I'm speaking only about irregular solutions. Thank you.
I know it's simple to construct the solution with a root in any given point x#0 (actually -x is a root too).
What do you mean by "irregular"? What sorts of domains and codomains are you using? If you restrict to linear spaces, you are focusing on bijective linear transformations. Using groups, you almost have isomorphisms. These are all fairly regular.
Your example "regular" bijection $f(x)=-x $ is continuous. Look up automorphisms of real numbers. But no, the only field automorphisms on the reals are the identity and the negative function. For order-preserving, it is just the identity.
Also, your terminology was not terribly clear, so if you were referring to automorphisms only on the additive group of the reals, that is very different, and there are many such automorphisms:
https://math.stackexchange.com/quest...sms-of-mathbbr
At first I thought it would suffice to use elementary math to prove or refute the existence of such bijection. But now I don't, so it's time to try using something more advanced.
Anyway, thank you for participation. In case of success I'll be back and put the solution here.
Suppose $f:\mathbb{R} \to \mathbb{R}$ is a bijection such that $\forall x,y \in \mathbb{R}, f(x+y) = f(x)+f(y)$. Then
$f(0) = f(0+0) = f(0)+f(0)$
Subtract $f(0)$ from both sides to get:
$0 = f(0)-f(0) = f(0) + f(0)-f(0) = f(0)$
So, $f(0) = 0$.
Next, suppose $f(1) = c \in \mathbb{R}$. Let $n\in \mathbb{N}$.
$f(n) = f\left( \underbrace{1+1+\cdots + 1}_{n\text{ times}} \right) = \underbrace{f(1)+f(1)+\cdots + f(1)}_{n\text{ times}} = nf(1) = cn$
$0 = f(0) = f(n+(-n)) = f(n)+f(-n) = nc+f(-n)$
Subtracting $nc$ from both sides gives $f(-n) = -nc$.
So, now we have for any $z\in \mathbb{Z}$, $f(z) = cz$.
Next, let's look at rational numbers. For any $a,b\in \mathbb{Z} $ such that $b\neq 0$, consider $f\left( \dfrac{a}{b} \right)$:
$ca = f(a) = f\left( b \dfrac{a}{b} \right) = f\left( \underbrace{ \dfrac{a}{b} + \dfrac{a}{b} + \cdots + \dfrac{a}{b} }_{b\text{ times}} \right) = \underbrace{f\left( \dfrac{a}{b} \right) + f\left( \dfrac{a}{b}\right) + \cdots f\left( \dfrac{a}{b} \right) }_{b\text{ times}} = bf\left( \dfrac{a}{b} \right)$
Since $b \neq 0$, we can divide both sides by $b$ to get:
$f\left( \dfrac{a}{b} \right) = \dfrac{a}{b}c$
Next, we can consider irrational numbers. This is where things get tricky. Suppose we consider the decimal expansion of $x\in \mathbb{R}\setminus \mathbb{Q}$ to be $x = \displaystyle \sum_{k\ge m} d_k10^{-k}$. For example, $\displaystyle \pi = \sum_{k\ge 0}d_k 10^{-k}$ where $d_0=3, d_1=1, d_2=4,d_3=1,d_4=5,d_5=9, \ldots$.
Now, we cannot guarantee that $\displaystyle f\left( \lim_{n \to \infty} \sum_{k=m}^n d_k10^{-k} \right) = \lim_{n \to \infty} f\left( \sum_{k=m}^n d_k 10^{-k} \right)$.
That is where things become problematic.
To continue where I left off, we can partition the reals. Define an equivalence relation ~ on the irrational numbers. For any irrational numbers $x,y \in \mathbb{R}\setminus \mathbb{Q}$, $x\sim y$ if and only if $\exists q\in \mathbb{Q}$ such that $y=qx$. It is easy to see that this is an equivalence relation. Let $\mathcal{B}$ be a basis for this partition. For each $b \in \mathcal{B}$, we have $f(b) = c_b\in \mathbb{R}$ and for any $q\in \mathbb{Q}, f(qb) = qc_b$.
That completes the bijection. So, there are uncountably many "irregular" bijections.
