Thread: Subspaces of functions

I finally figured out what subspaces are for vectors and matrices, well at least how to do the 3 axions to check anyways, but I'm not sure how to do it for functions since it just says f and doesn't give a function (if that makes any sense).

F[0,1] = {f | f:[0,1] -> R}

So for example, with a function with a domain of [0, 1] with all real numbers as the output...which of these would be subspaces?

1. f(0)f(1) = 0
2. f(0) + f(1) = 0
3. f(x) = -5f(x) such that x is any real number between 0 and 1

Here is what I figured, I don't know if I'm right or not...

1 is true as it contains a zero vector, and anything multiplied that is zero so it is closed under scalar multiplication, not sure about closed under addition though

3 is true because every number between 0 and 1 is a multiple of negative 5, -5(0) = 0 so it has the zero function, and by adding any two functions with values x between 0 and 1 you still get a number that is a multiple of -5.

I looked over my notes and I still don't quite understand how to prove things by checking if it's closed under addition, but am I right so far for those?

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.

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?

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.