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Does there exist a function $f:\Bbb R \to \Bbb R$ such that $f$ is discontinuous at a point and satisfies $f(x+y)=f(x)+f(y)$ for all $x,y$ in $\Bbb R$?

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Yes. The reason is that we have a lot of $\mathbb{Q}$-linear independence over $\mathbb{R}$, i.e. $\mathbb{R}$ is an infinite dimensional vector space over $\mathbb{Q}$.

First note we must have $f(0) = 0$. Then we can define $f(1)$ to be anything. Once we have $f(1)$ determined, it's easy to see we must have $f(k) = kf(1)$ for $k \in \mathbb{Z}$ and then $f(q) = qf(1)$ for $q \in \mathbb{Q}$. However, there are no other restrictions this puts on $f$! Note if $f$ were continuous everywhere, then we would know what $f$ is completely, since we know what it is on rationals.

Therefore, we can define $f(\sqrt{2})$ to be whatever we want, and then once again, we can deduce what $f(x)$ is for $x \in \mathbb{Q}[\sqrt{2}]$. And then we can define $f(\alpha)$ to be whatever we want, for some $\alpha \not \in \mathbb{Q}[\sqrt{2}]$. And we can keep doing this on and on, infinitely many times since $\mathbb{R}$ has infinite dimension as a vector space over $\mathbb{Q}$.

It should be clear that the resulting $f$ need not be continuous. To see this directly, note that $\sqrt{2}$ is a limit of rational numbers so once $f$ is defined on $\mathbb{Q}$, if it were continuous, it would be defined at $\sqrt{2}$ already. But we chose $f(\sqrt{2})$ to be whatever we wanted.

The moral of the story is that continuity is more of a metric property of the reals whereas the equation $f(x+y) = f(x)+f(y)$ only uses the algebraic properties of the real line and allows freedom metrically.

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  • $\begingroup$ And the graph would be dense as well as far as I know $\endgroup$ – CoffeeCCD Apr 15 '17 at 5:21
  • $\begingroup$ @RonitDebnath It can be dense. It doesn't have to be. Can keep the image of the function inside $(0,1)$ if you want. $\endgroup$ – Stella Biderman Apr 15 '17 at 5:22
  • $\begingroup$ @RonitDebnath, $f$ can be identically $0$ so no. This is the only counter example I believe, so you're basically correct. $\endgroup$ – mathworker21 Apr 15 '17 at 5:23
  • $\begingroup$ @StellaBiderman, I don't know if that's true... I think if some $x$ satisfies $f(x) \not = 0$, then $f(qx) = qf(x)$ which is dense in $\mathbb{R}$ since $f(x) \not = 0$. $\endgroup$ – mathworker21 Apr 15 '17 at 5:23
  • $\begingroup$ Yeah sorry you are right.Maybe upto its range it is graph is dense in R2.I am talking about the horizontal strip. $\endgroup$ – CoffeeCCD Apr 15 '17 at 5:24

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