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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 $a+b+c=0.$

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

1 Answer 1

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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.

$g(a)+g(b)=g((a+b))$.

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  • $\begingroup$ Also I still don't know how we can show $g(x)=-g(-x),$ help me, thanks a lot. (+1) $\endgroup$
    – NKellira
    Sep 16, 2021 at 4:29
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    $\begingroup$ 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 $f(x)=-18-f(-x)$. $\endgroup$ Sep 16, 2021 at 4:37

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