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\newcommand{\defn}{\bigskip\par Define~}
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\newcommand{\ex}{\bigskip\par\noindent Example.~~}

\begin{document}

\begin{flushright}
  A. Miller\\
  M521\\
  May 11, 1994
\end{flushright}

\defn $f_n$ converges to $f$ pointwise.

\defn $f_n$ converges to $f$ uniformly.

\ex Give an example of a sequence of functions $f_n$ which converges
    to a function $f$ pointwise, but not uniformly.

\thm If $f_n:D\arrow \reals^m$ converges uniformly to a $f:D\arrow \reals^m$
 and each $f_n$ is continuous, then $f$ is continuous.

\thm (Weierstrass) There exists a function $f:\reals\arrow \reals$ which
is continuous, but not differentiable at any $x\in \reals$.

\thm (Peano) There exists a continuous function $f:I\arrow I^2$ which
maps $I=[0,1]$ onto the unit square $I^2$.


\defn ${{\partial f}\over{\partial x}}(p_0)$

\defn $Df(p_0)$

\defn $f$ is $C^\prime$.

\defn $f$ is differentiable (for a function mapping an open subset
of $\reals^d$ into $\reals^m$). (p.345)

\lem Suppose ${{\partial f}\over{\partial x}}$
${{\partial f}\over{\partial y}}$ exist for every $p\in B(p_0,\delta)$.
Then
$\forall p\in B(p_0,\delta)\;\exists p_1,p_2\in B(p_0,\delta)$ such
that
$$f(p)-f(p_0)={{\partial f}\over{\partial x}}(p_1)\triangle x +
              {{\partial f}\over{\partial y}}(p_2)\triangle y $$

\thm If $f$ is $C^\prime$, then $f$ is differentiable.

\defn the directional derivative $D_uf(p_0)$.

\thm If $f$ is differentiable and $u$ a unit vector, then
$$D_uf(p_0)=Df(p_0)\bullet u.$$

\defn $||A||$ for $A$ a $m\cross d$ matrix.

\thm $||Ap||\leq ||A||\; ||p||$ for every $p\in \reals^d$.

\thm If $f$ is differentiable, then $f$ is continuous (for a function
mapping an open subset of $\reals^d$ into $\reals^m$).

\ex Give an example of a function $f:\reals^2\arrow\reals$ such
that for every $p\in\reals^2$ both ${{\partial f}\over{\partial x}}(p)$
and ${{\partial f}\over{\partial y}}(p)$ exist but $f$ is not
continuous at some point (hence it is not differentiable).

\ex Give an example of a function  $f:\reals^2\arrow\reals$ such
that for every $p\in\reals^2$ both ${{\partial f}\over{\partial x}}(p)$
and ${{\partial f}\over{\partial y}}(p)$ exist, and at some
point $p$ neither partial derivative is continuous, but $f$ is
differentiable at $p$.

\thm Suppose $f:D\arrow \reals^m$ where $D\subseteq\reals^d$ is open,
$p_0\in D$, $A$ is an $m\cross d$ matrix, and $R$ defined by:
$$f(p)-f(p_0)=A\bullet (p-p_0)+ R$$
and $R$ has the property that
$$lim_{p\arrow p_0}{||R||\over ||\triangle p||}=0$$
Then the partial derivatives of the component functions of
$f$ exist at $p_0$ and $Df(p_0)=A$.

\thm Chain rule. (p.136 and p.346)

\thm Suppose $g$ is $C^{(n+1)}$ on the interval $(-r,r)$, $x\in (-r,r)$,
and $R$ is defined by:
$$g(x)=g(0)+g^\prime(0)x+{{g^{(2)}(0)}\over2}x^2\cdots+
                         {{g^{(n)}(0)}\over {n!}}x^n + R.$$
Then for some $\zeta$ strictly between $x$ and $0$
$$R={{g^{(n+1)}(\zeta)}\over {(n+1)!}}x^{n+1}.$$


\thm If $f$ is $C^{(n+1)}$ on some $B(p_0,r)$ for $p_0\in \reals^2$,
$p\in B(p_0,r)$, and $R$ is defined by:
$$f(p)=\sum^{n}_{k=0}
{1\over{k!}}(\triangle x{{\partial }\over{\partial x}}+
\triangle y{{\partial }\over{\partial y}})^kf(p_0)\;\; + \;\;R.$$
Then for some $q$ on the line segment joining $p$ to $p_0$
$$R= {1\over{(n+1)!}}(\triangle x{{\partial }\over{\partial x}}+
\triangle y{{\partial }\over{\partial y}})^{n+1}f(q).$$

\thm Min-Max-Saddle point tests, see page 156-157.


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