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Here is the setup. Let $V$ be the vector space of functions $f: \mathbb{R} \to \mathbb{R}$, $U_o$ be the subspace of even functions, and $U_e$ the subspace of odd functions. I want to show that $V = U_e \oplus U_o$.

By some algebra, I was able to find $f(x) = f_e (x) + f_e (x)$ where $f_e (x) = \frac{f(x) + f(-x)}{2}$ and $f_o (x) = \frac{f(x) - f(-x)}{2}$. My question is the strategy of the proof write-up.

The right-hand side is certainly a subspace of the right-hand side by definition, so it suffices to show $V \subset U_e \oplus U_o$, that is, that every $f \in V$ can be written uniquely as a sum of an even and odd function.

The first strategy: I could start the proof by writing down these formulas that I found from algebra, proving $f(x) = f_e (x) + f_o (x)$, check that $f_e$ is even and $f_o$ odd. That establishes that $V = U_e + U_o$. To prove that the sum is direct, I can then prove that $U_o \cap U_e = \{0\}$.

The second strategy: if I show the algebra, I believe every step is reversible and I get uniqueness. This is the one I am most unsure about. Here is the algebra I used to derive this. I want to solve a system of equations in $f_e$ and $f_o$. If $f = f_e + f_o$, where $f_e$ is even and $f_o$ is odd, then for any $x \in \mathbb{R}$, $$ f (x) = f_e (x) + f_o (x), $$ so $$ f(-x) = f_e (-x) + f_o (-x) = f_e (x) - f_o (x), $$ Adding these equations, we get $$ f(x) + f(-x) = 2 f_e (x) \implies f_e (x) = \frac{f(x) + f(-x)}{2}. $$ Substituting into the first equation, we get $$ f_o (x) = f(x) - f_e (x) = \frac{2 f(x) - f(x) - f(-x)}{2} = \frac{f(x) - f(-x)}{2}. $$ I was unsure at this point whether I needed to then check that $f(x) = f_e (x) + f_o (x)$ using these formulas, but my understanding of solving a system of equations is that every step I've written down is reversible. That is, the first line is true if and only if the second line is true and so on. So I've established that $f(x) = f_e (x) + f_o (x)$ if and only if $f_e$ and $f_o$ have these formulas, so not only do they solve this equation, but they are the only such solutions, so the sum is direct.


My primary concern is on the second proof strategy. Is this reasoning correct?

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    $\begingroup$ You defined $f_e=f_o$. $\endgroup$
    – Paul Frost
    Commented Apr 3, 2022 at 8:03
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    $\begingroup$ You use the method of analysis-synthesis for your second proof, so yes it's valid. $\endgroup$
    – SacAndSac
    Commented Apr 3, 2022 at 8:08
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    $\begingroup$ @SacAndSac I have never heard of this method. Would you mind saying a bit more about it? $\endgroup$
    – JohnT
    Commented Apr 3, 2022 at 8:09
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    $\begingroup$ @JohnT Analysis-synthesis reasoning is a type of mathematical reasoning that demonstrates the existence and uniqueness of an object verifying given properties. It breaks down into two parts: -analysis : we assume that the object exists and we try to find the necessary conditions that this object must satisfy. By doing so, one proves that if the object exists, then it is necessarily equal to some object $O$ (this ensures uniqueness). - synthesis: we consider the object $O$ identified in the analysis part, and we check that it has the desired properties (this ensures existence). $\endgroup$
    – SacAndSac
    Commented Apr 3, 2022 at 8:15
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    $\begingroup$ My answer is here. $\endgroup$
    – ryang
    Commented Apr 5, 2022 at 6:45

2 Answers 2

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In your second proof, you show that if the decomposition of $f (x) = f_e (x) + f_o (x)$ exists, with $f_e$ even and $f_o$ odd, then $$ f_e(x) = \frac{f(x)+f(-x)}{2} $$ and $$ f_o(x) = \frac{f(x)-f(-x)}{2}. $$ But given that, you still need to check that these functions satisfy your assumption: that $f_e$ is even, $f_o$ is odd, and $f = f_e+f_o$. Showing that $f = f_e+f_o$ precisely is showing that your steps can be reversed.

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  • $\begingroup$ I think I'm a bit confused here, mostly on the logic, since at the start of the proof, $f_e$ and $f_o$ were assumed to be odd and to satisfy $f = f_e + f_o$. From this, it seems that you're saying that the steps I wrote down aren't themselves reversible unless I prove that they are. Is that correct? $\endgroup$
    – JohnT
    Commented Apr 3, 2022 at 8:13
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    $\begingroup$ @JohnT yes, they are reversible, but if you want your proof to be strictly complete, you have to show that they are. $\endgroup$
    – Steven
    Commented Apr 3, 2022 at 8:15
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Proof is quite simple:

  1. $U_e$ and $U_o$ are subspaces of $V$.

  2. $f(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2}$ is a sum of an even and odd function. Thus $V=U_e + U_o$.

  3. $\frac{f(x) + f(-x)}{2} = \frac{f(x) - f(-x)}{2}$ means that $f(x)=0$ and so the sum is direct.

Hence, $V = U_e\oplus U_o$.

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    $\begingroup$ I’m guessing you mean $f(x)=0$ in part 3? $\endgroup$
    – Milten
    Commented Apr 3, 2022 at 15:12
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    $\begingroup$ I think 3. should read "if $f$ is even and odd, then for all $x$, we have $f(x) = f(-x) = -f(x)$, and hence $f(x) = 0$". That this decomposition gives idempotent projections is what we're trying to prove! $\endgroup$ Commented Apr 3, 2022 at 22:01

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