$f:\mathbb{R}\to\mathbb{R}$ is a function such that for all $x,y$ in $\mathbb{R}$, $f(x+y)=f(x)+f(y)$. If $f$ is continuous, then of course it has to be linear. But here $f$ is NOT continuous. Then show that the set $\big\{\big(x,f(x)\big) : x \text{ in } \mathbb{R}\big\}$ is dense in $\mathbb{R}^2$.
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2$\begingroup$ A proof is given e.g. in Functional equations in several variables By J. Aczél, Jean G. Dhombres p.14. $\endgroup$ – Martin Sleziak Apr 13 '12 at 7:29
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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.
answered Apr 13 '12 at 6:51
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$\begingroup$ It is $\mathbb Q$-linear, but you have to prove it! $\endgroup$ – Mariano Suárez-Álvarez Apr 13 '12 at 7:05
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$\begingroup$ 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 $(0,0)$. $\endgroup$ – copper.hat Apr 13 '12 at 7:15
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1$\begingroup$ @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. $\endgroup$ – t.b. Apr 13 '12 at 8:43
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1$\begingroup$ @MarianoSuárez-Alvarez: Utilizing the linearly independent points is very slick. My functional analysis is very much at the introductory level. $\endgroup$ – copper.hat Apr 13 '12 at 22:36
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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.
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$\begingroup$ as we know if f is discontinous the kernel f must be dense, so how could it be capital delta dense? $\endgroup$ – Marso Apr 13 '12 at 6:49
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$\begingroup$ 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 $\mathbb{R}$. $\endgroup$ – copper.hat Apr 13 '12 at 7:04
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$\begingroup$ 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 $\endgroup$ – Dave L. Renfro Apr 13 '12 at 15:19
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1$\begingroup$ Not that it matters, but why unaccept the answer after, what, almost 8 years? $\endgroup$ – copper.hat Feb 14 '18 at 15:14
