Notice that if $f(x)=c$ then $c= (c+\frac 12)^2$, thus $c^2 + \frac 14 = 0$, absurd. So we may assume $f$ is not constant. If $y = -1$ we get $f(f(-1)) = \left ( f(-1)+\frac 12 \right ) \left ( f(y)+\frac 12 \right ) $. If $f(-1) + \frac 12 \neq 0$, that gives $f$ constant, which we know cannot happen. Then $f(-1) = -\frac 12$. Suppose $f(a) = -\frac 12$. For $y = a$ we get $f(x(a+1) +f(a)) = 0$. If $a \neq -1$, $x(a+1)+f(a)$ can take any value, so $f$ would be constant $0$, absurd. Then $f(a) = -\frac 12 \iff a=-1$. Now, if $x = -1$ we get $f(-1 -y +f(y)) = 0$. If $y = -1 -a + f(a)$, we get $f(-1 +1 +a - f(a) +f(-1 -a +f(a)))=0$, thus $f(a-f(a))=0$ for all $a$. Now, for $y \neq -1$, if $x = \frac{y-2f(y)}{1+y}$, we get $f(y-f(y)) = \left(f(x)+\frac{1}{2}\right)\left(f(y)+\frac{1}{2}\right)$. As $f(y-f(y)) = 0$, we have $f \left( \frac{y-2f(y)}{1+y}\right) = -\frac 12$ or $f(y) = -\frac 12$ for all $y \neq -1$. But as $f(-1) = -\frac 12$, the second case would imply $f$ constant. Then $f\left( \frac{y-2f(y)}{1+y}\right) = -\frac 12$. But that implies $\frac{y-2f(y)}{1+y} = -1$ for $y \neq -1$. It follows that $f(y) = y + \frac 12$ for all $y$, and it is easy to check that it satisfies the conditions.

Edit: fixed a typo

ocha wrote: [We choose this sequence so that $=...=f\left(g_a\left(\lim_{n\to \infty} x_n\right)\right) $ I'm afraid you need continuity to write this. And we dont have ...

pco wrote: ocha wrote: [We choose this sequence so that $=...=f\left(g_a\left(\lim_{n\to \infty} x_n\right)\right) $ I'm afraid you need continuity to write this. And we dont have ... i think we only need continuity of $g_a(x)$ and $g_b(x)$, which we do have

ocha wrote: pco wrote: ocha wrote: [We choose this sequence so that $=...=f\left(g_a\left(\lim_{n\to \infty} x_n\right)\right) $ I'm afraid you need continuity to write this. And we dont have ... i think we only need continuity of $g_a(x)$ and $g_b(x)$, which we do have Let $u_n=g_a(x_n)$ and let $c=f(g_a(x_0))$ Continuity of $g_a$ (that we indeed have) allows you to write $\lim g_a(x_n) = g_a(\lim x_n))$ (when such limits exist) So $\lim u_n=g_a(\lim x_n)$ and this is OK for me. But from $f(u_n)=c$ $\forall n$, you conclude then $f(g_a(\lim x_n))=f(\lim u_n)=c[=\lim(f(u_n))]$ And writing $f(\lim u_n)=\lim(f(u_n))$ demands continuity of $f$.

you are right, sorry

Very nice solution mithril, just wondering what prompted you to use the substitution $ x =\frac{y-2f(y)}{1+y} $?

manifestdestiny wrote: Very nice solution mithril, just wondering what prompted you to use the substitution $ x =\frac{y-2f(y)}{1+y} $? just try to solve the equation x+xy+f(y)=y-f(y) with respect to x. Since we need to find f(y-f(y)).

