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REAL ANALYSIS

FUNCTIONS OF SEVERAL VARIABLES

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\begin{description}

\item[]
We denote $(x_1\ldots x_2)$ by $X$ and $f(x_1\ldots x_n)$ by
$f(X)$. We may think of $X$ as a vector or a point.

If $A=(a_1\ldots a_n)\ B=(b_1\ldots B-n)$ then

$A-B=(a_1-b_1\ldots a_n-b_n)$

$A+B=(a_1+b_1\ldots a_n+b_n)$

$A.B=a_1b_1+\ldots+a_nb_n$ - a scalar

$||A||=\sqrt{a_1^2+\ldots+a_n^2}$ norm of $A$

$||A-B||=\sqrt{(1_a-a_b)^2+\ldots+(a_n-b_n)^2}$ is the distance
$AB$

$|A.B|\leq||A||\ ||B||$ Cauchy's inequality.

Suppose we have $m$ functions of $n$ variables
$^{(1)}f(X),^{(2)}f(X)\ldots^{(m)}F(X)$.

We shall denote by the vector function

$$F(X)=(^{(1)}f(x_1\ldots x_n),\ldots^{(m)}f(x_1\ldots x_n))$$

\item[Theorem 1]
If $f(X)\ g(X)$ are continuous at $A$ relative to $S$ then so are
$f(X)\pm g(X), f(X)g(X)$ and, if $g(A)\neq0, \frac{f(X)}{g(X)}$.

\item[Theorem 2]
Suppose that the components $^{(1)}f(X)\ldots^{(m)}f(X)$ of the
vector function $F(X)$ are continuous at $A$ relative to $S$. Let
$B=F(A)$ and let $T$ be the set of all points $F(X)$ with $X$ in
$S$. Then if $g(Y)=g(y_1\ldots y_m)$ is continuous at $B$ relative
to $T$, it follows that $g(F(x))$ is continuous at $A$ relative to
$S$.

\item[Differentiability]
$f(X)$ is differentiable at $X+A\Leftrightarrow\exists $ a vector
$G|\frac{f(X)-f(A)-G(X-A)}{||X-A||}\to0$ as $X\to A$.

If $f$ is differentiable then $\frac{\partial f}{\partial
x_1}\ldots\frac{\partial f}{\partial x_n}$ all exist and the
vector $G$ is $\left(\frac{\partial f}{\partial
x_1}\ldots\frac{\partial f}{\partial x_n}\right)$.

We call this $(grad\  f(X))_{X=A}$ or $(\nabla f)_{X=A}$.

Thus $f(X)$ is differentiable $\Leftrightarrow$ $\frac{\delta
f-\nabla f\delta X}{||\delta X||}\to0$ as $\delta X\to 0$.

\item[Theorem 3]
If $\frac{\partial f}{\partial x_1},\ldots\frac{\partial
f}{\partial f}{\partial x_n}$ are continuous at $X=A$, then $f(X)$
is differentiable at $X=A$.

\item[Proof]
Suppose $H\neq0$ and $||H||$ is sufficiently small. Consider

\begin{eqnarray*}
&&\frac{1}{||H||}\left|\sum_{r=1}^n\left\{f(a_1+h_1\ldots,a_r+h_r,a_{r+1}\ldots
a_n)\right.\right.\\ &&\left.\left.-f(a_1+h_1\ldots
a_{r-1}+h_{r-1}a_r\ldots a_n)-h_rf_r(A)\right\}\right|\\
&&(\textrm{This is }\frac{1}{||H||}\{f(A+h)-f(A)-A\nabla f\})\\
&\leq&\frac{1}{||H||}\left|\sum_{r=1}^n\left[\left\{f_r(a_1+h_1\ldots
a_{r-1}+h_{r-1}a_r+\theta_rf_{r_1}a_{r+1}-a_{n})-f_r(A)\right\}h_r\right]\right|
\\ && 0<\theta_r<1
\end{eqnarray*}

Let $V$ be the vector with components

$$f_r(a_1+h_1,\ldots a_r\theta_fh_r,a_{r+1}\ldots a_n)-f_rA\
(r=1,2,\ldots n)$$

Each component can be made as small as we please, provided only
that $||H||$ is sufficiently small since $f_r(X)$ is continuous at
$X+A$. Hence we can make $||V||<\varepsilon$ if $||H||$ is
sufficiently small. The above inequality then gives

$$\frac{1}{||H||}\left|f(A+h)-f(A)-\sum_{r=1}^nh_rf_rA\right|\leq\frac{|V.H|}{||H||}\leq||V||<\varepsilon.$$

\item[Theorem 4]
If $f(X)$ and $g(X)$ are both differentiable at $X+A$, then so are
$f(X)\pm g(X),\ f(X).g(X)$ and, provided $g(A)|neq0,\
\frac{f(X)}{g(X)}$.

$$\left.\begin{array}{c}\nabla(f\pm g)=\nabla f\pm\nabla g\\
\nabla(fg)=f\nabla g+g\nabla f\\
\nabla\frac{f}{g}=\frac{1}{g}\nabla f-\frac{f}{g^2}\nabla
g\end{array} \right\}\textrm{ at}X=A$$

