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Let $E,O$ $\subset$ $F(R,R)$ denote the sets of even and odd functions respectively. Prove that the $E$ and $O$ are subspaces.

My proof: (for simplicity, I am only showing the proof of set of even functions.

first, we prove that 0 exists in $E$. By definition of even function, a null function is an even function. Hence 0 exists in $E$. The same logic goes to the set odd function.

then we prove that the set of even functions is closed under addition. Let $x$ $\in$ $R$, $a$ and $b$ be two separate even functions, then $(a+b)(-x)$= $a(-x)$+$b(-x)$= $a(x)$+ $b(x)$= $(a+b)(x)$ $\in$ $E$. Hence it is closed under addition.

Next, we prove that it is closed under scalar multiplication. Let $r$ $\in$ $R$, then $(r*a)(x)$= $r* a(x)$= $r*a(-x)$= $-(r*a)(x)$ $\in$ $E$, hence it is closed under scalar multiplication.

Hence $E$ is a subspace.

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2 Answers 2

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How do you jump “$E$ is nonempty” to “there exists $0\in E$ so that $a(−x)=0= a(x)$”? That doesn't make sense. The set $\{1\}$ is also nonempty, but it doesn't contain the $0$ function. You can simply say that $0$ is an even function; in other words, $0\in E$.

The rest is correct.

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  • $\begingroup$ I was making that up from what I remember about the subspace. I edited it, does it look correct now? $\endgroup$
    – Beacon
    Commented Jan 17, 2020 at 6:12
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    $\begingroup$ No, it is not correct. You cannot say “Let $a$ be an even function in $E$” and conclude that $a$ is the null function. Just say that the null function is an even function. That's all you need. $\endgroup$ Commented Jan 17, 2020 at 6:16
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Your proof of the fact the set is closed under addtion and scalar multiplication are fine. But I don't quite understand what you are saying about the zero vector. The zero function $f$ is defined by $f(x)=0$ for all $x$. This function is even and this is the zero element of the space.

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  • $\begingroup$ I took your suggestion and edited now, does it look ok? $\endgroup$
    – Beacon
    Commented Jan 17, 2020 at 6:13

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