Non-existence of Almost Calibrator of Circle : Assume that $U$ is $\varepsilon$-open ball at origin in $\mathbb{R}^2$.
Prove that there does not exist the following function : $f: U\rightarrow \mathbb{R}$ s.t. $$ f(c(t))=t$$ where $c(t)=(1,0)+ (\cos\ t,\sin\ t)$ and $$ |{\rm grad}\ f|(x)\leq 1 + C\cdot {\rm dist}\ (x,c)^3 $$ for $x\in U$.
Proof : i) Polar Coordinate : Define $$ p :X:=(0,\infty)\times (\pi-\epsilon ,\pi+\epsilon )\rightarrow Y:= \mathbb{R}^2 $$ by $$ p(r,\theta)= r \bigg( \cos\ t,\sin\ t\bigg) + (1,0) $$
ii) $$ df = f_r \langle \partial_r ,\ \rangle +f_t\langle \partial_t,\ \rangle $$
Assume that $p^\ast \alpha =df$ so that $$ \alpha =A \langle V ,\ \rangle +B\langle \frac{1}{r}W ,\ \rangle$$
where $V=dp\ \partial_r,\ W=dp\ \partial_t$. Hence two condition on $\alpha$ are
1) $A^2+B^2 \leq 1+C|1-r|^3$ and 2) $f(1,t)=t$
which can be translated into :
1) $f_r^2+(f_t/r)^2 \leq 1+C|1-r|^3$ and 2) $f(1,t)=t$
iii) Consider a translation on $X$ : $T(a,b)=(a+1,b+\pi)=(r,t)$. If $F=f\circ T$, then
1) $F_r^2+(\frac{F_t}{1+r})^2 \leq 1+C|r|^3$ and 2) $F(0,t)=t$.
Hence $F=rH +t$ and $$ (H+rH_r )^2(1+r)^2 + (rH_t+1)^2 \leq 1+ 2r+r^2 + C|r|^3 $$
Hence $H=rU$ so that $$r^2 (2U +rU_r)^2(1+r)^2 + (r^2 U_t+1)^2 \leq 1+ 2r+r^2 + C|r|^3 $$
Here left side has no $r$-term so that it is a contradiction.
Question : Can we prove this intuitively ? ${\rm dist}\ (x,c)^2$ is possible ?
Note that $\nabla f=c'(t)$ along $c$. Hence we assume that $f^{-1}(t) =\{ (1,0)+r(\cos\ t,\sin\ t)|0<r<2 \}$. Fix $p=(1,0)+r(\cos\ t,\sin\ t)$ for some $r<1$ and $t$. At $p$, assume that integral curve $\alpha$ satisfies $\alpha(0)=p,\ \alpha(l)\in f^{-1}(l+t)$ and ${\rm length}\ \alpha|[0,l]=lr$.
Hence $lr=_{assumption}l|\alpha'| = \frac{l}{|\nabla f|} \geq \frac{ l}{1 + C|1-r|^3 }$ which is a contradiction.