# Graph of discontinuous additive function is dense in $\mathbb R ^ 2$

$$f : \mathbb R \to \mathbb R$$ is a function such that for all $$x , y \in \mathbb R$$, $$f ( x + y ) = f ( x ) +f ( y ) \text .$$ If $$f$$ is continuous, then of course it has to be linear. But here $$f$$ is not continuous. Then show that the graph of $$f$$, i.e. the set $$\left\{ \big( x , f ( x ) \big) : x \in \mathbb R \right\}$$, is dense in $$\mathbb R ^ 2$$.

• A proof is given e.g. in Functional equations in several variables By J. Aczél, Jean G. Dhombres p.14. Commented Apr 13, 2012 at 7:29

Let $\Gamma$ be the graph.

If $\Gamma$ is contained in a $1$-dimensional subspace of $\mathbb R^2$, then it in fact coincides with that line. Indeed, the line will necessarily be $L=\{(\lambda,\lambda f(1)):\lambda\in\mathbb R\}$, and for all $x\in\mathbb R$ the line $L$ contains exactly one element whose first coordinate is $x$, so that $\Gamma=L$. This is impossible, because it clearly implies that $f$ is continuous.

We thus see that $\Gamma$ contains two points of $\mathbb R^2$ which are linearly independent over $\mathbb R$, call them $u$ and $v$.

Since $\Gamma$ is a $\mathbb Q$-subvector space of $\mathbb R^2$, it contains the set $\{au+bv:a,b\in\mathbb Q\}$, and it is obvious that this is dense in the plane.

• Is our f linear? Commented Apr 13, 2012 at 7:04
• It is $\mathbb Q$-linear, but you have to prove it! Commented Apr 13, 2012 at 7:05
• Slick proof. You don't need closure to show the existence of two l.i. points. Discontinuity alone ensures that the graph cannot be contained in a line through $(0,0)$. Commented Apr 13, 2012 at 7:15
• @copper.hat: Banach proved that the open mapping theorem and the closed graph theorem also hold for Polish groups (second countable and metrizable with a complete metric) and homomorphisms, so linearity could be dispensed with, but of course it is serious overkill for the present question.
– t.b.
Commented Apr 13, 2012 at 8:43
• @MarianoSuárez-Alvarez: Utilizing the linearly independent points is very slick. My functional analysis is very much at the introductory level. Commented Apr 13, 2012 at 22:36

Let $\Gamma = \{ (x,f(x)) \}_{x \in \mathbb{R}}$. First show that the set $\Delta = \{ x | f(x) \neq 0 \}$ is dense in $\mathbb{R}$. Then show that $f$ is discontinuous at $0$, and that this implies that the closure of $\Gamma$ contains $\{0\}\times \mathbb{R}$. Then show that the closure of $\Gamma$ contains $\{x\}\times \mathbb{R}$, $\forall x \in \Delta$. Presumably the result will be obvious at this point.

• as we know if f is discontinous the kernel f must be dense, so how could it be capital delta dense? Commented Apr 13, 2012 at 6:49
• would you please elaborate? Commented Apr 13, 2012 at 6:51
• If $f(x_0) \neq 0$ for some $x_0$, then since $f(q x) = q f(x)$, $\forall q \in \mathbb{Q}$, clearly $\Delta$ is dense in $\mathbb{R}$. Commented Apr 13, 2012 at 7:04
• A proof with some more details is given at the first URL. (The second URL contains a minor correction to something else in the first post.) groups.google.com/group/sci.math/msg/98d0bb02228bd4bd and groups.google.com/group/sci.math/msg/4016347301a71140 Commented Apr 13, 2012 at 15:19
• Not that it matters, but why unaccept the answer after, what, almost 8 years? Commented Feb 14, 2018 at 15:14

