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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

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

1 Answer 1

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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

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

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