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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 $\biggl(O(x):=\dfrac{f(x)-f(-x)}{2}\biggr)$?

José Carlos Santos
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Skm
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  • See the linked dupe for some intuition and generalizations. – Bill Dubuque Aug 04 '19 at 15:04
  • The intuition and generalizations are definitely helpful. This question that I posted, on the other hand, wasn't really meant to go into any deep theorems in mathematics. Rather, it was meant as a self-exercise (and a possible exercise for others who may read the post) for constructing proofs. For example, by partitioning all the functions into the three "classes" even, odd, and neither, showing that $E(x)$ and $O(x)$ are indeed even and odd, respectively, means that the aforementioned step is to be shown relatively later in the incomplete proof that I've provided-regardless of its redundancy. – Skm Aug 04 '19 at 18:47
  • Again, this is all relative to other proofs that I've seen on this site so far. – Skm Aug 04 '19 at 18:51

3 Answers3

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

José Carlos Santos
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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 $$E(x)+O(x)=\frac {f(x)+f(-x)}2+\frac {f(x)+f(-x)}2=f(x)$$

Ross Millikan
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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 $(b-a) = - (a-b)$.

fleablood
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