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Homework Help: Determining if a subset W is a subspace of vector space V

  1. May 18, 2018 #1
    1. The problem statement, all variables and given/known data
    Let V = RR be the vector space of the pointwise functions from R to R. Determine whether or not the following subsets W contained in V are subspaces of V.

    2. Relevant equations
    W = {f ∈ V : f(1) = 1}
    W = {f ∈ V: f(1) = 0}
    W = {f ∈ V : ∃f ''(0)}
    W = {f ∈ V: ∃f ''(x) ∀x ∈ R}

    3. The attempt at a solution

    Note: I am doing Linear Algebra, but I have not yet done Calculus.

    I know that to prove a subset W is a subspace, I must prove that the 0 vector belongs to W, that vector addition is closed, and scalar multiplication as well.

    I am not sure how to go about proving these statements, however. My intuition tells me that the first subset is not a subspace and the second one is, and the other two I am not sure about. Nonetheless when I sit down to start proving, I'm very insecure about what I find.

    For example, given (f+g)(1) = f(1) + g(1) = 2 ≠ 1, does this result mean vector addition is not closed?

    In the case of the second problem, (f+g)(1) = f(1) + g(1) = 0 + 0 = 0, so addition is closed.

    Similarly, for the second proof, (λf)(1) = λ1 ≠ 1 in my first problem, and in the second (λf)(1) = λ0 = 0.

    Am I on the right track?

    If someone could really help me with the third and fourth problems, I have a rudimentary understanding of derivatives and how they work and what the first and second derivatives can indicate about a function, but I do not know how to relate the statements in my problem to the concepts I know about subspaces.
     
  2. jcsd
  3. May 18, 2018 #2

    fresh_42

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    Yes.
    That's a bit difficult if you cannot use the properties of differentiation. However, school level is sufficient here. All you need to know is what ##(f(x)+g(x))'## and ##(\lambda f(x))'## are.
     
  4. May 18, 2018 #3
    Thank you!

    But for example, for ##W = {f \in V : \exists f''(0)}##

    ##(f(0) + g(0))'' = f(0)'' + g(0)''##. If the only data I have is that ##f(0)''## exists, I'm not entirely sure where to go from here.

    Likewise, ##(\lambda f(x))''## exists if ##\exists f''(0)## ...
     
  5. May 18, 2018 #4

    fresh_42

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    Normally it is written in words: "##f''(0)## exists" means "twice differentiable at zero". So if ##f''(a)## and ##g''(a)## exist, then ##f''(a)+g''(a)## exists as well, which is equal to ##(f''+g'')(a)##, i.e. for ##f+g## there exists ##f''+g''## at ##x=a##. You have to show that the sum fulfills the condition, that there is a ##h## with ##h''(a)=(f+g)''(a)=f''(a)+g''(a)##. Now does ##h''(a)## exits? Yes, you've already written it down. Differentiation is linear.
     
  6. May 18, 2018 #5
    I understand. Then, following from what you've just told me, both ##\exists f''(0)## and ##\exists f''(x)\forall x## are closed under addition. It follows that ##\lambda f''(x) = (\lambda f''(x))## and both are also closed under scalar multiplication? I would also say that ##0_v## exists trivially in both of them as long as the second derivative of some real number is 0 in my second case, and as long as the second derivative of 0 is 0 in my first case?
     
    Last edited: May 18, 2018
  7. May 18, 2018 #6

    fresh_42

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    That's all right until here.
    This is a bit unluckily formulated. There is nothing like a derivative of a real number. What is meant in such cases, is a function that is constant everywhere: ##f(x)=c## for all ##x##. Here we have ##0_v(a)=0## for all ##x=a##. So it is the constant functions which are differentiated.
    What do you mean by "in my first case"? ##0\notin W## in your first example, but the derivatives of constant functions, regardless which constant, are always zero because they don't change by varying ##x##. In this sense, ##0\in W## in your cases three and four, as ##0(a)''=0##.
     
  8. May 18, 2018 #7
    I'm sorry, I was unclear, when I said "in my first case" I was referring to what you describe as my third case.

    I understand the derivative of a constant is 0 and that the null vector here is the scalar ##0## multiplying any given function. I do not understand what you mean by "So it is the constant functions which are differentiated," however. I understand that statement is true, but isn't the null vector present regardless of what kind of function is being differentiated here? If I multiply any function by the scalar ##0##, which I can do because ##W## is closed under scalar multiplication, I obtain the null vector.

    I apologize because I intuitively begin to grasp what you're trying to tell me but I want to make sure my comprehension is thorough.

    In any case, I include that my cases 3 and 4 are subspaces because the null vector belongs to ##W## in both cases, and both are closed under vector addition and scalar multiplication.

    Thank you for your help! I hope that it won't be a problem if I continue to post problems here as I get stuck, even though it's all low level linear algebra concepts.

    Is there anything like StackExchange's reputation that I can give you for helping me with this problem?
     
  9. May 18, 2018 #8

    fresh_42

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    When it comes to differentiation, then there are quickly many views involved. Let's take the last example of functions which are differentiable twice everywhere. The number 0 isn't part of the vector space ##W##, because ##W## only contains functions. So the ##0\in W## refers to the function ##f(x)=0## which is also denoted by ##0##. We can identify all numbers ##c## with constant functions ##f(x)=c## so in a way they become elements of ##W##, but strictly speaking not as numbers but as functions instead.

    The null vector is the null function, not the zero number. Derivatives (at ##x=a##) as results of differentiation are defined as ##\lim_{h\to 0} \frac{f(a+h)-f(a)}{h}##. Now if we have a constant function ##f(x)=c##, then ##f(a+h)-f(a)=c-c=0## and the limit at ##x=a## is always identically zero.
    Well, at least the functions have to be differentiable. The null vector is only present, because ##f(x)=0## is differentiable (twice) and we can take this as the null vector. It has to be part of ##W## in the first place, and as such it has to be a function, because ##W## contains only functions.
     
  10. May 18, 2018 #9

    fresh_42

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    Of course you're welcome to do so.
     
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