Determining if a subset W is a subspace of vector space V

  • #1
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Homework Statement


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.

Homework 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}

The Attempt at a Solution


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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.
 

Answers and Replies

  • #2
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Am I on the right track?
Yes.
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.
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.
 
  • #3
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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)## ...
 
  • #4
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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.
 
  • #5
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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?
 
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  • #6
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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...
That's all right until here.
... as long as the second derivative of some real number is 0 in my second case,...
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.
.... and as long as the second derivative of 0 is 0 in my first case?
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##.
 
  • #7
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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?
 
  • #8
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"So it is the constant functions which are differentiated," however. I understand that statement is true,...
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.
... but isn't the null vector present regardless of what kind of function is being differentiated here?
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.
 
  • #9
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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.
Of course you're welcome to do so.
 

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