What is the decomposition into even and odd function of a function equal to $x$ when $x \geq 0$ and equal to zero when $x <0$?

Question

Every function $f$ with domain in $\mathbb{R}$ can be written

$$f=E+O$$

where $E$ is an even function and $O$ is an odd function.

Proof

Assume $f(x) = E(x) + O(x)$.

Then

$$f(-x)=E(-x) + O(-x)=E(x)-O(x)$$

Therefore, given a function $f$,

$$f(x) = E(x) + O(x)$$ $$f(-x)=E(x)-O(x)$$

represent a system of two equations in two unknowns.

We can solve for

$$E(x) = \frac{f(x) + f(-x)}{2}$$

$$O(x) = \frac{f(x) - f(-x)}{2}$$

This concludes the proof.

Now consider a function

$$f(x) = \begin{cases} 0\text{ if } x < 0 \\ x\text{ if } x \geq 0 \end{cases}$$

What do $E(x)$ and $O(x)$ look like?

$$E(x) = \begin{cases} \frac{0+x}{2}=\frac{x}{2}\text{ if } x<0 \\ \frac{x+0}{2}=\frac{x}{2}\text{ if } x \geq 0 \end{cases}$$

Isn't the function $E(x)=\frac{x}{2}$ odd?

Similarly, we reach

$$O(x) = \begin{cases} \frac{0-x}{2}=-\frac{x}{2}\text{ if } x<0 \\ \frac{x-0}{2}=\frac{x}{2}\text{ if } x \geq 0 \end{cases}$$

Isn't $O(x)$ even?

I must be missing something very silly here.

Answer 1

True, I felt a bit weird about it too. I should have made the proof something like as follows. Let $f$ be any function. Let $E(x)=\frac{f(x)+f(-x)}{2}$ and let $O(x)=\frac{f(x)-f(-x)}{2}$. Then, I easily show that $E$ is even, $O$ is odd, and $f=E+O$. Thus we can say $f=E+O$ with $E$ even and $O$ odd, for any function $f$.

Yes, exactly. You'd originally (in the Question) proved the required theorem's converse instead. The theorem and its converse together show that every function has a unique decomposition into a pair of even and odd functions.

$$E(x) = \begin{cases} \frac{0+x}{2}=\frac{x}{2}\text{ if } x<0 \\ \frac{x+0}{2}=\frac{x}{2}\text{ if } x \geq 0 \end{cases}$$ Isn't the function $E(x)=\frac{x}{2}$ odd?

Correction: $$E(x) = \begin{cases} \frac{0+(-x)}{2}=-\frac{x}{2}\text{ if } x<0 \\ \frac{x+0}{2}=\frac{x}{2}\text{ if } x \geq 0 \end{cases}.$$

$$O(x) = \begin{cases} \frac{0-x}{2}=-\frac{x}{2}\text{ if } x<0 \\ \frac{x-0}{2}=\frac{x}{2}\text{ if } x \geq 0 \end{cases}$$

Correction: $$O(x) = \begin{cases} \frac{0-(-x)}{2}=\frac{x}{2}\text{ if } x<0 \\ \frac{x-0}{2}=\frac{x}{2}\text{ if } x \geq 0 \end{cases}.$$