September 21st, 2015, 04:38 PM  #1 
Member Joined: Jan 2012 Posts: 51 Thanks: 1  vector spaces
If V is the real vector space of all functions f from R to R, is the set, W, of functioncs such that f(x^2)=f(x)^2 a subspace of V ? Why ? I think its not, because if I take f(x1), f(x2) E W then (f(x1)+f(x2))^2 = f(x1)^2 +2(fx1)f(x2)+ f(x2)^2 which doesnt seem to belong to W. Is it right ? Last edited by bonildo; September 21st, 2015 at 04:49 PM. 
September 21st, 2015, 10:19 PM  #2  
Senior Member Joined: Aug 2012 Posts: 2,044 Thanks: 584  Quote:
The key to the notation is that f is a function, and f(x) is the value of the function at the point x. So if I have two different functions, I'd notate them f1 and f2, and their respective values at the point x are f1(x) and f2(x). What you wrote, f(x1) and f(x2) are the values of a single function f at two different points. Not what you want. You're right that W doesn't seem like it would be closed under addition. To prove it you need to find a pair of functions in W whose sum isn't in W. Offhand I don't see an obvious solution so I share your intuition but don't have the proof either. Last edited by Maschke; September 21st, 2015 at 10:28 PM.  
September 21st, 2015, 11:55 PM  #3 
Senior Member Joined: Aug 2012 Posts: 2,044 Thanks: 584 
I think my Edit capability timed out. Hint: This is a cute problem because the temptation is to try to think of complicated functions in W. But the trick is to think of the simplest functions there are. There are two really simple functions in W and one of them is the counterexample you need. To get going, see if you can write down some functions in W. Think really really simple. Last edited by Maschke; September 21st, 2015 at 11:57 PM. 
September 22nd, 2015, 07:14 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
f(x)= x is obviously in W. What about multiples of f?


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