Either show that the set is a subspace of C(R) or explain why it is not a subspace. note: R in () is real numb

1. {f in C(R) | f(x) %26lt; or = 0, All x in R},


2. {f in C(R) | f(0)=0}


3. {f in C(R) | f(0)=2}

Either show that the set is a subspace of C(R) or explain why it is not a subspace. note: R in () is real numb
#1: not a subspace. For let f(x) = -1, then f is in the set {f∈C(R) : ∀x∈R f(x) ≤ 0}, but -1*f(x) = 1 is not, so this set is not closed under scalar multiplication





#2: a subspace. Note that the zero function is in the space. If f and g are two functions such that f(0) = 0 and g(0) = 0, then (f+g)(0) = f(0) + g(0) = 0 + 0 = 0, so it is closed under addition, and finally if f is a function such that f(0) = 0 and k is any scalar, then (kf)(0) = kf(0) = k*0 = 0, so it is closed under scalar multiplication.





#3: not a subspace. Every subspace must contain the zero function, and this one does not (it is also closed neither under addition nor scalar multiplication).
Reply:Recall (one definition of) a subspace is a set such that if x,y are in it, and a is a scalar, then ax + y is also in the set. Consequently, 0 is a member of the subspace (show it). Also, if x is in the set, then ax is also (just take y =0)-- that is, the set is closed under multiplication by a scalar.





3 is eliminated because 0 isn't in it. 1 is eliminated because say the constant function -1 is in that set, but 1 is not, so it isn't closed under multiplication by a scalar. 2 is fine.