video

$\gdef\pa{\partial}$

Last time we have given the definition what it means for a map $f: U \to \R^m$ where $U \In \R^n$ to be differentiable at a point $p$, and we proved a bunch of nice properties about $(Df)_p$, like linear dependence on $f$, chain rules etc.

• definition of higher order derivatives. $C^r$ class means $r$-th derivative exists and it is continuous.
• skip thm 15
• thm 16: 2nd order derivatives is symmetric. $\pa_i \pa_j f = \pa_j \pa_i f$. (of course, you say,…., but why?)

If $X$ is complete and $f: X \to X$ is a contraction, namely, there is $0\leq c<1$, such that $d(f(x), f(y)) < c d(x,y)$, then $f$ has a unique fixed point $p$. and iterates of $f$ with any starting point tends to $p$.

Let $f: U \to \R^m$ be a $C^r$ function, where $U \In \R^n \times \R^m$. If for any $(x_0,y_0) \in U$, the $B = \pa f(x_0,y)/\pa y$ is invertible, then near $(x_0, y_0)$, the level set $\{(x,y): f(x,y) = f(x_0, y_0)\}$ is the graph of a function $g(x)$ from a neighborhood of $x_0$ to a neighorbhood of $y_0$.

Take linear approximation of $f$ near $(x_0, y_0)$, then define $g(x)$ as solution to some equation. Find solution by contraction principle. Need to show $g(x)$ is differentiable at $x_0$(hard). Then, same argument shows $g(x)$ is differentiable on the level set near $x_0$. We get explicit expression about $(Dg)_p$. Finally, upgrade the regularity of $(Dg)_p$ to $C^{r-1}$.