Let $X$ be a subset of $\mathbb{C}$ auch that if $x\in X$ then also $-x\in X$. A function $f:X\rightarrow \mathbb{C}$ is called even if $f(x)=f(-x)$, a function is called uneven if $-f(x)=f(-x)$. Show that for every function $\varphi:X\rightarrow \mathbb{C}$ there exists unique $e$ and $u$ such that $\varphi=e+u$ where $e$ is even and $u$ is uneven.
I did not understand the solution, can somebody explain me the chain of thoughts so I could derrive to the solution myself and also there are specific parts of the uniqueness proof that I could not understand, I hope somebody can explain them to me.
The existence solution is given without further comment that $e(x)=\frac{1}{2}(\varphi(x)+\varphi(-x))$ and $u(x)=\frac{1}{2}(\varphi(x)-\varphi(-x))$
I have already checked that they already fullfill every condition but I have no clue how the solution was built.
The uniquenes proof is:
The uniqueness of the decomposition $\varphi = e + u$ results from the fact that from an analogous presentation $0=e+u$ necessarily $e=u=0$ follows. Latter follows from the both identities $e(z)+u(z)=0$ and $e(z)-u(z)=e(-z)+u(-z)=0$.
I did not understand anything about the uniqueness proof. I hope that the Translation of the solution text was done correctly.
What is the analogos presentation ? Analogous to what? And what identites does he mean? And how do we know that those identities are true?
If $e_1$ and $e_2$ are even functions, and $u_1$ and $u_2$ are uneven, let us assume that $e_1 + u_1 = e_2 + u_2$. Then $e_1 - e_2 = u_2 - u_1$. $e_1 - e_2$ is an even function, and $u_2 - u_1$ is an uneven function. But we know that he only function that is both even and uneven is the constant $0$. So $e_1 = e_2$ and $u_1 = u_2$.
EDIT: Lemma: If a function $f$ is both even and uneven, then it is the constant $0$.
Proof: Let $x \in \mathbb{R}$. Then $f(-x) = f(x)$ because $f$ is even, and then $f(-x) = -f(x)$ because $f$ is uneven. So $f(x) = -f(x)$, and so $f(x) = 0$.