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Let $f:\mathbb{R}\to \mathbb{R}$ such $f(x)$ at $x=0$ continuous, and for any $x,y\in \mathbb{R}$ such $$f(x+2f(y))=f(x)+y+f(y)$$ Find $f(x)$.
Let $x=0,y=0$ then we have $$2f(2f(0))=f(0)++f(0)$$ Let $y=2f(0)$ then we have $$f(x+2f(2f(0))=f(x)+2f(0)+f(2f(0))$$ so we have $$f(x+4f(0))=f(x)+4f(0)$$ on the other hand we have $$f(x+4f(0))=f(x+2f(0))+0+f(0)=2f(0)+f(x)$$ so we have $f(0)=0$ Then I can't deal with this problem
Substitute $x\mapsto y$ to get $$ f(y+2f(y))=y+2f(y). $$ Now substitute $y\mapsto y+2f(y)$: $$ f(x+2f(y+2f(y)))=f(x)+y+2f(y)+f(y+2f(y)). $$ Thus $$ f(x+2y+4f(y))=f(x)+2y+4f(y). $$ Substituting $x\mapsto x+2y+2f(y)$, $$ f(x+2y+4f(y))=f(x+2y+2f(y))+y+f(y). $$ Substituting $x\mapsto x+2y$, $$ f(x+2y+2f(y))=f(x+2y)+y+f(y). $$ Combining the last 3 equations, $$ f(x+2y)=f(x)+2f(y). $$ Thus $$ f(x)=f(0+2x/2)=f(0)+2f(x/2)=2f(x/2), $$ so $$ f(x+y)=f(x+2y/2)=f(x)+2f(y/2)=f(x)+f(y). $$ Since you are given that $f$ is continuous at $0$, it must be continuous everywhere. Now there is a standard argument to show that $f(x)=f(1)x$ (prove it on rationals and extend by continuity). Finally using the original equation, we see $f(1)=1$ or $-\frac12$, so the solutions are $f(x)=x$ or $f(x)=-x/2$.
Setting $x=y$ gives $f(x+2f(x))=x+2f(x)$, and so $f(y)=y$ whenever $y$ is of the form $x+2f(x)$ for some $x$. Note also that if $f(y)=y$, then for any $x$, $f(x+2y)=f(x)+2y$.
Now consider the function $g(x)=f(x)-x$. From the previous paragraph, we see that if $f(y)=y$ then $g(x+2y)=g(x)$. In particular, every number of the form $2x+4f(x)$ is a period of $g$. Let $P\subseteq\mathbb{R}$ be the group of periods of $g$, i.e. the set of $p$ such that $g(x)=g(x+p)$ for all $x$. There are now two cases.
The first case is that $P$ is not cyclic. Then $P$ is dense in $\mathbb{R}$, and so each coset of $P$ is dense in $\mathbb{R}$, and in particular $0$ is in the closure of each coset. But by definition of $P$, $g$ is constant on each coset of $P$, and so by continuity of $g$ at $0$ this constant value must be equal to $g(0)$. Thus $g(x)=g(0)$ for all $x$, and $g$ is constant everywhere. Thus $f(x)$ has the form $f(x)=x+c$ for some constant $c$. Plugging this into the functional equation gives $$x+2(y+c)+c=x+c+y+y+c,$$ so $c=0$. Thus $f(x)=x$.
The second case is that $P$ is cyclic, generated by some $p\in\mathbb{R}$ (possibly $0$). Then every number of the form $2x+4f(x)$ is an integer multiple of $p$. But $2x+4f(x)$ is continuous at $0$, so this means $2x+4f(x)$ is constant in a neighborhood of $0$. As you've shown, $f(0)=0$, so $2x+4f(x)=0$ for all $x$ in some neighborhood of $0$, so $f(x)=-x/2$ for all $x$ in some neighborhood of $0$.
Now suppose $\epsilon>0$ is such that $f(x)=-x/2$ whenever $|x|\leq\epsilon$. If $0\leq a\leq \epsilon$, the functional equation with $x=a$ and $y=-\epsilon$ then gives $$f(a+\epsilon)=-\frac{a+\epsilon}{2}.$$ That is, $f(x)=-x/2$ is also valid if $\epsilon\leq x\leq 2\epsilon$. Similarly, using $y=\epsilon$ gives that $f(x)=-x/2$ is also valid if $-2\epsilon\leq x\leq -\epsilon$.
That is, if $f(x)=-x/2$ whenever $|x|\leq\epsilon$, $f(x)=-x/2$ whenever $|x|\leq 2\epsilon$ as well. Repeating this argument over and over, we get that $f(x)=-x/2$ for all $x$.
Thus the only possibilities for $f$ are $f(x)=x$ and $f(x)=-x/2$, and you can easily check that both of these work.
