How do I show that a function which is neither even nor odd can be written as the sum of an even and odd function?

Question

I've seen quite a few proofs already that demonstrate that any function $f$ can be written as the sum of an even and odd function. I'm trying to prove this result by dividing into cases where $f$ is even, $f$ is odd, and $f$ is neither even nor odd. But I'm having trouble with the case where $f$ is neither even nor odd.

This is the construction of the proof so far:

By definition a function $f$ is even if $f(x)=f(-x)$ and odd if $f(x)=-f(-x).$ From a previous result, the sum of two even functions is even, the sum of two odd functions is odd, and the sum of an even and odd function is neither even nor odd.

$f$ is even: $f(x)=\dfrac{f(x)+f(x)}{2}=\dfrac{f(x)+f(-x)}{2}+0=\dfrac{f(x)+f(-x)}{2}+\dfrac{f(x)-f(-x)}{2}$

$f$ is odd: $f(x) = \dfrac{-f(x)-f(x)}{-2}=0 +\ \dfrac{-f(x)+f(-x)}{-2}=\dfrac{f(x)+f(-x)}{2}+\dfrac{f(x)-f(-x)}{2} $

How would I go about showing that a function $f$ which is neither even nor odd can be written as the sum of an even function $\biggl(E(x):=\dfrac{f(x)+f(-x)}{2}\biggr)$ and an odd function $\biggl(O(x):=\dfrac{f(x)-f(-x)}{2}\biggr)$?

Answer 1

You simply use the fact that$$f(x)=\frac{f(x)+f(-x)}2+\frac{f(x)-f(-x)}2.$$

Answer 2

Once you have defined $E(x)$ and $O(x)$ you need to show that they are respectively even and odd and that $f(x)=E(x)+O(x)$ So, $$E(-x)=\frac {f(-x)+f(x)}2=E(x)$$ is even and the proof for $O(x)$ is similar.
Then $$E(x)+O(x)=\frac {f(x)+f(-x)}2+\frac {f(x)+f(-x)}2=f(x)$$

Answer 3

If we assume it can be so we can attempt to "solve" for the functions.

$f(x) = e(x) + o(x)$

And $f(-x) = e(-x) +0(-x)=e(x) -o(x)$.

So $f(x) + f(-x) = 2e(x)$ and $e(x) = \frac {f(x) + f(-x)}2$. That's clearly an even function $f(-(-x))= f(x)$ and addition is commutative.

And $f(x) - f(-x)=2o(x)$ so $o(x)= \frac {f(x) -f(-x)}2$. That's clearly an odd function as $f(-(-x)) = f(x)$ and $(b-a) = - (a-b)$.