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

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1. May 18, 2018 at 6:57 AM

### iJake

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. May 18, 2018 at 7:18 AM

### Staff: Mentor

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.

3. May 18, 2018 at 7:24 AM

### iJake

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. May 18, 2018 at 7:36 AM

### Staff: Mentor

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. May 18, 2018 at 8:03 AM

### iJake

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 at 8:15 AM
6. May 18, 2018 at 8:24 AM

### Staff: Mentor

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 $0(a)''=0$.

7. May 18, 2018 at 8:40 AM

### iJake

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. May 18, 2018 at 8:57 AM

### Staff: Mentor

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.

9. May 18, 2018 at 8:58 AM

### Staff: Mentor

Of course you're welcome to do so.