A Q-linear transformation!

a) $f(0) = f(\alpha(0 + 0)) = f(0) + f(0) \Rightarrow f(0) = 2f(0) \Rightarrow f(0) = 0$ (This statement is always true)

b) $0 = f(0) = f(\alpha(x + (-x))) = f(x) + f(-x) \Rightarrow f(x) = - f(-x), \space \forall \space x \in \mathbb{R}$ (This statement is always true)

c) If $\alpha \neq 1$ , then $x = \alpha(x + (\frac{1}{\alpha} - 1)x) \Rightarrow f(x) = f(x) + f((\frac{1}{\alpha} - 1)x) \Rightarrow f((\frac{1}{\alpha} - 1)x) = 0$ . Due to $g: \mathbb{R} \to \mathbb{R}$ such that $g(x) = (\frac{1}{\alpha} - 1)x$ is a bijective function, this implies that $f = 0$ . (This statement is always true)

d) $\mathbb{R}$ is a vectorial space over $\mathbb{Q}$ , and assuming axiom of choice , $\mathbb{R}$ has a basis over $\mathbb{Q}$ (in this case, it is an uncountable basis). Let's call this basis $\{(e_i)\}_{i \in I}$ with $I$ a set of indices. In this section, we are supposing that $\alpha = 1$ . We are going to prove that if $f:\mathbb{R} \to \mathbb{R}$ fullfills $f(x +y) = f(x) + f(y)$ , then this function is a $\mathbb{Q}$ -linear transformation, thus,furthemore this function fulfills $f(q\cdot x) = qf(x), \space \forall q \in \mathbb{Q}$ and $x \in \mathbb{R}$ . Given arbitrary $i \in I$ $f(2e_i) = f(e_i) + f(e_i) = 2f(e_i) \Rightarrow f(ne_i) = nf(e_i), \space \forall n \in \mathbb{N}$ applying induction. $f(-2e_i) = f(-e_i) + f(-e_i) = -2f(e_i) \Rightarrow f(ne_i) = nf(e_i), \space \forall n \in \mathbb{Z}$ applying induction and $f$ is an odd function. $f((2/3)e_i) = f((1/3)e_i) + f((1/3)e_i) = 2f((1/3)e_i) \Rightarrow f(e_i) = 3f((1/3)e_i) \Rightarrow f((1/3)e_i) = (1/3)f(e_i)$ $\Rightarrow f((n/3)e_i) = (n/3)f(e_i), \space \forall n \in \mathbb{N}$ applying induction and keeping on like this $f(q\cdot e_i) = qf(e_i), \space \forall q \in \mathbb{Q}$ and due to the linearity of the sum, this function sattisfies $f(q\cdot x) = qf(x), \space \forall q \in \mathbb{Q}$ .

Conclusion: When $\alpha = 1$ ,with these constraints, $f$ is a $\mathbb{Q}$ -linear transformation.

Let $I = J \cup L$ with $J = \mathbb{N}$ and $L$ an uncontable set.

Defining $f(e_{1}) = 1$ and $f(e_{2}) = 2$ , then $f(x) \neq Cx$ using reductio ad absurdum and the function $y = Cx$ is a bijective function when $C \neq 0$ . ( (This statement is not always true)

e) For this section I'm going to use going the next theorems (what I'm not going to prove, but if someone wants me to prove them, he/she can ask it to me):

$\boxed{1.-}$ A linear transformation is a continuous function $\iff$ this linear transformation is continuous at $\vec{0}$

$\boxed{2.-}$ A linear transformation $g: E \to E'$ (where $E,E'$ are normed vectorial spaces with norm $|| \cdot ||$ and $|| \cdot ||'$ respectively.(In this case, the norm is the absolute value) is continuous at the vector $\vec{0}$ $\iff$ there exists a number $\beta > 0$ such that $|| g(x) || < \beta ||x||', \space \forall \space x \in E.$ .

The basis $\{(e_i)\}_{i \in I}$ can be taken such that $|e_i| \leq 1$ and we can define $f(e_j) = j, \space \forall \space j \in J \Rightarrow$ $f$ is not a continuous function because of 2 ... (This statement is not always true)