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Let us consider a bounded, Borel function $F\colon \mathbb R^d \to \mathbb R^d$. Assume it satisfies the following Osgood-like condition: $$\tag{O} \boxed{\langle F(x) - F(y), x-y \rangle \le \Vert x-y \Vert \rho( \Vert x-y \Vert)} \qquad \forall x,y \in \mathbb R^d, $$ 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)?

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)?

Let us consider a bounded, Borel function $F\colon \mathbb R^d \to \mathbb R^d$. Assume it satisfies the following Osgood-like condition: $$\tag{O} \boxed{\langle F(x) - F(y), x-y \rangle \le \Vert x-y \Vert \rho( \Vert x-y \Vert)} \qquad \forall x,y \in \mathbb R^d, $$ 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)?

Let us consider a bounded, Borel function $F\colon \mathbb R^d \to \mathbb R^d$. Assume it satisfies the following Osgood-like condition: $$\tag{O} \boxed{\langle F(x) - F(y), x-y \rangle \le \Vert x-y \Vert \rho( \Vert x-y \Vert)} \qquad \forall x,y \in \mathbb R^d, $$ 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)?

Let us consider a bounded, Borel function $F\colon \mathbb R^d \to \mathbb R^d$. Assume it satisfies the following Osgood-like condition: $$\tag{O} \boxed{\langle F(x) - F(y), x-y \rangle \le \Vert x-y \Vert \rho( \Vert x-y \Vert)} \qquad \forall x,y \in \mathbb R^d, $$ 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 $d \ge 2$. Some sparse thoughts: condition $(O)$ immediately yields $$ \left \vert \left\langle F(x) - F(y), \frac{x-y}{\Vert x - y \Vert} \right\rangle \right\vert \to 0 $$ as $x \to y$, however I am not able to derive anything useful from this. I honestly believe that the answer to Q is false, but I am not able to work out a counterexample. Thanks.

Let us consider a bounded, Borel function $F\colon \mathbb R^d \to \mathbb R^d$. Assume it satisfies the following Osgood-like condition: $$\tag{O} \boxed{\langle F(x) - F(y), x-y \rangle \le \Vert x-y \Vert \rho( \Vert x-y \Vert)} \qquad \forall x,y \in \mathbb R^d, $$ 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)?