Let $\mathbb H$ - Heisenberg group, $$ X=\partial_x - \frac12y\partial_t,\quad Y=\partial_y + \frac12x\partial_t,\quad T=\partial_t $$

• vector fields, and $U\subset\mathbb H$ - open set.

I am looking for function $$ f:U\to\mathbb R $$ such that derivatives $Xf$ and $Yf$ exist in $U$, but $Tf$ doesn't.

It is easy to find $f$ which doesn't have $Tf$ for a pint in $U$ (say $f=\sqrt{t}$). But we need a function such that $Tf$ does't exist for every point from $U$.

remark1: If not thinking about Heisenberg group, we could consider directional derivatives. Existence derivative along one direction shouldn't imply existence along another. But in the case those directions are changed from point to point.

remark2: Since $T=[X,Y]$, any function that is twice differentiable for X and Y will also be differentiable for $T$. So the function will be quite irregular. (Mike Benfield)

remark3: Some more sophisticated examples on "lack of regularity of functions in $C^1_{\mathbb H} (\mathbb R^3)$" http://link.springer.com/article/10.1007%2Fs002080100228. See Example 2 and 3 there on page 43 (521).

It almost obvious that there should be the example, but still I haven't saw any mention.

Let $\mathbb H$ be Heisenberg group with vector fields $$ X=\partial_x - \frac12y\partial_t,\quad Y=\partial_y + \frac12x\partial_t,\quad T=\partial_t $$ and $U\subset\mathbb H$ is an open set.

I am looking for a function $$ f:U\to\mathbb R $$ such that derivatives $Xf$ and $Yf$ exist in $U$, but $Tf$ doesn't.

It is easy to find $f$ which doesn't have $Tf$ for a point in $U$ (say $f=\sqrt{t}$). But we need a function such that $Tf$ does't exist for every point from $U$.

remark1: If not thinking about Heisenberg group, we could consider directional derivatives. Existence derivative along one direction shouldn't imply existence along another. But in the case those directions are changed from point to point.

remark2: Since $T=[X,Y]$, any function that is twice differentiable for X and Y will also be differentiable for $T$. So the function will be quite irregular. (Mike Benfield)

remark3: Some more sophisticated examples on "lack of regularity of functions in $C^1_{\mathbb H} (\mathbb R^3)$" http://link.springer.com/article/10.1007%2Fs002080100228. See Example 2 and 3 there on page 43 (521).

It almost obvious that there should be the example, but still I haven't saw any mention.

Let $\mathbb H$ - Heisenberg group, $$ X=\partial_x - \frac12y\partial_t,\quad Y=\partial_y + \frac12x\partial_t,\quad T=\partial_t $$

• vector fields, and $U\subset\mathbb H$ - open set.

I am looking for function $$ f:U\to\mathbb R $$ such that derivatives $Xf$ and $Yf$ exist in $U$, but $Tf$ doesn't.

It is easy to find $f$ which doesn't have $Tf$ for a pint in $U$ (say $f=\sqrt{t}$). But we need a function such that $Tf$ does't exist for every point from $U$.

remark1: If not thinking about Heisenberg group, we could consider directional derivatives. Existence derivative along one direction shouldn't imply existence along another. But in the case those directions are changed from point to point.

remark2: Since $T=[X,Y]$, any function that is twice differentiable for X and Y will also be differentiable for $T$. So the function will be quite irregular. (Mike Benfield)

It almost obvious that there should be the example, but still I haven't saw any mention.

Let $\mathbb H$ - Heisenberg group, $$ X=\partial_x - \frac12y\partial_t,\quad Y=\partial_y + \frac12x\partial_t,\quad T=\partial_t $$

• vector fields, and $U\subset\mathbb H$ - open set.

I am looking for function $$ f:U\to\mathbb R $$ such that derivatives $Xf$ and $Yf$ exist in $U$, but $Tf$ doesn't.

It is easy to find $f$ which doesn't have $Tf$ for a pint in $U$ (say $f=\sqrt{t}$). But we need a function such that $Tf$ does't exist for every point from $U$.

remark1: If not thinking about Heisenberg group, we could consider directional derivatives. Existence derivative along one direction shouldn't imply existence along another. But in the case those directions are changed from point to point.

remark2: Since $T=[X,Y]$, any function that is twice differentiable for X and Y will also be differentiable for $T$. So the function will be quite irregular. (Mike Benfield)

remark3: Some more sophisticated examples on "lack of regularity of functions in $C^1_{\mathbb H} (\mathbb R^3)$" http://link.springer.com/article/10.1007%2Fs002080100228. See Example 2 and 3 there on page 43 (521).

It almost obvious that there should be the example, but still I haven't saw any mention.