Every function on reals a sum of two surjective real functions?

Question

From this question, and the answer thereof, we can see that every real valued function on reals is a sum of two injective functions. Is the same true if we replace injectivity by surjectivity.

For continuous functions, we could use weierstrass polynomial approximation to get our desired answer (I hope), or some harmonic analysis. For discontinuous, much worse, for non-measurable functions, is it possible?

Extending this, is it possible to write a function as a sum of two bijective functions. Any hints?

Answer 1

Here is a simple proof that every function is the sum of two surjective functions.

Pick any onto functions $g:[0, \infty) \to \mathbb R$ and $h: (-\infty, 0) \to \mathbb R$. Extend them to onto functions $\mathbb R\to\mathbb R$ as follows: $$ g(x)= f(x)-h(x) \qquad \forall x < 0\\ h(x)= f(x)-g(x) \qquad \forall x \geq 0\\ $$

Note: It is actually possible to show any function $f$ can be written as $f=g+h$, where $g,h$ are surjective and satisfy the Intermediate Value Property. This can be done via the standard argument that each function on $\mathbb R$ can be written as the sum of two functions with the IVP.

Answer 2

The bijective case is impossible. As a counterexample consider

$$ f(x)= \begin{cases} 1 & \text{if } x = 0 \\ 0 & \text{otherwise.} \end{cases} $$

Now consider the $x$ such that $h(x)=-g(0)$.

$x\neq 0$ since otherwise $f(0)=g(0)+h(0)=g(0)-g(0)=0≠1$.

Thus

$$f(x)=g(x)+h(x)=0$$

$$g(x)=-h(x)$$

$$g(x)=g(0)$$

But this implies $x=0$ by bijectivity of $g$. We have a contradiction.

Answer 3

The answer is yes.

Theorem. Every function $f:\newcommand\R{\mathbb{R}}\R\to\R$ is the sum of two surjective functions, $f=g+h$. Indeed, we can find such $g$ and $h$ that are surjective on every nontrivial interval.

Proof. Enumerate the reals in order type continuum $\R=\{x_\alpha\mid\alpha<\frak{c}\}$. We define $g$ and $h$ in stages, so that at every stage, fewer than continuum many values have been specified. To ensure surjectivity, we look at $x_\alpha$, and define $g(a)=x_\alpha$ for some $a$ for which $g(a)$ is not yet specified, and $h(a)=f(a)-g(a)$, and similarly $h(b)=x_\alpha$ and $g(b)=f(b)-h(b)$ for some $b$ similarly not yet used. And we also make sure $g(x_\alpha)$ and $h(x_\alpha)$ are defined, if they haven't yet been, by using $f(x_\alpha)/2$ for each. At each stage, the domains of the approximations to $g$ and $h$ are the same.

This defines two functions $g$ and $h$, and they will both be surjective, because at stage $\alpha$ we specifically placed $x_\alpha$ into their ranges, and we will have $g(x)+h(x)=f(x)$ for every $x$, since we ensured that property every time we defined another value of $g$ or $h$.

A simple modification of the argument will enable $g$ and $h$ to be surjective even when restricted to any interval, no matter how small. The reason is that we can enumerate the $x_\alpha$ along with rational intervals, and find $a$ and $b$ within the desired interval. $\Box$

The argument uses the axiom of choice, but perhaps there is an effective construction as in the injectivity question.