Prove linear functions that are not multiplications by a constant are unbounded on every interval

Question

An exercise in a book on logic and set theory is as follows:

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be such that $\forall x, y : f(x + y) = f(x) + f(y)$, but $f$ is not a multiplication by a constant. Prove that it is unbounded on every interval.

While looking for hints I've come across Herrlich's "Axiom of Choice", where (Theorem 5.5) he proves a related result that non-continuous solutions to this functional equation are non-measurable. This, coupled with continuous solutions being necessarily a multiplication by a constant and some real analysis (which is rusty in my case), should yield the required result if I understand correctly.

But considering whether the functions are measurable requires adding a lot more "structure" to $\mathbb{R}$ that the book I'm using typically considers, so perhaps there is a more concise and straightforward solution. So, my question is: is there one?

The exercise is in a section on Hamel basis, so I guess it's fine to prove this just for Hamel-related functions, like $f(x)$ being a projection of $x$ on a basis vector in a Hamel basis of $\mathbb{R}$ considered a vector space over $\mathbb{Q}$.

Answer 1

First of all we can observe that an $f$ of this kind is not continuous. In fact we can put $f(1) = a$ and from the additivity property it follows that $f(q) = aq$ for every $q \in \mathbb{Q}$. If then $f$ would be continuous, considered that $\mathbb{Q}$ is dense on $\mathbb{R}$, we could prove that $f(x) = ax$ for every $x \in \mathbb{R}$, but that is a contradiction under our hypothesis.

From that we can prove that $f$ is discontinuous in $0$. If $f$ had been continuous in $0$ then for every $x \in \mathbb{R}$ we would have: $$ \lim_{h \to 0} [f(x +h) -f(x)] = \lim_{h \to 0} f(h) = 0$$ And we would have $f$ continuous on every point, which is an absurd.

Now we can prove our thesis. Let $[a, b]$ be an interval of $\mathbb{R}$. Since $f$ is discontinuous in $0$ there exists an $\epsilon > 0$ such that for every $\delta > 0$ there exists an $x_{\delta} < \delta$ such that $f(x_{\delta}) > \epsilon$ (It could also be $f(x_{\delta}) < -\epsilon$, but the proof would be analogue and to avoid confusion let us restrict at the first case).

Now taken any $n \in \mathbb{N}$ let $c$ be a positive integer such that $c \epsilon > n - f(a)$ and put $\delta < \frac{b - a}{c}$. So we have an $0 < x_0 < \delta$ such that $f(x_0) > \epsilon$. From that it follows that $f(cx_0) > c \epsilon > n - f(a)$ and $cx_0 < b - a$.

Now we can see that $a + cx_0 \in [a,b]$ and $$f(a + cx_0) = f(a) + f(cx_0) > f(a) + n - f(a) = n$$ And we proved our proposition since for every $n \in \mathbb{N}$ we have found an $x \in [a, b]$ such that $f(x) > n$.

Hope I helped you, I know my english isn't perfect, sorry me if I did some mistake.

Answer 2

Just to simplify.

As $f$ is not continious at $0$ there exists a sequence $x_n$ which tends to $0$ with $|f(x_n)|>n.$

Indeed there exists $\varepsilon>0$ s.t. for all $\delta$, say $\delta = \frac {1}{n^3}$, there exists $x$, s.t. $|x|<\frac {1}{n^3}$ and $|f(x)|>\varepsilon$. Then for $n>\frac1\varepsilon$ and $x_n=n^2x$ we have $|x_n|<\frac1n$ and $|f(x_n)|=|f(n^2x)|=|n^2f(x)|>n^2\varepsilon>n$

It remains to drag the unboundeness near zero to any place. Let $c$ be center of interval. Then $|f(c+x_n)|= |f(c)+f(x_n)|\geqslant |f(x_n)|-|f(c)|>n-|f(c)|$ tends to $\infty$ and $c+x_n$ belong to given interval for sufficiently large $n$.