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I am working on the question below, which involves subsets and subspaces. It has four parts - hopefully it is alright if I post all of them. They seem to be straightforwards, but I have some questions and want to make sure I understand what I am doing.

Prove that the given subset $W$ is a subspace of $V$, or show why it is not a subspace of $V$.

(a) $V=C^{\infty}(\mathbb{R})$, $W$ is the set of functions $f \in V$ for which $\lim_{x \to \infty} f(x) =0.$

So, $V$ is the vector space of functions from $\mathbb{R}$ to $\mathbb{R}$ with infinitely many derivatives. My first question: are there functions that tend to zero as they approach infinity that also have infinitely many derivatives? If there is, let $h(x)=g(x)+f(x)$ where $g(x) \in W$, and we have

$$\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \left ( g(x)+f(x) \right )= \lim_{x \to \infty} g(x) + \lim_{x \to \infty} f(x) = 0+0 = 0$$

so $g(x)+f(x)$ is in $W$. Now, let $c$ be any scalar, then

$$\lim_{x \to \infty} c \cdot f(x) = c \cdot \lim_{x \to \infty} f(x)= c \cdot 0=0$$

and we see that $c\cdot f(x)$ is in $W$.

(b) $V=C^{\infty}(\mathbb{R})$, $W$ is the set of functions $f \in V$ for which $f(4)=1.$

Let $h(x)=g(x)+f(x)$ where $g(x) \in W$, then we have

$$h(4)=g(4)+f(4)=1+1=2 \neq 1,$$

showing that $g(x) + f(x)$ is not in $W$ and as such $W$ is not a subspace of $V$.

(c) $V=M_{2 \times 2} (\mathbb{R})$, $W$ is the set of matrices whose square is the zero matrix (i.e, those matrices $A$ and $A^2$ the zero matrix.

$M_{2 \times 2} (\mathbb{R})$ is the set of matrices where addition is addition of matrices and scalar multiplication is multiplying all entries by that number. Let $B$ be a $2 \times 2$ matrix in $W$.

$$(A+B)^2 = (A+B)(A+B)=A(A+B)+B(A+B)=A^2 +AB+BA+ B^2=AB+BA$$

Is this sufficient to show that $(A+B)$ is not in $W$? If not, I suppose I would find what the entries of $A$ and $B$ would have to be in order to have $A^2= \mathbf{0}$ and $B^2= \mathbf{0}$, and then show that $AB+BA \neq \mathbf{0}$.

(d) $V= \mathbb{R}^{\infty}$, $W$ is the subset of vectors $(x_1, x_2, x_3,...)$ for which $x_3 = x_2 + x_1$, $x_4 = x_3 + x_2$, $x_5 = x_4 + x_3$,..., and in general $x_{n+2}=x_{n+1}+x_n$ for all $n \geq 1$.

Let $z_{n+2}=x_{n+2} + y_{n+2}$ where $(y_1,y_2,y_3,...) \in W$. Then we have

$$z_{n+2}=x_{n+2} + y_{n+2} = x_{n+1} + x_{n} + y_{n+1} + y_{n} = x_{n+1} + y_{n+1} + x_{n} + y_{n} = z_{n+1} + z_{n}$$

and $(x_{n+2} + y_{n+2}) \in W$. Now, let $c$ be any scalar.

$$c \cdot x_{n+2} = c \cdot (x_{n+1}+x_{n}) = c \cdot x_{n+1} c \cdot x_{n}$$

divide each side by $c$, which yields $x_{n+2}=x_{n+1} + x_{n}$. Multiplying by a scalar has no effect. $W$ is a subspace.

Thanks for any input.

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3 Answers

up vote 0 down vote accepted

To show that $W$ is a subspace of $V$ you have to show the three following conditions:

  1. $0\in W$,
  2. $\forall u,v\in W, u+v\in W$, and

  3. $\forall u\in W, c\cdot u\in W$.

For (a), you have to first show that $0\in W$ (it should be ok :D). And with the rest of you answer your are done.

For (b), you have shown that condition 2 doesn't hold so $W$ is not a subspace. (showing that 1 doesn't hold seems easier but your answer is fine).

For (c), what about the matrix: $$ A=\begin{pmatrix}0&0\\1&0 \end{pmatrix},B =\begin{pmatrix}0&1\\0&0 \end{pmatrix}$$

For (d), you are missing conditions 1 again. But otherwise you are right

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Your proofs of a) and d) are correct. For b) and c) you could simply present a counterexample. I.e., for b) you pick any function $f \in W$ to build a counterexample. For c) you have to find matrices $A$ and $B$ such that the condition is not satisfied. Maybe $AB + BA$ is zero for all $A, B \in V$ (for a reason you didn't see yet).

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My first question: are there functions that tend to zero as they approach infinity that also have infinitely many derivatives?

yes, take $\exp(-\frac{x^2}{2})$

Certainly theis hold for the 0 function. (If there are no other functions of this type, it does not matter. $\{0\}$ is a subspace)

For $AB+BA\neq 0$ is sufficient. Try something with 3 zero's, including the 2 zeros on the diagonals. set one of the remaining to be 0 and the other to be 1.

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