) \\[0.35em] &= f\big(f(y)(1-f(x))\big) + f(y)(f(x)-1) + f(y) \\[0.35em] &= f(y) + f(y)f(x) - f(y) + f(y) \\[0.35em] &= f(y)(1+f(x)) \end{align*}

If we take $x = -1$ in last equality and note that $f(x) \not\equiv 0$ we get that $f(-1) = -1$. From $(7)(1)$ follows that \begin{gather} f(1) = 0 \tag*{$(10)(y)$} \end{gather}

$(1)(1, -1)$ implies $f(-2) = -1$, then $(7)(2)$ implies $f(2) = 1$ and $(9)(2)$ leads to contradictions $f(1)=1$, so there are no functions with $f(0) = 0$.

Case $f(0) \neq 0$

From $(5)(y)$ and $(6)(y)$ we have that: \begin{gather} f(-f(0)y) + f(-f(-y)) = f(0) \\[0.35em] f(-f(0)y) - f(-f(y)) = f(0) \\[0.35em] f(-f(0)y) + f(f(0)y) = 2f(0) \\[0.35em] f(-y) + f(y) = 2f(0) \tag*{$(11)(y)$} \end{gather}

Denote set of zeros of $f$ as $\mathrm{U}$. $\mathrm{U}$ nonempty due $(2)$. For any $u \in \mathrm{U}$ substitution $(1)(u,u)$ together with $(11)(u)$ shows us that \begin{align*} f(0) &= f(-u) + u \\[0.35em] &= 2f(0) + u \\[0.35em] u &= -f(0) \\[0.35em] \mathrm{U} &= \Big\lbrace -f(0) \Big\rbrace \tag*{$(12)$} \end{align*}

We denote by $\mathrm{Z}$ the set $y \in \mathbb{R}$ such that $f(y) = f(0)$. From $(11)(y \in \mathrm{Z})$ we have that \begin{gather} y \in \mathrm{Z} \Rightarrow -y \in \mathrm{Z} \tag*{$(13)(y)$} \end{gather} Now take any $y \in \mathrm{Z}$. Then substitution $(5)(f(0)^{-1}y)$ and $(12)$ leads to \begin{gather} f(-f(f(0)^{-1}y)) = 0 \\[0.35em] -f(f(0)^{-1}y) = -f(0) \\[0.35em] f(f(0)^{-1}y) = f(0) \\[0.35em] y \in \mathrm{Z} \Rightarrow f(0)^{-1}y \in \mathrm{Z} \tag*{$(14)(y)$} \end{gather}

Let $x, y \in \mathrm{Z}$. Then $(1)(f(0)^{-1}x, y)$ with $(3)(y)$ implies \begin{align*} f(x + y) &= f\big(y(f(0)-1)\big) + y \\[0.35em] &= f(y) \\[0.35em] &= f(0) \\[0.35em] x, y \in \mathrm{Z} &\Rightarrow x + y \in \mathrm{Z} \tag*{$(15)(x, y)$} \end{align*}

$(3)(-f(0))$ equivalent to $-f(0)(f(0)-1) \in \mathrm{Z}$. Then by $(13)$ and $(14)$ both $f(0)(f(0)-1)$ and $1-f(0)$ also lies in $\mathrm{Z}$. By $(15)$ we have $(f(0)-1)^2 \in \mathrm{Z}$. If we put $y=f(0)-1$ in $(3)$, we conclude that $f(0) = 1$. Then $(3)(y)$ simplifies to $$ f(y) = 1 + y $$ Function $f(y) = 1 + y$ satisfies equation $(1)$, so this is the only solution.

P. S. English is not my native language. I apologize for any possible mistakes when writing the answer in English.