Suppose that $a$ is non-zero number such that $f(a)=0$ $(a,a)$ $\Rightarrow $ $f(2a)=0$ $\Rightarrow$ $f(4a)=0$ $\Rightarrow $ .... Use surjectivity we have $(a,y)$ and $(2a,y)$ $\Rightarrow$ $f(x+a)=f(x)$ for all $x$ . (1) So $f(0)=f(a)=0$ . $(x,0)$ and $(0,x)$ $\Rightarrow $ $f(x+f(x)) = f(2x)=f(2f(x))$ (2) $(x , f^{-1} (-f(x)) )$ $\Rightarrow $ $-f(x)=f(-x)$ (3) $2f(y)$ covers all the $\mathbb R$ so (2) and main equation $\Rightarrow $ $f(x+f(x) +y)=f(2x)+f(y)$ (4) Let $P(x,y)$ be the assertion of $ f(x+f(x)+y)=f(2x)+f(y) $ $P(x,-x)$ and (3) gives $f(f(x))+f(x)=f(2x)=f(2f(x))$ (#) So $f(x)+x$ covers $\mathbb R$ Now (2) tell us $P(x,y)$ can change to $f(x+y)=f(x)+f(y)$ so $f(2x)=2f(x)$ (#) , surjectivity $ \Rightarrow $ $f(x)+x=f(2x)$ Hence $f(x)=x$ is unique solution.

I`ve edited above post .

mahanmath wrote: ... It means from $(5)$ we can make its period as small as we want which is contradiction with surjectivity. Hello, I've not checked the other parts (neither the second post) but I just want to say that this sentence is likely wrong. I dont think there is a contradiction between "periodic with period as small we want" and "surjectivity" : Look at the equivalence relation $x-y\in\mathbb Q$ and let $c(x)$ a choice function associating to each real a representant (unique per class) of its class. $c(\mathbb R)$ is likely equinumerous to $\mathbb R$ (at least it's not countable) and so $\exists$ a bijection $h(x)$ from $c(\mathbb R)\to\mathbb R$ Then $f(x)=h(c(x))$ is a surjection from $\mathbb R\to\mathbb R$ such that $f(x+q)=f(x)$ $\forall x\in\mathbb R$, $\forall q\in\mathbb Q$

Thank you pco for reading my post . Actually I wrote second solution because I had doubt in this part . mahanmath wrote: ... It means from $(5)$ we can make its period as small as we want which is contradiction with surjectivity. Is the second solution correct ?

Because f is surjective so there existed a which $f(a)=0$ +,Let x=y=a we have $f(2a)=0,f(2^na)=0...$ +,Let x=a,y=y we have $f(a+2f(y))=f(2y)$ and x=2a,y=y we have $f(2a+2f(y))=f(2y)$ so $f(a+2f(y))=f(2a+2f(y))$ and then we have $f(x)=f(x+a)$ and $f(x)=f(x-a)$ (1) +,for every $x\in R$ there existed $y_0$ which $f(y_0)=\frac{x-f(x)}{2}$ and let $x=x,y=y_0$ we have $f(2y_0)=0$ because $0=f(2y_0)=f(a+2f(y_0))=f(x-f(x))$ so $f(x-f(x))=0$ for every $x\in R$ do the same thing with (1) we have $f(x)=f(x-(x-f(x)))=f(f(x))$ because f is surjective so $f(x)=x $ for every $x\in R$

