I need to prove the following is a subspace:

Let $V$ be a set of vectors over $F=\mathbb{R}$, $V=\operatorname{Functions}(\mathbb{R} ,\mathbb{R})$ and $W$ is a subgroup of $V$ such that $$W=\{f\in V| \, f(x)=f(-x)\}$$

I'm not sure about the "close under vector addition" part.

My solution: Let be $w_1 , w_2 \in W. w_1=\{f\in V| \, f(x_1)=f(-x_1)\},\,\,\,w_2=\{f\in V| \, f(x_2)=f(-x_2)\}$. Therefore: $$w_1+w_2=f(x_1+x_2)=f(x_1)+f(x_2)=f(-x_1)+f(-x_2)=f\big(-(x_1+x_2)\big)\\ $$

Please tell me if something is wrong here (I feel like something is wrong...). Many thanks for you time and effort.


Your thinking about $W$ is muddled, as evidenced when you refer to $w_1$ and $w_2$ as sets. You may as well call them $f_1$ and $f_2$ because they are functions, and they are the elements of $W$ (vectors) that you are working with.

There is no reason to index the $x$: it is just an arbitrary element of $\Bbb R$. I suspect you have fallen into a common misunderstanding among beginners. The fact is that the functions are the elements of the vector space. (Whereas beginners sometimes are overly attached to the $x$ being an element of a vector space.)

By virtue of being in $W$, both of them have the property that $f_i(x)=f_i(-x)$. Then

$$ (f_1+f_2)(x)=\dots=(f_1+f_2)(-x) $$

proving that $f_1+f_2$ is also a member of $W$. (I omitted the middle computation so you could puzzle it out :) )

See if you can do the closure under scalars now: the goal is to show that if $\lambda\in \Bbb R$, if $f\in W$, then $\lambda f\in W$. Finally, show that $0\in W$, "0" here denoting the function that is constantly 0 on $\Bbb R$.

In words, what you are doing is showing that the even functions on $\Bbb R$ form a subspace. As a followup exercise, you could additionally show that the odd functions also form a subspace of $V$.

  • $\begingroup$ Is that the missing part?: $$(f_1 + f_2)(x)=f_1(x)+f_2(x)=f_1(-x)+f_2(-x)=(f_1+f_2)(-x)$$ Also, for the scalar ($\alpha \in \mathbb{R}$): $$\alpha f(x)=f(\alpha x) = f(-\alpha x)=\alpha f(-x)$$ $\endgroup$ – yuvalz Apr 17 '13 at 18:32
  • $\begingroup$ @yuvalz You are perfectly right on the first part... but the second part has a mistake. There is no reason to believe $\alpha f(x)=f(\alpha x)$. It should look like this: $(\alpha f)(x):=\alpha \cdot f(x)=\dots=(\alpha f)(-x)$. $\endgroup$ – rschwieb Apr 17 '13 at 18:48

Your understanding is a little off. Here, a vector in $V$ is a function $f:\mathbb{R}\to\mathbb{R}$, and those in $W$ are exactly the functions satisfying $f(x)=f(-x)$. Such functions are called "even." Now suppose that $f_1$ and $f_2$ are two even functions. See if you can show that the function $f_1+f_2$ defined by $(f_1+f_2)(x)=f_1(x)+f_2(x)$ is also even.


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