# Does this Osgood-like condition imply continuity?

Let us consider a bounded, Borel function $F\colon \mathbb R^d \to \mathbb R^d$. Assume it satisfies the following Osgood-like condition: $$\tag{O} \boxed{\vert \langle F(x) - F(y), x-y \rangle\vert \le \Vert x-y \Vert \rho( \Vert x-y \Vert)} \qquad \forall x,y \in \mathbb R^d,$$ where $\rho \colon [0,1) \to \mathbb [0,+\infty)$ is an Osgood modulus of continuity, i.e. a continuous, non-decreasing function with $\rho(0)=0$ and $$\int_0^1 \frac{1}{\rho(s)} \, ds = +\infty.$$

Q. Is it true that $F$ is continuous? Or, more precisely, is it true that $F$ is equivalent, up to null sets, to a continuous function?

Beside the link above, something (mildly) related can also be found here.

Of course the case $d=1$ is trivial and the question makes sense essentially for $d \ge 2$. Thanks.

Update (the "linear" case). Quoting from here (pag. 2)

[...] in the case when the modulus $\rho$ is linear, (O) implies that the symmetric part of the distributional derivative is bounded, hence Korn’s inequality gives that $F$ is equivalent, up to Lebesgue negligible sets, to a continuous function.

I honestly do not see how to show rigorously this claim in the case when $\rho$ is linear. How to use Korn's inequality (I suspect a variant of this $L^2$ inequality is needed, with estimates in $L^\infty$ but do not know)?

• Can you prove this in the special case when $C$ is a constant function? Or perhaps separate continuity in one the variables? – erz Jan 20 '18 at 1:31
• @erz Thanks for the interest. Actually, you are right, time-dependence is not so important, I have edited the question accordingly. Do you have any hints on how to approach it? – Y.B. Jan 20 '18 at 10:06

You do not need such heavy high-tech as Korn's inequality or even Lebesgue measure theory for an elementary geometry homework. Let's say $F(0)=0$.
The first claim is that $F$ is bounded in some neighborhood of the origin. Indeed, just choose finitely many points $x_j$ such that $x_i-x_j$ span the space. Then for any $x$ in some fixed ball we have $\langle F(x)-F(x_j), x-x_j\rangle$ bounded by some constant whence $\langle F(x), x-x_j\rangle$ are bounded by a constant, whence $\langle F(x), x_i-x_j\rangle$ are bounded too and that is the end of this story.
Now suppose $x_j\to 0$ with $|F(x_j)|\ge\delta>0$. Then WLOG $F(x_j)\to u\ne 0$. Take any $x\ne 0$ close to $0$. We have $$|\langle F(x)-0,x-0\rangle|\le |x|\rho(|x|)$$ and $$|\langle F(x)-F(x_j),x-x_j\rangle|\le |x-x_j|\rho(|x-x_j|)$$ which, after passing to the limit, becomes $$|\langle F(x)-u,x\rangle|\le |x|\rho(|x|)$$ Subtracting, we get $$|\langle u,x\rangle|\le 2|x|\rho(|x|)$$ but this is absurd if $x$ is collinear with $u$ and $2\rho(|x|)<|u|$.