Real Function is Expressible as Sum of Even Function and Odd Function

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Theorem

Let $f: \R \to \R$ be a real function which is neither an even function nor an odd function.

Then $f$ may be expressed as the pointwise sum of an even function and an odd function.


Proof

Let:

\(\ds \map g x\) \(=\) \(\ds \dfrac {\map f x + \map f {-x} } 2\)
\(\ds \map h x\) \(=\) \(\ds \dfrac {\map f x - \map f {-x} } 2\)


We note that:

\(\ds \map g {-x}\) \(=\) \(\ds \dfrac {\map f {-x} + \map f {-\paren {-x} } } 2\)
\(\ds \) \(=\) \(\ds \dfrac {\map f {-x} + \map f x} 2\)
\(\ds \) \(=\) \(\ds \map g x\)

Thus $g$ is an even function.


Then:

\(\ds \map h {-x}\) \(=\) \(\ds \dfrac {\map f {-x} - \map f {-\paren {-x} } } 2\)
\(\ds \) \(=\) \(\ds \dfrac {\map f {-x} - \map f x} 2\)
\(\ds \) \(=\) \(\ds -\dfrac {\map f x - \map f {-x} } 2\)
\(\ds \) \(=\) \(\ds -\map h x\)

Thus $h$ is an odd function.


Then:

\(\ds \map g x + \map h x\) \(=\) \(\ds \dfrac {\map f x + \map f {-x} } 2 + \dfrac {\map f x - \map f {-x} } 2\)
\(\ds \) \(=\) \(\ds \dfrac {\map f x } 2 + \dfrac {\map f {-x} } 2 + \dfrac {\map f x } 2 - \dfrac {\map f {-x} } 2\)
\(\ds \) \(=\) \(\ds \map f x\)

Hence the result.

$\blacksquare$


Examples

Arbitrary Function

Let $f: \R \to \R$ denote the real function:

$\map f x = e^{2 x} \sin x$

$f$ can be expressed as the pointwise sum of:

the even function $\map g x = \dfrac {\paren {e^{2 x} - e^{-2 x} } \sin x} 2$

and:

the odd function $\map h x = \dfrac {\paren {e^{2 x} + e^{-2 x} } \sin x} 2$


Sources