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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.

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    $\begingroup$ It doesn't seem relevant that you are working in the Heisenberg group. Since $T = [X,Y]$, any function that is twice differentiable for $X$ and $Y$ will also be differentiable for $T$. If the function you're seeking does exist, it will not be very regular. $\endgroup$
    – Michael Benfield
    Commented May 3, 2015 at 22:10
  • $\begingroup$ @MikeBenfield You are right. It is (if one can say) nonsmooth analysis on Carnot group. And we assume that functions could have only first derivatives (of derivatives in weak sense). $\endgroup$
    – Nikita Evseev
    Commented May 4, 2015 at 5:20
  • $\begingroup$ If you can find a $C^1$ diffeomorphism $g:U\to V$ pulling $\partial_x,\partial_y,\partial_t$ back to $X,Y,T$, taking as $h:\mathbb R\to\mathbb R$ any continuous nowhere differentiable function, then you are done with $f=h\circ p\circ g$ where $p$ is given by $(x,y,t)\mapsto t$ . Since $X,Y,T$ are locally linearly independent, at least locally this can be done around every point. $\endgroup$
    – TaQ
    Commented May 4, 2015 at 5:55
  • $\begingroup$ Since you are not imposing any continuity condition on $f$ , you may even take $h(t)=0$ for $t\in\mathbb Q$ and $h(t)=1$ for $t\in\mathbb R\setminus\mathbb Q$ . So the only problem is the construction of a diffeomorphism $g$ . $\endgroup$
    – TaQ
    Commented May 4, 2015 at 6:08
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    $\begingroup$ @TaQ: Won't any diffeomorphism $g$ preserve the Lie bracket? So $g$ can't pull back $\partial_x, \partial_y$ to $X,Y$ since the former commute and the latter do not. I think this is true even if $g$ is only $C^1$. $\endgroup$
    – Nate Eldredge
    Commented May 4, 2015 at 6:55

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