OK, the relation ~ is an equivalence on irrationals: it's reflexive, symmetric and transitive.
But I don't see how the existence of "nontrivial" (i.e. discontinuous) bijection $\displaystyle f()$ follows from the equivalence you've built.
You are correct. It doesn't. I have been thinking about it and I realized that I need to treat the real numbers as a vector space. We need a rational basis for the reals. You form a rational basis as follows:
Let $\mathcal{B}_1=\{1\} $. Obviously, the span of $\mathcal{B}_1$ over rational numbers is $\mathbb{Q}$.
Next, choose any $x\in \mathbb{R}\setminus \mathbb{Q} $. Let $\mathcal{B}_2=\{1,x\} $. It is easy to show by a Cantor Diagonalization argument that the rational span of $\mathcal{B}_2$ is not bijective with the reals.
Using the Axiom of Choice, we can choose a rational basis (following the same procedure uncountably many times). A bijection requires $f $ send every basis element to a different subspace (a bijection of subspaces). Doing this rigorously is tedious, but hopefully obvious how one would go about setting it up.
A sketch of proof (we're looking only for "irregular", or discontinuous, solutions of $\displaystyle f(x+y)=f(x)+f(y)$):
1)
If our solutions are based on Hamel basis on $\displaystyle \mathbb{R}$ over $\displaystyle \mathbb{Q}$, so that $\displaystyle \underline{f(x) \in \mathbb{Q}}$ at any $\displaystyle x \in \mathbb{R}$, then the bijection sought-for is impossible: the map $\displaystyle x \mapsto f(x)$ can't be a bijection, as the cardinalily of $\displaystyle x$ set, which is $\displaystyle 2^{\aleph_0}$, is strictly more than the cardinality of $\displaystyle f(x)$ set, or $\displaystyle \aleph_0$. It's the first part of the proof.
2)
The second part is a bit more tedious. Let's call the solutions specified in 1) as the "primary" solutions (based on Hamel basis on $\displaystyle \mathbb{R}$ over $\displaystyle \mathbb{Q}$, so that $\displaystyle \underline {f(x) \in \mathbb{Q}}$ at any $\displaystyle x \in \mathbb{R}$).
It's obvious that if $\displaystyle f(x), f_i(x)$ are "primary" solutions of Cauchy equation, then $\displaystyle f(\alpha x)$ and $\displaystyle \sum_i^{N<\infty} \alpha_i f_i(x)$ are the solutions too, if $\displaystyle \alpha, \alpha_i \in \mathbb{R}$. These solutions aren't "primary", because the range of the functions isn't necessary the $\displaystyle \mathbb{Q}$ itself. On the other hand, these operations on the "primary" solutions do not alter the cardinality of the range of the function, so it is the same $\displaystyle \aleph_0$. So the bijection is impossible here too. $\displaystyle \blacksquare$
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I understand that there is smth left unsaid in the second part of the "proof".
Absolutely true (I deliberately deleted your "No" from the quote).
But it's not what I'm speaking about in the "proof". Please look at the underlined clause. It's an essence of the discontinuous solution built with Hamel basis: the range of values of function $\displaystyle f$ is some injection in $\displaystyle \mathbb{Q}$, and so is true for any element of Hamel basis $\displaystyle \mathcal{B}$.
(Actually, as I see now, the second part is not true. But now I definitely know how to construct the bijection, and I'll put it shortly.)
Here we're speaking about any solution $\displaystyle f()$ of Cauchy equation, if not stated otherwise.
When we speak about "primary" solutions (in the Theorem) we mean the discontinuous solutions $\displaystyle f()$ constructed with Hamel basis such that $\displaystyle \forall x \in \mathbb{R}$ we have $\displaystyle f(x) \in \mathbb{Q}$.
Lemma 1. $\displaystyle f()$ is a bijection $\displaystyle \iff \exists ! x: f(x)=0$.
Proof:
$\displaystyle \Rightarrow$: Obvious.
