10
$\begingroup$

$f:\mathbb{R}\rightarrow\mathbb{R}$ is a function such that for all $x,y$ in $\mathbb{R}$, $f(x+y)=f(x)+f(y)$. If $f$ is cont, then of course it has to be linear. But here $f$ is NOT continous. Then show that the set $\{{(x,f(x)) : x {\rm\ in\ } \mathbb{R}\}}$ is dense in $\mathbb{R}^2$.

$\endgroup$
  • 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
  • 1
    $\begingroup$ Did you replace the question? Please don't do that. Ask a new question instead. $\endgroup$ – Asaf Karagila Apr 13 '12 at 9:49
  • $\begingroup$ oops i am sorry!! $\endgroup$ – Ding Dong Apr 13 '12 at 9:54
  • $\begingroup$ oops! I made a mistake, I am so sorry! thank you. $\endgroup$ – Ding Dong Apr 13 '12 at 9:55
  • 3
    $\begingroup$ Don't just add a new, unrelated question to this question. Ask a new, separate question by clicking "ask question" at the top of the page. $\endgroup$ – Chris Eagle Apr 13 '12 at 10:06
16
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Is our f linear? $\endgroup$ – Ding Dong Apr 13 '12 at 7:04
  • $\begingroup$ It is $\mathbb Q$-linear, but you have to prove it! $\endgroup$ – Mariano Suárez-Álvarez Apr 13 '12 at 7:05
  • $\begingroup$ 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
  • 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
  • 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
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ as we know if f is discontinous the kernel f must be dense, so how could it be capital delta dense? $\endgroup$ – Ding Dong Apr 13 '12 at 6:49
  • $\begingroup$ would you please elaborate? $\endgroup$ – Ding Dong Apr 13 '12 at 6:51
  • $\begingroup$ 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
  • $\begingroup$ 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
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.