# find all continuous functions $f(x+y)+f(y+z)+f(z+x)=f(x)+f(y)+f(z)+f(x+y+z)$

QUESTION -

Find all continuous functions $$f: \mathbb{R} \rightarrow \mathbb{R}$$ such that $$f(x+y)+f(y+z)+f(z+x)=f(x)+f(y)+f(z)+f(x+y+z)$$

MY TRY -

i proved that $$f(0)=0$$ then $$f_{o}$$ satisfies $$f_{o}(x+y)+f_{o}(x-y)=2 f_{o}(x)$$ and $$f_e$$ satisfies $$f_{e}(x+y)+f_{e}(x-y)=2 f_{e}(x)+2 f_{e}(y)$$..

where $$f(x)$$=$$f_e$$+$$f_o$$ ...(odd and even parts of f)

so now using above eqaution for $$f_o$$ i am able to find $$f_o$$ ...but not able to find $$f_e$$ by using the above equation of $$f_e$$...

any help will be helpful..... thankyou

• @KaviRamaMurthy Every quadratic polynomial function with zero constant term is a solution. Apr 7, 2020 at 6:15
• I got it myself ...by putting y=x and using induction and continuity of fe we get fe(x)= ax^2 ....hurray!!! Apr 7, 2020 at 6:48

The PCO's soution:

Let $$P(x,y,z)$$ be the assertion $$f(x+y)+f(y+z)+f(z+x)=f(x)+f(y)+f(z)+f(x+y+z)$$

$$P(0,0,0)$$ $$\implies$$ $$f(0)=0$$ $$P(x,y,-x-y)$$ $$\implies$$ $$f(x+y)-f(-x-y)=f(x)-f(-x)+f(y)-f(-y)$$ and so $$g(x+y)=g(x)+g(y)$$ where $$g(x)=f(x)-f(-x)$$ is continuous Fo $$f(x)-f(-x)=cx$$ and so $$f(-x)=f(x)-cx$$

$$P((n+1)x,x,-x)$$ $$\implies$$ $$f((n+2)x)=2f((n+1)x)-f(nx)+(2f(x)-cx)$$ Considering this as a sequence $$a_{n+2}=2a_{n+1}-a_n+b$$, we easily get $$f(px)=p^2f(x)-cx\frac {p(p-1)}2$$

So $$f(x)=q^2f(\frac xq)-cx\frac {(q-1)}2$$

And so $$f(\frac pqx)=\frac{p^2}{q^2}f(x)-\frac 12 cx\frac pq(\frac pq-1)$$

And so $$f(x)=x^2f(1)-\frac 12 cx(x-1)$$ $$\forall x\in\mathbb Q^+$$ and the equation $$f(-x)=f(x)-cx$$ shows that this must be true $$\forall x\in\mathbb Q$$

Continuity implies then $$\boxed{f(x)=ax^2+bx}$$ $$\forall x$$ which indeed is a solution

• Yes, i see it before but many steps i don't understand in his proof.. Apr 7, 2020 at 6:36
• @User88463 Ask your question. The first step it's a solution of the known functional equation. Apr 7, 2020 at 6:50