$\displaystyle \Leftarrow$: For $\displaystyle \forall x,y \in \mathbb{R}: x \ne y$ we have $\displaystyle f(y)=f(x)+f(y-x) \ne f(x)$, because $\displaystyle f()$ has the only one "trivial" root $\displaystyle x=0$. So $\displaystyle f()$ is bijective. $\displaystyle \blacksquare$
Lemma 2. The functions $\displaystyle f(x)$ and $\displaystyle g(x)=f(x)+x$ have no common roots except for the only "trivial" root $\displaystyle x=0$.
Proof:
Both $\displaystyle f()$ and $\displaystyle g()$ are the solutions of Cauchy equation if $\displaystyle f()$ is a solution.
If there exists a non-trivial root for $\displaystyle f()$ such that $\displaystyle f(x_0)=0, x_0 \ne 0$, then $\displaystyle g(x_0)=f(x_0)+x_0=x_0 \ne 0$, so that $\displaystyle x_0$ is not the root of $\displaystyle g()$.
And, vice versa, if $\displaystyle g()$ has a non-trivial root $\displaystyle x_0 \ne 0: g(x_0)=0$, then $\displaystyle g(x_0)=f(x_0)+x_0=0$, and therefore $\displaystyle f(x_0)=-x_0 \ne 0$. So $\displaystyle x_0$ is not the root of $\displaystyle f()$. $\displaystyle \blacksquare$.
Theorem. If $\displaystyle f(x)$ is a "primary" solution of Cauchy equation, and $\displaystyle g(x)=f(x)+x$ (as in Lemma 2), then $\displaystyle \forall \alpha, \beta \in \mathbb{R}$ that are linearly independent over $\displaystyle \mathbb{Q}$ the function $\displaystyle h(x)=\alpha f(x) + \beta g(x)$ is bijective.
Proof:
First, it's obvious that $\displaystyle h()$ is a solution of Cauchy equation, if $\displaystyle f()$ is a solution. Then there are three cases which are to be analyzed:
- some $\displaystyle x_0 \ne 0$ is a nontrivial root of $\displaystyle f()$: then $\displaystyle h(x_0)=\alpha f(x_0) +\beta g(x_0)=\beta g(x_0) \ne 0$, because of Lemma 2 ($\displaystyle x_0$ is the non-trivial root of $\displaystyle f(x)$ so it isn't the root of $\displaystyle g(x)=f(x)+x)$.
- some $\displaystyle x_0 \ne 0$ is a nontrivial root of $\displaystyle g()$: then $\displaystyle h(x_0)=\alpha f(x_0) +\beta g(x_0)=\alpha f(x_0) \ne 0$, because of Lemma 2.
- some $\displaystyle x_0 \ne 0$ is not a root of both $\displaystyle f()$ and $\displaystyle g()$: then $\displaystyle h(x_0)=\alpha f(x_0) +\beta g(x_0)=\alpha r_1+\beta r_2 \ne 0 (r_1,r_2 \in \mathbb{Q})$, because of linear independence of $\displaystyle \alpha$ and $\displaystyle \beta$. $\displaystyle \blacksquare$
The discontinuity of $\displaystyle h()$ is obvious. As for $\displaystyle \alpha$ and $\displaystyle \beta$, you may take, f.e., $\displaystyle 1$ and $\displaystyle \sqrt {2}$.
A lot of what you wrote makes no sense. Lemme 1, for example, is false. The function that tales 0 to 0 and every other element to 1 is a counterexample. You have a unique x such that $f(x)=0$ but you definitely do not have a bijection.
I don't understand why you are trying to make things so complicated. Let $\mathcal{B}_1,\mathcal{B}_2$ be Hamel bases for the reals such that $f|\mathcal{B}_1 \to \mathcal{B}_2$ is a bijection (that is the restriction of $f $ to the first basis forms a bijection to the second basis). Now, for any $b\in \mathcal{B}_1, r\in \mathbb{Q}$, we have $f(rb)=rf (b) $ (see my post above for the explanation).
Also, for any $x\in \mathbb{R} $ with $x=r_1b_1+\ldots +r_kb_k $, we have $f(x) = r_1f(b_1)+\ldots + r_kf (b_k) $
This is obviously a bijection and can easily be made discontinuous since the choice of bases is arbitrary, as is the choice of objections between bases.