let $P(x,y)$: $f(x+f(x)+2f(y))=f(2x)+f(2y)$, since $f$ is surjective, there exist a real number $d$ so that $f(d)=0$, put $c=f(0)$. $P(x,d)$ : $f(x+f(x))=f(2x)+f(2d)$. $P(d,d)$ : $f(2d)=0$ and then $f(x+f(x))=f(2x)$ and so $f(c)=c$ $P(0,x)$ : $f(c+2f(x))= c + f(2x)$, so whenever $f(x)=f(y)$ we will have $f(2x)=f(2y)$. $P(0,0)$ : $f(3c)=2c$ so $2c=f(3c)=f(0+f(0)+2f(c))=c+f(2c)$ and then $f(2c)=c=f(c)$ this implies from the previous result, that $f(4c)=f(2c)=c$. But $P(x,x)$ : $f(x+3f(x))=2f(2x)$ and then $f(4c)=2f(2c)=2c$. We deduce then that $c=0$. And thus $f(2f(x))=f(2x)=f(x+f(x))$. Considering $y$ the real number for which $f(y)=-\frac{1}{2}(x+f(x))$ we get that $f(-x-f(x))=-f(2x)$. From another hand we have: $P(x+f(x),y)$ : $f(2x)=f(4x)+f(-x-f(x))$ and so $f(2x)=2f(x)$ for all real $x$, it follows directly that $f(x)=x$ for all real $x$.

vladimir92 wrote: ... Let $a$ and $b$ be any real numbers, so there exist numbers $x$ and $y$ so that $a=f(2x)$ and $b=f(y)$ hence $a+f(2b)=f(2x)+f(2f(y))=f(2x)+f(2y)=f(x+f(x)+2f(y))=f(a+2b)$, ... Why $f\left(x+f(x)+2f(y)\right)=f(a+2b)$ ?

Potla wrote: Find all surjective functions $f: \mathbb R \to \mathbb R$ such that for every $x,y\in \mathbb R,$ we have \[f(x+f(x)+2f(y))=f(2x)+f(2y).\] A solution: Let $P(x,y)$ be the assertion $f\left(x+f(x)+2f(y)\right)=f(2x)+f(2y)$. There exist a real number $\alpha$ such that $f(\alpha)=0$. $P(\alpha,\alpha)\implies f(2\alpha)=0$ $P(\alpha,2\alpha)\implies f(4\alpha)=0$ $P(x,\alpha)\implies f\left(x+f(x)\right)=f(2x)$ There exist a real number $\beta$ such that $f(\beta)=-\frac{\alpha}{2}$. $P(\alpha,\beta)\implies f(0)=f(2\beta)$ $P(2\alpha,\beta)\implies f(2\beta)=0$ Therefore $f(0)=0$. $P(0,x)\implies f\left(2f(x)\right)=f(2x)$ Therefore if $a,b$ be real numbers such that $f(a)=f(b)$, then $f(2a)=f(2b)$. Hence $f\left(2x+2f(x)\right)=f(4x)$. For each real number $u$, there exist a real number $v$ such that $f(v)=\frac{u}{2}$. $P(x,v)\implies f\left(x+f(x)+u\right)=f(2x)+f(2v)$ Therefore $f\left(x+f(x)+u\right)=f(2x)+f\left(2f(v)\right)=f(2x)+f(u)$. Let $Q(x,y)$ be the assertion $f\left(x+f(x)+y\right)=f(2x)+f(y)$. $Q(x+f(x),y)\implies f\left(x+f(x)+f\left(x+f(x)\right)+y\right)=f\left(2x+2f(x)\right)+f(y)$, therefore $f\left(x+f(x)+f(2x)+y\right)=f(4x)+f(y)$. $Q(x,f(2x)+y)\implies f(x+f(x)+f(2x)+y)=f(2x)+f(f(2x)+y)$, therefore $f(4x)+f(y)=f(2x)+f\left(f(2x)+y\right)$. $Q(2x,y-2x)\implies f\left(f(2x)+y\right)=f(4x)+f(y-2x)$ Therefore $f(4x)+f(y)=f(2x)+f(4x)+f(y-2x)$, hence $f(y)=f(y-2x)+f(2x)$, substitution $x\rightarrow\frac{x}{2}$, $y\rightarrow2x$ we have $f(2x)=2f(x)$. For each real number $t$, there exist a real number $s$ such that $f(s)=\frac{t}{2}$, hence $f(t)=f\left(2f(s)\right)=f(2s)=2f(s)=t$. Hence the answer is $f(x)=x$ for all $x\in\mathbb{R}$.