$$\def \Q {\mathbb Q} \def \R {\mathbb R} \def \Rp {\mathbb R _ +} \def \Rt {\mathbb R ^ 2}$$ Assume that $$f : \R \to \R$$ is additive, and we have $$f ( x _ * ) \ne k x _ *$$ for some $$x _ * \in \R$$, where $$k = f ( 1 )$$. Define $$g : \R \to \R$$ with $$g ( x ) = f ( x ) - k x$$. It's then straightforward to verify that $$g$$ is additive, $$g ( 1 ) = 0$$ and $$y _ * \ne 0$$, where $$y _ * = g ( x _ * )$$. We know that an additive function is linear over $$\Q$$, and therefore $$f$$ and $$g$$ are linear over $$\Q$$. In particular, we have $$g ( q ) = 0$$ for all $$q \in \Q$$.

Now, note that for any $$\alpha , a , b \in \Q$$, If we define $$X = \alpha + b ( x _ * - a )$$ and $$Y = g ( X )$$, we have $$Y = g \big( \alpha + b ( x _ * - a ) \big) = g ( \alpha ) + b g ( x _ * ) - b g ( a ) = b y _ * \text .$$ Therefore, given any $$\alpha , \beta \in \Q$$ and any $$r \in \Rp$$, if we choose $$a , b \in \Q$$ such that $$b \ne 0$$, $$\left| \frac \beta { y _ * } - b \right| < \frac r { \sqrt 2 | y _ * | }$$ and $$| x _ * - a | < \frac r { \sqrt 2 | b | }$$, then we will have $$d \big( ( X , Y ) , ( \alpha , \beta ) \big) < r$$, where $$d$$ is the Euclidean metric on $$\R ^ 2$$. This proves that the graph of $$g$$ is dense in $$\Rt$$, as there is a ball centered at a point with rational coordinates contained in any given ball in $$\Rt$$.

Finally, note that denseness of the graph of $$g$$ in $$\Rt$$ implies that of $$f$$, since $$x \mapsto k x$$ is a continuous function. To elaborate, assume that $$\epsilon \in \Rp$$ is given. For any $$\alpha \in \R$$, there is some $$\delta \in \Rp$$ such that for all $$X \in \R$$, if $$| X - \alpha | < \delta$$ then $$| k X - k \alpha | < \frac \epsilon 2$$. As the graph of $$g$$ is dense in $$\Rt$$, given any $$\alpha , \beta \in \R$$, there is some $$X \in \R$$ such that $$d \Big( \big( X , g ( X ) \big) , ( \alpha , \beta - k \alpha ) \Big) < \min \left( \delta , \frac \epsilon 2 \right)$$. As this implies $$| X - \alpha | < \delta$$, we have $$| k X - k \alpha | < \frac \epsilon 2$$, and therefore \begin {align*} d \Big( \big( X , f ( X ) \big) , ( \alpha , \beta ) \Big) & \le d \Big( \big( X , f ( X ) - k X \big) , ( \alpha , \beta - k \alpha ) \Big) + d \Big( \big( 0 , k X \big) , ( 0 , k \alpha ) \Big) \\ & = d \Big( \big( X , g ( X ) \big) , ( \alpha , \beta - k \alpha ) \Big) + | k X - k \alpha | \\ & < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon \text . \end {align*} Hence, we can find a point on the graph of $$f$$ in any given ball in $$\Rt$$; i.e. the graph of $$f$$ is dense in $$\Rt$$.

This proof may seem to be too much involved and unnecessarily complicated, for example in comparison with those on Wikipedia or ProofWiki. The idea is using continuity of linear functions to reduce the problem to the more restricted case of $$g$$, for which many calculations are simpler, due to the fact that its value is zero at rational points. This idea can become handy in more complicated cases, where there are many more terms to handle. For example, see my answer to the question "Paralellogram law functional equation: $$f ( x + y ) + f ( x - y ) = 2 \big( f ( x ) + f ( y ) \big)$$", in which I'm essentially trying to do the same thing with biadditive functions.