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Let $\mathbb D^n$ be the closed $n$-dimensional unit disk. Let $f:\mathbb D^n \to \text{SO}(n)$ be a smooth map. Let $R \in \text{SO}(n)$ be a fixed rotation. I am trying to prove a quantitative bound in the spirit of $$ \|R-f\|_2 \ge C\|df\|_2, \tag{1}$$

where $\|\cdot\|_2$ are the appropriate $L^2$-norms, and I would like the constant $C$ to be independent of $f$. I am not sure if an inequality in this exact form should really hold;

Any non-trivial bound which will roughly imply that if $f$ "is close on average" to a single rotation, then its derivative must be "small on average" would be nice. Again, what I am really looking for is something like

$$ \int_{\mathbb D^n}|R-f(x)|^2dx \ge C\int_{\mathbb D^n} \psi(df(x))dx, $$ where $\psi:\text{Hom}(\mathbb R^n , \mathbb R^{n^2} ) \to \mathbb R$ gives some "weight" to $df(x)$.

Note that we can think of $f$ as a map $\mathbb D^n \to \mathbb R^{n^2}$, so $df(x) \in \text{Hom}(\mathbb R^n , \mathbb R^{n^2} ) $. Since $\text{Image}(df(x)) \subseteq T_{f(x)}\text{SO}(n)$, maybe we don't need $\psi$ to be defined on all $\text{Hom}(\mathbb R^n , \mathbb R^{n^2} )$, but only on the relevant subset of it which is "attainable" by $df$.

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  • $\begingroup$ Stop looking. You are wasting your time. Derivatives are not so easily bounded. You can find easy counter-examples with $n = 1$. Consider for example the function $\epsilon \sin\left(\frac x{\epsilon^2}\right)$ for very small $\epsilon$. $\endgroup$ – Paul Sinclair Mar 18 at 23:02
  • $\begingroup$ Thank you. I agree with you. $\endgroup$ – Asaf Shachar Mar 20 at 17:37
  • $\begingroup$ It is a lesson I also once had to learn. $\endgroup$ – Paul Sinclair Mar 20 at 17:39

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