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?
Best Answer
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.