abch42 wrote: vladimir92 wrote: ... Let $a$ and $b$ be any real numbers, so there exist numbers $x$ and $y$ so that $a=f(2x)$ and $b=f(y)$ hence $a+f(2b)=f(2x)+f(2f(y))=f(2x)+f(2y)=f(x+f(x)+2f(y))=f(a+2b)$, ... Why $f\left(x+f(x)+2f(y)\right)=f(a+2b)$ ? Thank's for detecting the mistake in my solution, it's fixed now.

mahanmath wrote: $(x , f^{-1} (-f(x))) \Rightarrow -f(x)=f(-x)$ (3) I do not correctly understand how you came to this result. substituting $(x , f^{-1}(-f(x)))$ gives: $f(x+f(x)+2f(y))=f(x+f(x)-2f(x))=f(x-f(x))=f(2x)+f(2y)=f(2x)+f(2f(y))=f(2x)+f(-2f(x))$ so:$f(x-f(x))=f(2x)+f(-2f(x))\Rightarrow f(-x)=-f(x)$???

ArefS wrote: mahanmath wrote: $(x , f^{-1} (-f(x))) \Rightarrow -f(x)=f(-x)$ (3) I do not correctly understand how you came to this result. substituting $(x , f^{-1}(-f(x)))$ gives: $f(x+f(x)+2f(y))=f(x+f(x)-2f(x))=f(x-f(x))=f(2x)+f(2y)=f(2x)+f(2f(y))=f(2x)+f(-2f(x))$ so:$f(x-f(x))=f(2x)+f(-2f(x))\Rightarrow f(-x)=-f(x)$??? As you mention it gives us $f(x-f(x))=f(2x)+f(-2f(x))$ but it`s obvious $f(x-f(x))=0$ so $-f(2x)=f(-2f(x))$ also (2) tell us $f(2x)=f(2f(x))$ it means : $-f(2f(x)) =-f(2x)=f(-2f(x))$ and it proves my claim in (3) .

now we have: f(2x)=f(2f(x))=f(x+f(x)) and f(x+f(x)+y)=f(2x)+f(y) +) (x,-x) => f(f(x)) = f(2x) + f(-x) then replace x by f(x) we have f(f(f(x))) = f(2f(x)) + f(-f(x)) For every real number u there exists x such that u=f(x) so we have: f(f(u)) = f(2u) + f(-u) (1) On the other hand (x, -f(x)) => f(x) = f(2x) + f(-f(x)) = f(2f(x)) + f(-f(x)). hence, u = f(2u) + f(-u) (2) From (1) and (2) we have f(f(u))=u for every u. so we have 2x = f(f(2x)) = f(f(2f(x))) = 2f(x) => f(x) = x

