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Question: Let $V$ be the vector space of all functions $\Bbb R\to \Bbb R$. Show that $V=U \oplus W$ for $$U=\{f\ | \ f(x)=f(-x)\ \ \forall x\}, \quad W=\{f \ |\ f(x)=-f(-x) \ \ \forall x\}$$

What I did:

I did prove that $U \cap W$={$0$}. But proving that any function from $\mathbb{R}$ to $\mathbb{R}$ can be displayed as a sum of odds and evens wasn't a success. I tried saying that for $v \in V, w \in W: v=v-w+w$ and proving that $v-w \in U$ but that didn't work (that trick worked with some linear transformations we saw, but this isn't a linear transformation).

Math Attack
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jreing
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1 Answers1

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Hint: $$f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}$$

Alex Becker
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  • should I use this as continuation to my path or as a new path? – jreing May 21 '13 at 16:38
  • @user1685224 This shows that every function is a sum of even and odds. Your work showing $U\cap W={0}$ still applies. – Alex Becker May 21 '13 at 16:41
  • Okay it took me about 30 minutes, to get the hint, but just to verify: What needs to be done is defining 2 functions (one equals the first factor, and the second equals the second factor), and then prove that one belongs to U, and the other to W. – jreing May 21 '13 at 16:57
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    @user1685224 Yes. – Alex Becker May 21 '13 at 17:01
  • By the way, @AlexBecker, do you know where this idea came from? It is hard to explain how we come up with it. – Alice Apr 17 '23 at 21:18
  • I've found a great answer from Ivo Terek here -> https://math.stackexchange.com/questions/4286284/direct-sum-of-even-and-odd-functions-make-whole-vector-space – Alice Apr 17 '23 at 21:33