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Suppose $f:\mathbb R\to\mathbb R$ is a strictly decreasing function which satisfy the relation $$f(x+y) + f( f(x) + f(y) ) = f( f( x+f(y) ) + f( y+f(x) ) ) , \quad \forall x , y \in\mathbb R $$

Find a relation between $f( f(x))$ and $x$.

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I think I have got hold of something , putting y=x in the functional equation we get $$ f(2x) + f(2f(x)) = f(2f(x+f(x))) $$ Changing $x$ to $f(x)$ we also get $$ f(2f(x)) + f(2f(f(x))) = f(2f(f(x)+f(f(x)))) $$ Subtracting the former from the later equation we get $$ f(2f(f(x))) - f(2x) = f(2f(f(x)+f(f(x)))) - f(2f(x+f(x))) $$ Now since $f(x)$ is strictly decreasing $x>y$ if and only if $f(x) < f(y)$ . Assume that $f(f(x)) > x$ , for some $x$, then $$ \begin{align} 2f(f(x)) > 2x&\Longleftrightarrow f(2f(f(x))) < f(2x)\\ &\Longleftrightarrow f(2f(f(x)+f(f(x))))< f(2f(x+f(x)))\\ &\Longleftrightarrow 2f(f(x)+f(f(x))) > 2f(x+f(x))\\ &\Longleftrightarrow f(f(x)+f(f(x))) > f(x+f(x))\\ &\Longleftrightarrow f(x)+f(f(x)) < x+f(x)\\ &\Longleftrightarrow f(f(x)) < x \end{align} $$ contradiction! So , $f(f(x)) = x$ , for all real $x$.

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You just got contradiction with assumption $f(f(x))>x$. Isn't it? –  Norbert Oct 17 '12 at 10:58
@ Norbert:- Yeah right , so could be got by assuming f(f(x)) < x .Hence the conclusion that f(f(x)) = x for all x . And thanks very much for the edit. –  Souvik Dey Oct 17 '12 at 11:01
you are welcome! –  Norbert Oct 17 '12 at 11:02
But can't $f$ be a constant function ? e.g $f(x)=1$ for all $x$. but then $f(f(x))=1$ and not $x$... –  Belgi Nov 10 '12 at 8:43
@Belgi, $f$ is strictly decreasing –  Norbert Nov 10 '12 at 9:52
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