Potla wrote: Find all surjective functions $f: \mathbb R \to \mathbb R$ such that for every $x,y\in \mathbb R,$ we have \[f(x+f(x)+2f(y))=f(2x)+f(2y).\quad (1) \] Since $f $ is surjective, there exists a real number $a$ such that $f(a)=0.$ Replacing $x=y=a$ in $(1),$ we get $f(2a)=0.$ Now, replacing $x=a$ in $(1),$ we get \[f\big(2\cdot f(y)+a\big)=f(2y),\quad \forall y \in \mathbb R.\quad (2)\] Thus, the equation $(1)$ can be written as \[f\big( x+f(x)+2\cdot f(y)\big)=f(2x)+f\big(2\cdot f(y)+a\big),\quad \forall x ,\, y \in \mathbb R.\quad (3)\] Since $f$ is surjective, it follows that \[f\big(x+y+f(x)\big)=f(2x)+f(y+a),\quad \forall x,\,y \in \mathbb R.\quad (4)\] Replacing $x=2a$ in $(4),$ we get \[f(y+a)=f(y),\quad \forall y \in \mathbb R.\quad (5)\] From this, it follows that \[f(2y)=f\big(2\cdot f(y)+a\big)=f\big(2\cdot f(y)\big),\quad \forall y \in \mathbb R.\quad (6)\] Now, using $(5)$ and $(6),$ we can rewrite $(4)$ as follow: \[f\big(x+y+f(x)\big)=f\big(2\cdot f(x)\big)+f(y),\quad \forall x ,\,y \in \mathbb R.\quad (7)\] Replacing $y=f(x)-x$ in $(7),$ we get \[f\big(f(x)-x\big)=0,\quad \forall x \in \mathbb R.\quad (8)\] Replacing $x$ by $f(x)-x$ and $y=x$ in $(7),$ we get \[f\big(f(x)\big)=f(x),\quad \forall x \in \mathbb R.\quad (9)\] Since $f$ is surjective, it follows that $f(x)=x,\, \forall x \in \mathbb R.$

Why from $f\big(f(x)\big)=f(x),\quad\forall x\in\mathbb{R}$ and surjectivity of $f$, we obtain $ f(x)=x,\,\forall x\in\mathbb{R}. $?

The former statement says, "$f$ acts as the identity on its image." Surjectivity is the assertion "the image of $f$ is $\mathbb{R}$."

I know this is an old thread but i dicieded to pick a random problem and solve it becuase i did not solve math problems for a long time : Because $f$ is surjective we know that there exists $a$ such that : $f(a) = 0$. Now we set $x = a$ and $y = a$ in our ecuation and we have $f(2a) = 0$ . Setting $x = a$ : $f(a + 2f(y)) = f(2y)$ so now we know that if $f(x) = f(y)$ then $f(2x) = f(2y)$ $(1)$ Let now $y = a$ : $ f(x + f(x)) = f(2x)$ . $(2)$ Let now $b = f(0)$ Setting $x = 0$ and $y = 0$ in our ecuation we have that $f(3b) = 2b$ Nou using $(2)$ with $x = 0$ we have : $f(f(0)) = f(0)$ so $f(b) = b$ so $f(b) = f(0)$ and using $(1)$ we get that $f(2b) = f(0)$ and $f(4b) = f(0)$ . Let $x = b$ and $y = b$ in our ecuation : $f(4a) = 2f(2a)$ so $a = 2a$ so $a = 0$ so $f(0) = 0$ . Now let $x = 0$ so we have $f(2f(y)) = f(2y)$ $(3)$ Now we can write our functional ecuation like this : $f(x + f(x) + 2f(y)) = f(2x) + f(2f(y))$ but because of the fact that f is surjective we have: $f(x + f(x) + y) = f(2x) + f(y)$ Now using $(2)$ we have that $f(x + f(x) + y) = f(x + f(x)) + f(y)$ . $(4)$ Set $y = f(x) - x$ Then we have : $f(2f(x)) = f(2x) + f(f(x) - x)$ But by $(3)$ we have that $f(2f(x)) = f(2x)$ so $f(f(x) - x) = 0$ Set $y = -2x$ then $f(f(x) - x) = f(2x) + f(-2x)$ so $f(2x) = -f(-2x)$ so f is odd function . Now set $y =- x$ $f(f(x)) = f(2x) + f(-x)$ so $f(f(x)) + f(x) = f(2x)$ so we can easy see that $f(x) + x$ is surjective . so our function is aditive form $(4)$ . So $f(x + f(x) +y) = f(x) + f(f(x)) + f(y) = f(2x) + f(y) = 2f(x) + f(y)$ so $f(f(x)) = f(x)$ but by surjectivity we have that $f(x) = x $

zamfiratorul wrote: ... this is cauchy function so $f(x) = cx$ .Easy to se $c = 1$ Very few Cauchy functions are linear. You need some quite strong conditions more to take this conclusion (continuity, or local bound for example).

