Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find all continuous functions $ f:\mathbb{R}\to\mathbb{R}$, that satisfy $f(xy)=f(\frac{x^2+y^2}{2})+(x-y)^2$

share|improve this question
    
What have you tried so far? –  Argon Aug 19 '12 at 3:55
    
"Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. Titles should be informative. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ( "Find") to be rude when asking for help; please consider rewriting your post." –  Pedro Tamaroff Aug 19 '12 at 3:58
add comment

1 Answer

Let $x=0$ first, we see $f(0)=f(\frac{y^2}{2})+y^2$, hence $f(x)=f(0)-2x$ if $x\ge 0$. Then let $y=-x$, we get $f(-x^2)=f(x^2)+4x^2$, hence $f(x)=f(-(-x))=f(-x)+4(-x)=f(0)-2x$ when $x\le 0$.

And it's easily checked that all those functions satisfy the equation. So all you want is $f(x)=-2x+c$ for any constant c.

share|improve this answer
    
Very nice. You've shown that if there IS a function that obeys this equation, then it must have the form that you've given. But to be a complete solution, you must ALSO show that all functions that HAVE this form obey the equation. –  user22805 Aug 19 '12 at 4:07
    
@DavidWallace:Yes, you're right! So I must add a sentence reads"And it's easily checked that all those functions satisfy the equation." –  Y.Z Aug 19 '12 at 4:17
2  
or better still, add a sentence that actually checks it. –  user22805 Aug 19 '12 at 4:19
    
@DavidWallace: I've checked that, and I leave this as an exercise for tangkhaihanh. :) –  Y.Z Aug 19 '12 at 4:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.