Re: Subspaces of functions

each of the three sets given are defined by properties. for example:

1. S = {f in F[0,1] : f(0)f(1) = 0}.

this means that S is the set of all real-valued functions defined on the interval [0,1], for which either f(0) or f(1) (or both) is 0.

the 0-vector of F[0,1] is the function z, where z(x) = 0 for all x in [0,1]. clearly this function is 0 at x = 0 and x = 1.

now suppose f(x) = x. since f(0) = 0, f(0)f(1) = 0, so f is in S. suppose further, that g(x) = 1-x. then g(0)g(1) = (1)(0) = 0, so g is in S.

but (f+g)(x) = x + 1 - x = 1, so (f+g)(0)(f+g)(1) = (1)(1) = 1.

Re: Subspaces of functions

Ok I think I understand how to prove the closed under addition now, but how did you suppose g(x) = 1-x? Or do we just pick __any__ random function there?

Re: Subspaces of functions

my point is it is NOT closed under addition. to disprove something, you only need exhibit one counter-example.

how did i find it?

the condition f(0)f(1) = 0 is equivalent to: either f(0) = 0 or f(1) = 0. now there is no logical reason to assume that if f and g satisfy this, f+g will. so i picked some function with f(0) = 0, and f(1) = 1. there are lots of functions i could have chosen with this property, (for example f(x) = sin(pi*x/2) is one) the identity function on [0,1] is just a function for which it is obvious that f(0) = 0, f(1) =1.

then i found another function g that satisfied g(0) = 1, and g(1) = 0. again, there are lots of possibilities (for example g(x) = cos(pi*x/2) would have also worked), g(x) = 1 - x is just easy to work with.

the point is f+g is defined point-wise, as (f+g)(x) = f(x) + g(x). there is no reason at all to assume that a property that holds for products, also holds for sums. finding counter-examples isn't a "formulaic" procedure, you have to be adaptable. in this case, since only one term in the expression f(0)f(1) has to be 0, i chose functions that had different terms be 0 (f is 0 only at 0, g is 0 only at 1, f+g is 0 at neither). i cant tell you how to do this for every set-definition that might come up, all i can tell you is that if you can't prove something, it might be that what you are trying to prove simply isn't true, so you might try to find an exception that disproves it.