Which Subspaces of the Vector Space of Real Valued Functions Are Valid?

[VinnyCee]

V = F(R, R), the vector space of all real valued functions f(x) of a real variable x. Which are subspaces of V?

(A) {f | f(0) = 0}

(B) {f | f(0) = 1}

(C) {f | f(0) = f(1)}

(D) C^0(R) = {f | f is continous}

(E) C^1(R) = {f | f is differentiable and f' is continous}

(F) P = {f | f is a polynomial}

(G) P_d\,\,\,\,=\,\,\,\,{f\,\in\,P\,|\,deg(f)\,\le\,d}

(H) {f\,\in\,C^1(R)\,|\,f'\,=\,f}

I have no idea what the last five of these instances mean. Case (C) is not a subspace because it only satifies the first rule that the set {0} be in the space before it can be considered a subspace, right?

Please help, I don't understand the terminology of the last five examples!

 

[AKG]

You don't know what "polynomial", "differentiable", and "continuous" mean? If f is a polynomial, you don't know what deg(f), the degree of f, is?

 

[VinnyCee]

Yes, I know those terms, but not the C^0,\,C^1 terms or the P or P_d terms.

 

[quasar987]

Well that's why they are defined for you

For instance, C^1(R) denoted the set of all continuously differentiable (meaning f' is continuous) functions.

 

[Daverz]

(G) would be polynomials that look like

<br /> f(x) = \sum_{i=0...d} a_i x^i<br />

 

[VinnyCee]

Both (A) and (B) are subspaces of V, but (C) is not because each element must be unique, right? f(0) = f(1) means that it is not unique and not 1-1. But, how would I prove that (A) and (B) are subspaces of V?

 

[Daverz]

If f(0) = f(1), it's still a function (think of a horizontal line), and there's no requirement for it to be 1-1.

For (B), what is the 0 element of the set?

 

[VinnyCee]

There will be no 0 element because it always equals 1. So (B) is NOT a subset. (C) could be a horizontal line. But how would I prove these items? what would an arbitrary f(x) function look like?

f(x)\,=\,a_1\,x^{n\,+\,2}\,+a_2\,x^{n\,+\,1}\,+\,...\,+a_n\,x^n

 

[Daverz]

It's just like the matrix case. You have functions in the set f and g and a real number a. Are af, f + g,and the zero function in the set?

 

[VinnyCee]

0 function would be f = 0 or f = 0 x^2 + 0 x + 0?

How would I write it out (prove) for case (A)?

{f | f(0) = 0}

1.) 0 \in V.

2.) a f(0) = 0 \in V.

3.) f(0) + g(0) = 0 + 0 = 0 \in V.

 

[Daverz]

0 function would be f = 0 or f = 0 x^2 + 0 x + 0?

How would I write it out though for case (A)?

{f | f(0) = 0}

1.) 0 \in V.

2.) a f(0) = 0 \in V.

3.) f(0) + g(0) = 0 + 0 = 0 \in V.

Those are all pretty clear, right? You just need a sentence after each of those computations saying "so <function> is in the set <whatever>." You may need a sentence explaining that the sum of a C^0 functions is also C^0. The same for C^1 functions. And similarly for a real number times a function (should be able to look this up in your calculus text.) For the polynomials you show directly by combining terms that the new function is a polynomial of whatever type is required.

 

[VinnyCee]

What about case (B)?

It says that a function with zero plugged-in as the variable is equal to one. But this does not mean that the function cannot equal zero at some other plugged-in constant, right? So rule 1 checks out.

However, rule 3 (additive) is not satisfied becuase f(0) + g(0) = (1) + (1) = 2. Which is not 1, obviously.

Does that sound right?

 

[radou]

Yes, it does. Further on, for scalar multiplication, i.e. a real scalar a \neq 1, you'd have a\cdot f(0) = a.

 

[VinnyCee]

So, is every case except (B) and (H) a subspace of V?

 

[HallsofIvy]

What about case (B)?

It says that a function with zero plugged-in as the variable is equal to one. But this does not mean that the function cannot equal zero at some other plugged-in constant, right? So rule 1 checks out.

No, it doesn't. In function spaces with f+ g defined by (f+g)(x)= f(x)+ g(x) (the usual definition), the 0 vector is the function f(x)= 0 for all x. Is that function in this set?

However, rule 3 (additive) is not satisfied becuase f(0) + g(0) = (1) + (1) = 2. Which is not 1, obviously.

Yes, that is true.