# If a rotational family is close to a single rotation must its derivative be small?

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 $$df$$.

• 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 $\epsilon$. – Paul Sinclair Mar 18 at 23:02
• Thank you. I agree with you. – Asaf Shachar Mar 20 at 17:37
• It is a lesson I also once had to learn. – Paul Sinclair Mar 20 at 17:39