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

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.

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, $$0\in E$$.
• 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. Commented Jan 17, 2020 at 6:16
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.