Is it possible to construct an odd function and an even function so that their sum is a constant?

Question

Is it possible to construct an odd function and an even function so that their sum is a constant (let's say in interval $(-1, 1)$)? Personally, I feel it is impossible.

Answer 1

In fact every single function can be written as the sum of an even and an odd function. Here's the construction: take a function $f: \mathbb R \longrightarrow \mathbb R$. Then we can write $f(x) = \frac{1}{2}(f(x) + f(-x)) + \frac{1}{2}(f(x) - f(-x))$. It's clear that the first summand is even and the latter is odd. In the special case of $f(x) = c$ a constant, this reduces to the trivial decomposition $c = c + 0$ as indicated by @player3236 in the comments.

EDIT: To be a bit more informative, this is actually the unique way to decompose a function into an even and an odd part. In more technical terms, let $V$ be the vector space of all functions $\mathbb R \longrightarrow \mathbb R$, and let $W_-, W_+$ be the subspaces of odd and even functions respectively. Then $V = W_- \oplus W_+$, as witnessed by the decomposition I gave above.

EDIT 2: To answer the question OP gave in the comments, I'll do all of this with some more generality. Let $A, B$ be vector spaces over the reals (more generally over any field of characteristic not equal to $2$) and let $C$ be the collection of all functions $A \longrightarrow B$. Then we can decompose $C = C_- \oplus C_+$, where $C_-$ consists of odd functions and $C_+$ consists of even functions via the same procedure: $f(a) = \frac{1}{2}(f(a) + f(-a)) + \frac{1}{2} (f(a) - f(-a))$. In the case you wrote in the comments, $A = \mathbb R^2$ and $B = \mathbb R$, and the exact same result holds with the exact same proof.