Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a vector space equipped with an inner product $\langle .,.\rangle$.

A function $f:X\rightarrow X$ is said to be monotone if, for all $x,y$, $\langle f(x)-f(y),x-y\rangle\geq 0$.

On $\mathbb{R}$, such functions can have only atmost countable number of discontinuities. Further, they have to be jump discontinuities.

Can we say something similar about other spaces? I have been trying this for $\mathbb{R}^n$, without any results so far.

To make the problem simpler for infinite dimensions, maybe consider just linear monotone operators on Hilbert spaces.

share|cite|improve this question
In higher dimensions we cannot expect "at most countable number of discontinuities". For example, consider $ f(x_1,x_2)=(x_1+x_1/|x_1|,x_2)$; $f(0,x_2)=(0,x_2)$ which is a monotone map on $\mathbb R^2$. Every point of the form $(0,x_2)$ is a point of discontinuity, and this is an uncountable set. ¶ But you could ask if the set of discontinuities is • of zero Lebesgue measure (in finite dimensions); or • of 1st category (which makes sense in general). I think the answer will be yes in both cases, and already exists in the literature, but I do not have a reference yet. – 75064 May 10 '13 at 16:59
of course! i dont know what i was thinking.... any monotone function $f$ on $\mathbb{R}$ with a discontinuity will give you a function $(x_1,x_2)\rightarrow (f(x_1),x_2)$ with uncountably many discontinuities. – Amudhan May 12 '13 at 13:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.