# Relation of Hamel basis with the equation $f(x + y) = f(x) + f(y)$? [duplicate]

I am reading "Linear and Nonlinear Functional Analysis with Applications by Philippe G. Ciarlet ", which explains the origin of hamel basis by a problem:

Describe the set $$F$$ of all functions: $$f: \mathbb{R} \rightarrow \mathbb{R}$$ that satisfying the functional equation $$f(x + y) = f(x) + f(y)$$ for all $$x,y\in \mathbb{R}$$

Actually I don't know how to describe this in a right way, and I always foucs on some concrete quality (for example, $$f(0) = 0, f(x) = f(-x)$$). Therefore, I don't figure out the connection between this and Hamel basis.

The connection is the following one. A linear function from $$\mathbb{R}$$ to $$\mathbb{R}$$, where $$\mathbb{R}$$ is considered as a vector space over itself, is a function $$f$$ that satisfies the following two equations: $$$$\forall x,y \in \mathbb{R},\qquad f(x+y) = f(x) + f(y)\;,\qquad (1)$$$$ $$$$\forall x,\lambda \in \mathbb{R},\qquad f(\lambda \cdot x) = \lambda \cdot f(x)\;. \qquad (2)$$$$ Notice that $$(2)$$ is equivalent to $$$$\forall x \in \mathbb{R}, \qquad f(x) = x \cdot f(1)\;. \qquad (3)$$$$ We may wonder if $$(1)$$ implies $$(3)$$. After all, if we only assume that $$(1)$$ holds, we have

• Using induction, $$\forall n \in \mathbb{N}, \forall x \in \mathbb{R}, f(n \cdot x) = f(\sum_{k=1}^n x) = \sum_{k=1}^n f(x) = n \cdot f(x)$$.
• $$\forall n \in \mathbb{N}, f(1) = f(n \cdot \frac{1}{n}) = n \cdot f(\frac{1}{n})$$, which implies $$\forall n \in \mathbb{N}, f(\frac{1}{n}) = \frac{f(1)}{n}$$.
• $$\forall m,n \in \mathbb{N}, f(\frac{m}{n}) = f(m\cdot\frac{1}{n}) = m\cdot f(\frac{1}{n}) = \frac{m}{n}\cdot f(1)$$.
• $$f(0) = f(0+0) = f(0)+f(0)$$, which implies $$f(0) = 0$$.
• $$\forall m,n \in \mathbb{N}, f(-\frac{m}{n}) = f(-\frac{m}{n})+f(\frac{m}{n})-f(\frac{m}{n}) \\= f(-\frac{m}{n}+\frac{m}{n}) -f(\frac{m}{n}) = f(0) -f(\frac{m}{n}) = 0 - f(\frac{m}{n}) = -f(\frac{m}{n}) = -\frac{m}{n}\cdot f(1)\;.$$

It follows that $$\forall q \in \mathbb{Q}, f(q) = q \cdot f(1)$$, so $$(3)$$ is "almost" implied by $$(1)$$. But there's a catch. Pick $$\sqrt{2}$$. Then for each $$q \in \mathbb{Q}$$ we have that $$q \neq -\sqrt{2}$$. This implies that $$1$$ and $$\sqrt{2}$$ are linearly independent over $$\mathbb{Q}$$. Then, we can build $$\tilde{f} \colon \mathbb{Q}[\sqrt{2}] \to \mathbb{R}$$ as the unique linear function on $$\mathbb{Q}[\sqrt{2}]$$ (as a vector space over $$\mathbb{Q}$$) such that $$\tilde{f}(1) = 1 = \tilde{f}(\sqrt{2})$$. This function satisfies $$\forall x,y \in \mathbb{Q}[\sqrt{2}], \tilde{f}(x+y) = \tilde{f}(x)+\tilde{f}(y)$$ but fails to satisfy $$\forall x \in \mathbb{Q}[\sqrt{2}], \tilde{f}(x) = x\cdot \tilde{f}(1)$$, since $$\tilde{f}(\sqrt{2}) = 1 \neq \sqrt{2} = \sqrt{2} \cdot 1 = \sqrt{2} \cdot \tilde{f}(1)$$.

Now, we may repeat the process adding another number $$\alpha \in \mathbb{R}$$ such that $$\alpha \notin \mathbb{Q}[\sqrt{2}]$$ and extending the previous function so to preserve the property $$\forall x,y \in \mathbb{Q}[\sqrt{2}, \alpha], f(x+y) = f(x)+f(y).$$ Ideally, we want to continue this process of extension adding numbers until we "fill" the whole $$\mathbb{R}$$, so as to obtain in the end a function that satisfies $$(1)$$ but not $$(3)$$.

However, it is not clear how this can be formally done.

It's here that Hamel bases come to our aid: they provide a formal way to do that.

Consider $$\mathbb{R}$$ as a vector space over the field $$\mathbb{Q}$$. Since $$1$$ and $$\sqrt{2}$$ are linearly independent over $$\mathbb{Q}$$, we can build a Hamel basis for $$\mathbb{R}$$ over $$\mathbb{Q}$$ containing them, say $$(e_i)_{i \in I}$$, where $$I$$ is some set. Then, we can define a linear function (thinking $$\mathbb{R}$$ as a vector space over $$\mathbb{Q}$$), say $$\bar{f} \colon \mathbb{R} \to \mathbb{R}$$, that extends our previously defined linear function $$\tilde{f} : \mathbb{Q}[\sqrt{2}] \to \mathbb{R}$$. We can do this specifying its values over the base $$(e_i)_{i \in I}$$. For example, $$\bar{f}(1) = 1 = \bar{f}(\sqrt{2})$$, and $$\forall i \in I \backslash \{1,\sqrt{2}\}, \bar{f}(e_i) = e_i$$.

This way, by construction $$\bar{f}$$ satisfies $$(1)$$, but we already know that $$(3)$$ is violated since $$\bar{f}$$ is an extension of $$\tilde{f}$$, so that $$\bar{f}(\sqrt{2}) = \tilde{f}(\sqrt{2}) \neq \sqrt{2} \cdot \tilde{f}(1) = \sqrt{2} \cdot \bar{f}(1)$$. Some final (exotic) remarks:

• $$\bar{f}$$ is discontinuous everywhere.
• If we impose that $$(1)$$ holds and that $$f$$ has to be continuous at some point $$x\in\mathbb{R}$$, then we can prove that $$(3)$$ (or, equivalently, $$(2)$$) holds.
• The continuous linear functions from $$\mathbb{R}$$ to $$\mathbb{R}$$ where $$\mathbb{R}$$ is considered as a vector space over $$\mathbb{Q}$$ coincide with the linear functions from $$\mathbb{R}$$ to $$\mathbb{R}$$ where $$\mathbb{R}$$ is considered as a vector space over $$\mathbb{R}$$.