uglysolutions wrote: Find all functions $f: \mathbb R \rightarrow \mathbb R$ such that $f(x+xy+f(y)) = \left(f(x)+\frac{1}{2}\right) \left(f(y)+\frac{1}{2}\right)$ holds for all real numbers $x,y$. $x = 0 \Rightarrow f\left( {f\left( y \right)} \right) = \left( {f\left( 0 \right) + \frac{1}{2}} \right)\left( {f\left( y \right) + \frac{1}{2}} \right),\forall y \in R$ $y = - 1 \Rightarrow f\left( {f\left( { - 1} \right)} \right) = \left( {f\left( x \right) + \frac{1}{2}} \right)\left( {f\left( { - 1} \right) + \frac{1}{2}} \right) = \left( {f\left( 0 \right) + \frac{1}{2}} \right)\left( {f\left( { - 1} \right) + \frac{1}{2}} \right)$ We consider the following two cases a) $f\left( { - 1} \right) \ne - \frac{1}{2} \Rightarrow f\left( x \right) = f\left( 0 \right),\forall x \in R$, Try again we see conflict b) $f\left( { - 1} \right) = - \frac{1}{2}$ If there exists $a$ such that $f\left( a \right) = - \frac{1}{2},a \ne - 1$. $y = a \Rightarrow f\left( {x + ax + f\left( a \right)} \right) = \left( {f\left( x \right) + \frac{1}{2}} \right)\left( {f\left( a \right) + \frac{1}{2}} \right) = 0 \Rightarrow f\left( x \right) = 0,\forall x \in R$ Try not satisfied $ \Rightarrow f\left( a \right) = - \frac{1}{2} \Leftrightarrow a = - 1$ If there exists $a$ such that $f\left( a \right) = 0,a \ne - \frac{1}{2}$ We ha ve $f\left( {f\left( { - 1} \right)} \right) = \left( {f\left( x \right) + \frac{1}{2}} \right)\left( {f\left( { - 1} \right) + \frac{1}{2}} \right) = 0 \Rightarrow f\left( { - \frac{1}{2}} \right) = 0$ $y = - \frac{1}{2} \Rightarrow f\left( {\frac{x}{2}} \right) = \frac{1}{2}\left( {f\left( x \right) + \frac{1}{2}} \right),\forall x \in R$ $y = a \Rightarrow f\left( {x + ax} \right) = \frac{1}{2}\left( {f\left( x \right) + \frac{1}{2}} \right),\forall x \in R$ Hence $f\left( {x + ax} \right) = f\left( {\frac{x}{2}} \right),\forall x \in R\,\,\,\left( 1 \right).$ $x = \frac{{ - 1}}{{a + 1}} \Rightarrow f\left( { - 1} \right) = f\left( { - \frac{1}{{2\left( {a + 1} \right)}}} \right) = - \frac{1}{2} \Leftrightarrow - \frac{1}{{2\left( {a + 1} \right)}} = - 1 \Leftrightarrow a = - \frac{1}{2}$ conflict. Hence $f\left( a \right) = 0 \Leftrightarrow a = - \frac{1}{2}$ $x = - 1 \Rightarrow f\left( { - 1 - y + f\left( y \right)} \right) = \left( {f\left( { - 1} \right) + \frac{1}{2}} \right)\left( {f\left( y \right) + \frac{1}{2}} \right) = 0 \Rightarrow f\left( { - 1 - y + f\left( y \right)} \right) = 0$ $ \Rightarrow - 1 - y + f\left( y \right) = - \frac{1}{2},\forall y \in R \Rightarrow f\left( y \right) = y + \frac{1}{2},\forall y \in R$ Try to be satisfied. So $f\left( y \right) = y + \frac{1}{2},\forall y \in R$

See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2154206&sid=fe7e79ac1f19199836adc92f1e310dfb#p2154206

$f(f(-1)) = (f(x)+\frac{1}{2})(f(-1)+\frac{1}{2}) \implies f(-1) = \frac{-1}{2} , f(\frac{-1}{2}) = 0$ Suppose that $f(a) = \frac{-1}{2}$ for some $a$ , plug in $y = a$ then $a = -1 $ Suppose that $f(a) = 0$ for some $a$ $P(x,a):f(x(a+1)) = \frac{1}{2}(f(x)+\frac{1}{2})$ $P(-1,a) : f(-a-1) = 0$ $P(x,-a-1) : f(-ax) = \frac{1}{2}f(x)+\frac{1}{2}) = f(x(a+1))$ , then , $f(-ax) = f(x(a+1)) $. Plug in $x = \frac{-1}{a+1}$ we have $\frac{a}{a+1} = -1$ so $a = \frac{1}{2}$ . Then , $f(x) = 0$ if and only if $a = \frac{-1}{2}$ $P(-1,y) : f(f(y)-y-1) = 0 \implies f(y) = y+\frac{1}{2}$