# Find all functions $f:\mathbb{R} \to \mathbb{R},$ which is continuous in $\mathbb{R}$ then $f(x-y)+f(y-z)+f(z-x)+27=0.$

Find all functions $$f:\mathbb{R} \to \mathbb{R},$$ which is continuous in $$\mathbb{R}$$ then $$f(x-y)+f(y-z)+f(z-x)+27=0.$$

I actually don't have any ideas to deal with it, but here is some tries:

Let $$x=y,y=z,z=x$$ then we have $$3f(0)=-27\Rightarrow f(0)=-9.$$

Let $$z=y$$ then $$f(x-y)+f(y-x)+18=0.$$

Let $$y=0$$ then $$f(x)=-18 -f(-x).$$

Also, if we let $$x-y=a;y-z=b;z-x=c\Rightarrow y-x=b+c.$$

Thus $$f(a)+f(b+c) +18 =0; f(a)+f(b)+f(c)+27=0,$$ and $$a+b+c=0.$$

• This is the Cauchy functional equation, but dressed up a little bit. Sep 16, 2021 at 3:25
• I still can't see how I can use Cauchy functional equation, sorry. Sep 16, 2021 at 3:38
• Fixed tags for you. FYI functional analysis is the study of infinite dimension vector spaces.
– Alan
Sep 16, 2021 at 4:00

First, do a translation $$f(x)=g(x)-9$$ then you will get $$g(x-y)+g(y-z)+g(z-x)=0$$. Now let $$x-y=a, y-z=b$$ then $$z-x=-(a+b)$$, putting this we get $$g(a)+g(b)=-g(-(a+b))$$. Now from $$f(x)=-18-f(-x)$$ we can show $$g(x)=-g(-x)$$ So we get the cauchy equation.
$$g(a)+g(b)=g((a+b))$$.
• Also I still don't know how we can show $g(x)=-g(-x),$ help me, thanks a lot. (+1) Sep 16, 2021 at 4:29
• This is the same, you solve for g and get $g(x)=cx$ then you will find $f(x)=cx-9$ for the second comment just put $f=g-9$ in $f(x)=-18-f(-x)$. Sep 16, 2021 at 4:37