Monday, January 18, 2016

functional equations - Show that f is linear



Let f:RR be a solution of the additive Cauchy functional equation satisfying the condition
f(x)=x2f(1/x)xR{0}.


Then show that f(x)=cx, where c is an arbitrary constant.


Answer



Let F(x)=f(x)xf(1)




For some x0, F(1x)=f(1x)1xf(1).



Hence for x0 x2F(1x)=x2f(1x)xf(1)=f(x)xf(1)=F(x)

and of course F(1)=0 and F is additive.






Let us prove that xR,F(x)=F(x)



Indeed, 0=F(1)=F(x+1x)=F(x)+F(1)+F(x)=F(x)+F(x)







Also, for some x1,



F(x)=F(x+1)=(x+1)2F(1x+1)=(x+1)2F(1xx+1)=(x+1)2F(xx+1)=(x+1)2(xx+1)2F(x+1x)=x2F(1+1x)=x2F(1x)=F(x)



This also holds for x=1 since F(1)=F(1).



Hence 2F=0.



Hence F=0 and we're done.



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