Upss , fixed now , i had on my paper this solution but when i writed it i was brainless and i taught that this is a faster way to end it , thancks for pointing this out !!

What about this approach: If the degree of \(f(x)\) is \(k\ge 2\) I have in the \(RHS\) the variable \(x\) and \(y\) at the degree of the function. While in the \(LHS\) I have \(x\) and \(y\) at degree \(k^2\). So I have \(k=k^2\) only when \(k=1\). So my function is in the form \(\alpha x+\beta\) applying substitution we notice that the only right function is when \(\beta=0 \land \alpha=1\) thus \(f(x)=x\). Is this approach right?

If i understand correctly you supposed that f is a polinomial , but this is not the case.

zamfiratorul wrote: If i understand correctly you supposed that f is a polinomial , but this is not the case. What do you mean with "it's a polinomial". What is it in this case? I want to say that if for example \(f(x)=\alpha x^2 + \beta x + \gamma\) I would have something that has different degree in the two sides.

If you speak about "degrees", it means you reduce the problem to polynomials and you are not allowed to do this. Simple example : let functional equation $g(x+y)=g(x)g(y)$ If you look at "degrees" for $g(x)$ you find degree $0$ and so only solutions $g(x)\equiv 0$ and $g(x)\equiv 1$ And you miss all solutions $g(x)=e^{a(x)}$ where $a(x)$ is any additive function (solutions which indeed are not polynomials, and so for which it is senseless to speak about degrees).

I haven't understood well last part. Anyway for that function \(g(x+y)=g(x)g(y)\) The degree is not a problem. Every degree is good. Because In this case in both sides the variables would have the same maximum degree. What I want to say is that if I had \(f(x+f(y))=f(x)+f(y)+1\) and \(f(x)=x^2\) then is easy to see that the \(y\) is at degree \(2\) on the \(RHS\) while in the \(LHS\) It is also at degree \(4\). So there is no identidy between the two sides. This is the reasoning, I don't think that there are some cases that make the identity true with degree \(\ge 2\).

In my example, equation implies $g(2x)=g(x)^2$ and so, with your reasonment, only degree $0$ fits. And so you miss soilution $e^x$ for example

Ok. Got it. So basically I forgot the case when \(x\) is an exponent. Anyway in our main equation (the one of the topic) can we say that \(e^x\) is not solution?

Not so simple. Imagine functional equation $f(2x)=2f(x)^2-1$ If you look at degrees, only degree $0$ fits. And so you miss $f(x)=\cos x$ (for example) So, looking at degrees, you dont miss only functions where $x$ is an exponent, you miss nearly ALL possible functions. As everybody tried to explain you, speaking of degrees is meaningful only if you know that you are dealing with polynomials. Is this is not given in the problem statement, you can not suppose it.

Thank you. Now it's more clear.

CanVQ wrote: Replacing $x$ by $f(x)-x$ and $y=x$ in $(7),$ we get \[f\big(f(x)\big)=f(x),\quad \forall x \in \mathbb R.\quad (9)\] There is a calculation mistake in this part? Substituting $x$ $\mapsto$ $f(x)-x$ and $y \mapsto x$ in $(7)$ gives, \begin{align*} f \left( f(x)-x +x+ f \big( f(x)-x \big) \right)= f \left( f(x) \right)=f \big( 2f( f(x)-x) \big) +f(x) \implies f(f(x))=f(x)+f(0) \end{align*}How do we reach $f(f(x))=f(x)$ from here? Or else we can have an alternative finish using $f \big( f(x)-x \big)$ $=$ $0$ and using, \begin{align*} f(x)= f(x+a) \implies f(x) = f \big( x+f(x)-x \big)=f\big(f(x)\big) \end{align*}