Let $f:U\rightarrow V$ , $U$ and $V$ open subsets of $\mathbb{R}^2$, be a smooth function. Let $Jf_p$ be the jacobian of $f$ in the point $p\in U$ and set $M_p:=\sup\{|df_pv|:\|v\|=1\}$ and $m_p:=\inf\{|df_pv|:\|v\|=1\}$. If fear I'm missing something quite basic, but I can't understand why is it true $Jf_p=M_pm_p$. Can you give some hints?
Thank you!
Let $A$ be an arbitrary $2\times 2$ matrix. It is well known that $A$ has a singular value decomposition $A = U D V$, where $U,V$ are orthogonal and $D = \rm{diag}(\sigma_1, \sigma_2)$ is a diagonal matrix with nonnegative entries and $\sigma_1 \geq \sigma_2$.
Now, since orthogonal matrices have determinant $\pm 1$, we get $|\det A| = \sigma_1 \sigma_2$.
But since $U,V$ preserve norms, we also have $$ \sup_{|v|=1} |Av| =\sup |Dv| = \sigma_1. $$ If you replace sup by inf, you get$\sigma_2$.
This easily yields the claim.