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The hint given is that we can assume the set of real numbers has a basis. I've seen several other questions of this form, but all of them use terminology which I am unfamiliar with ($Q$-linear, extend linearly to $\mathbb{R}$ etc).

Edit : I know what a Hamel basis is, and I don't have to use the axiom of choice to show that it exists or anything. I'm essentially told that I can assume it exists.

John
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3 Answers3

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Consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$ and let $H$ be a Hamel basis. Then for any real $x$ there exist a unique finite collection (the cardinality depends on $x$) $h_1,\dots,h_n\in H$ and $q_1,\dots,q_n\in \mathbb{Q}$ such that $$x=\sum_{k=1}^nq_k h_k.$$ Now for $h\in H$, define $f(h)\in\mathbb{R}$ in some way, but be sure that $f(g_1)/f(g_2)\not=g_1/g_2$ for at least two distinct $g_1,g_2\in H$ (this will be used to prove discontinuity). Then for all $x\in \mathbb{R}$, let $$f(x):=\sum_{k=1}^nq_k f(h_k).$$ Now verify that $f$ is additive and show that $f$ is not continuous at the origin.

P.S. Since $f(g_1)/f(g_2)\not=g_1/g_2$, there exist $\epsilon>0$ and rational sequences $(q_n)_n$, $(p_n)_n$ such that $|f(q_ng_1)-f(p_ng_2)|=|q_nf(g_1)-p_nf(g_2)|>\epsilon$ and $\lim_{n\to\infty}|q_ng_1-p_ng_2|=0$.

Robert Z
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  • Do you mean $f(h)$ must be defined in an intelligent way, or will any $f(h)$ work? Also, I've no idea how to show this is discontinuous.. – John Feb 15 '18 at 07:25
  • @Saad See my revised answer. – Robert Z Feb 15 '18 at 07:27
  • How do we know what $h$ is? – John Feb 15 '18 at 07:30
  • We just need that a Hamel basis EXISTS (this is implied by the axiom of choice). – Robert Z Feb 15 '18 at 07:42
  • Okay, so I can take $f(g_1) = 0$, $f(g_2) = 1$. Then as none of the $h$ in $H$ can be 0 (else linearly dependent), $f$ is of the desired form. We have that $f(0)$ = 0, but how do I show that it's discontinuous? – John Feb 15 '18 at 07:49
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    Here is an argument which makes Robert's construction discontinuous and does not use the fact that continuous additive functions are multiples of the identity: instead of the condition imposed on $f(g_1)/f(g_2)$ observe that H is uncountable. Otherwise $\mathbb R$ would become countable. Any uncountable set has a fi nite limit point. Let ${h_n}$ be a convergent sequence in H. When you define f on H take $f(h_n)=n$. Then f is not continuous because $\lim f(h_n)=\infty$ – Kavi Rama Murthy Feb 15 '18 at 07:53
  • I don't know what a finite limit point is, and this is a linear algebra class, so I'm not sure we're just allowed to use that.. – John Feb 15 '18 at 07:58
  • For (dis)continuity, you need know what a limit point is... – Robert Z Feb 15 '18 at 08:11
  • Sorry, I don't see how the last sentence follows from the way we defined $f$ – John Feb 15 '18 at 21:32
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Well, not disregarding the other answers, I suggest that it is better understood in the following order. Firstly, there is a

Lemma Let $f:\mathbb{R}\to \mathbb{R}$ additive ($f(x_1+x_2)=f(x_1)+f(x_2)$ identically) and continuous at zero, then $f$ is linear i.e. $$ (\exists a\in \mathbb{R})(f(x)=a.x) $$

Then assuming the existence of a Hamel $\mathbb{Q}$-basis $(b_i)_{i\in I}$ of $\mathbb{R}$, remark that the indexing set $I$ cannot be finite (otherwise $\mathbb{R}$ would be countable, or if $I=\emptyset$, $\mathbb{R}$ would be a singleton !) and pick $i_0\not= i_1$ in $I$. Then the mapping defined by $f(b_i)=b_i$ if $i\notin \{i_0,i_1\}$ and $f(b_j)=b_{1-j}$ (exchange $i_0,i_1$) is discontinuous. The final argument goes as follows.

If $f$, constructed as above, were continuous, we would have $f(x)=ax$ for some $a\in \mathbb{R}$. From the fact the basis is not finite, one can choose $i_2\notin\{i_0,i_1\}$ and by definition $f(b_{i_2})=b_{i_2}$ which entails $a=1$ ($b_{i_2}\not=0$ as element of a basis). Now (still supposing $f$ continuous) $f$ must be the identity which contradicts $f(b_{i_0})=b_{i_1}$. Hence this $f$ is not continuous.

Final remark The lemma shows that among additive functions $\mathbb{R}\to \mathbb{R}$ the class of ($\mathbb{R}$-) linear functions is exactly the same as the class of continuous ones. That's why answers of this question are the same even if their wording is different.

Duchamp Gérard H. E.
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Let $f:\mathbb R\to \mathbb R$ be additive. The form any integer $n$ you get $f(n)=f(1+\ldots+1)=nf(1)$ so that values on integers are determined by value at $1$. Furthermore, for any integer $q$ you also have:

$$ f(1)=f(\frac 1 q+\ldots+\frac 1 q)=f(\frac 1 q)+\ldots+f(\frac 1 q)=qf(\frac 1 q) $$

from which you get that $f(\frac 1 q)=\frac 1 q f(1)$ and putting the two together you end up with $$ f(\frac p q)=\frac p q f(1) $$ which tells you that this function is determined by its values at $1$. This means exactly that the function is $\mathbb Q$-linear. Now if this function is also continuous, once its value on $\mathbb Q$, which is dense in $\mathbb R$ is given all its values regimen and by continuity $f(a)=af(1)$ for every real $a$. But if the function is not continuous then if you want to extend it to $\mathbb R$ you can do it in a variety of possible ways, essentially depending on a choice of a basis of $\mathbb R$ as a $\mathbb Q$-vector space.

N. Ciccoli
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