Checking if the solution to the functional equation is correct $f(2x+f(y))=x+y+f(x)$

Question

Find all functions $f:\Bbb R\rightarrow \Bbb R$ such that

$$f(2x+f(y))=x+y+f(x)$$

for all real $x, y$.

Ok, so I started off by proving that the function is surjective. Let us choose a number $c$ in the range of the function. Now set $x=c$ and $y=-f(c)$. From the equation above we obtain that

$$f(2c+f(-f(c))=c$$

therefore to obtain any $c$ in the range we just have to input $2c+f(-f(c))$ in the function argument.

Now since the function is surjective there exists an $a$ such that $f(a) = 0$. Let us set $y = a$. Then we get that

$$f(2x)=x+a+f(x)$$

From the above we obtain that $a$ must be $0$. From the original equation we get that $$f(f(y))=y+f(0)$$

$$f(f(y)) = y$$

That means that the function is it's own inverse. This implies that $f$ is the identity function or $f(x)=x$.

And indeed it does satisfy the functional equation.

I just want to know if I made any critical errors.

Answer 1

The only critical error you made is $$f(f(x))=x \implies f(x)=x$$ This is a huge mistake, try $f(x)=c-x$ or $f(x)=\frac{c}{x}$ and you'll get also that $f(f(x))=x$

The step you were missing is putting $x=f(x)$ in the original equation to get: $$f(2f(x)+f(y))=f(x)+y+f(f(x))=f(x)+y+x=f(2x+f(y))$$ and since $$f(a)=f(b)\implies f(f(a))=f(f(b)) \implies a=b$$ then $$f(2f(x)+f(y))=f(2x+f(y)) \implies2f(x)+f(y)=2x+f(y)\implies f(x)=x$$ That's it!

Another solution that would be useful for others is this: $$P(x,y)\implies f(2x+f(y))=x+y+f(x)$$ $$P(0,0) \implies f(f(0))=f(0)$$ $$P(0,x) \implies f(f(x))=x+f(0)\implies f(f(f(x)))=f(x+f(0))=f(x)+f(0)$$ So putting $x=0$ in the last equation, we get $$f(f(0))=2f(0) \implies f(0)=2f(0) \implies f(0)=0 \implies f(f(x))=x$$ Now, again, combining $P(f(x),y)$ and $P(x,y)$ , we get, by injectivity, $$f(2f(x)+f(y))=f(2x+f(y)) \implies f(x)=x \ \ \Box.$$ The trick I used here to get that $f(0)=0$ is called tripling an involution: $$f(f(x))=g(x) \implies f(f(f(x))=f(g(x))=g(f(x))$$hopefully, you see how that works. It's a very useful trick indeed.

Answer 2

You're not quite done. You've successfully concluded that $f(f(x))=x$ for all $x$, and $f(0)=0$. For a hint as to how to continue, note that firstly $f$ must be injective and secondly $$x+f(x)=f(f(x))+f(x).$$ Can you see how this helps?

Answer 3

You can continue with $f(f(x))=x$. So $f$ is bijective. For any $x$ there exists one and only one $y_x$ such that $f(y_x)=-x$. Replace $y_x$ in original equation give you $y_x=-x$ so $f(-x)=-x$ for any $x$, i.e., $f(x)=x$.

Answer 4

I was checking my answer to this problem, but I found another very interesting solution: $$P(x,y) := f(2x+f(y))=x+y+f(x)$$ $$P(-f(x),x) \implies f(-f(x))=-f(x)+x+f(-f(x)) \iff f(x)=x$$