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

Solution 1:

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.

Solution 2:

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.

Solution 3:

$ \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.