\item[Proof of (iii)]
Take $f\equiv1$ and suppose $g(A)\neq0$. Consider

\begin{eqnarray*}
&&\frac{1}{||H||}\left|\frac{1}{g(A+h)}-\frac{1}{g(A)}-\left\{\frac{\nabla
g_A}{g^2(A)}\right\}.H\right|\\
&=&\frac{1}{||H||}\left|\frac{g(A)\{g(A)-g(A+H)+\nabla
g_A.H\}+\{g(A+H)-g(A)\}\{\nabla g_A.H\}}{g^2(A)g(A+H)}\right|\\
&\leq&\left|\frac{g(A+h)-g(A)-\nabla
g.H}{||H||}\right|\frac{1}{g(A)g(A+H)}+\frac{|g(A+H)-g(A)|}{|g^2(A).g(A+H)|}||\nabla
g|| \end{eqnarray*}

using $|\nabla g.H\\leq||\nabla g||\ ||H||\to 0$ as $||H||\to0$.

\item[Theorem 5]
Function of a function rule.

Let $^{(1)}f(X),^{(2)}f(X),\ldots^{(n)}f(X)$ be differentiable at
$X=A$. Let $g(Y)=g(y_1\ldots y_m)$ be differentiable at $Y=B$
where $B=F(A)=( ^{(1)}f(A)\ldots^{(m)}f(A))$.

Then $h(X)=g(F(x))$ is differentiable at $X+A$ and

$$\left(\begin{array}{c}h_1(x)\\ \vdots\\
h_n(X)\end{array}\right)_{X=A}=\left(\begin{array}{c}^{(1)}f_1(X)\ldots^{(m)}f_1(X)\\
\\
\\ ^1f_n(X)\ldots^mf_n(X)\end{array}\right)_{X=A}\left(\begin{array}{c}g_1(Y)\\
\vdots \\ g_m(Y)\end{array}\right)_{Y=B}$$

\item[Proof]
Let RHS of the expression be $D$. Let $(g_1(Y)\ldots
g_m(Y))^T=G'$.

We have the following results

\begin{enumerate}

\item
Since $f(X)$ is differentiable at $X+A$
$\frac{F(A+H)-f(A)}{||H||}$ is bounded for $0<||H||<\delta$.

Thus, if each component of $F(X)$ is differentiable at $X=A\
\frac{||F(A+H)-F(A)||}{||H||}$ is bounded in $0<||H||<\delta$

\item
Since $g(Y)$ is differentiable at $Y+B$

$$g(B+\Omega)=g(B)-G'.\Omega=\varepsilon(\omega)||\Omega||$$

where $\varepsilon(\Omega)\to0$ as $||\omega||to0$.

\end{enumerate}


Consider

\begin{eqnarray*}
&&\frac{1}{||H||}\left|g(F(A+H))-g(F(A))-D.H\right|\\
&=&\frac{1}{||H||}\left|g(F(A+H))-g(F(A))-G'.(F(A+H)-F(A))\right.\\
&&+\left.G'.(F(A+H)-F(A))-D.H\right|\\
&\leq&\frac{1}{||H||}\left|g(B+\Omega)-g(B)-G'.\Omega\right|+\frac{1}{||H||}\left|G'.(F(A+H)-F(A))-D.H\right|
\end{eqnarray*}

(writing $F(A)+B\ F(A+H)-F(A)=\Omega$.

First term $=\varepsilon(\Omega)\frac{||\Omega||}{||H||}$ by (2)
where $\varepsilon\to0$ as $\Omega\to0$.

By (1) $\frac{||\Omega||}{||H||}$ is bounded for $0<||H||<\delta$
also since $\Omega\to0$ as $H\to0,\ \varepsilon(\Omega)\to0$ as
$H\to0$.

Second term

\begin{eqnarray*}
&=&\frac{1}{||H||}\left|\sum_{r=1}^mg_r(B)\left\{^{(r)}f(A+H)-^{(r)}f(A)-H.\nabla^rf(A)\right\}\right|\\
&\leq&\sum_{r=1}^n|g_r(B)|\frac{1}{||H||}\left|^{(r)}f(A+H)-^{(r)}f(A)-H.\nabla^rf(A)\right|
\end{eqnarray*}

$\to0$ as $||H|||to0$.

Hence $\frac{1}{||H||}\left|g(F(A+H))-g(F(A)-D.H\right|\to0$ as
$||H|||to0$. Hence the result.

\item[Corollary]
In The special case when $n=1$ we get, when $h(x)=g(F(X))$ that

$$h'(x)=F'(a).(\nabla g)_B$$

\item[Theorem 6 First Mean Value Theorem]
Suppose $d(X)$ is differentiable at all points of the open line
segment $(A,A+H)$ and continuous on the closed segment. Then for
some %\theta$ with $0<\theta<1$ we have

$$g(A+H)-g(A)=H.\nabla g(A+\theta H)$$

\item[Proof]
Suppose $0<t_0<1$. Then $h(t)=g(A+tH)$ is differentiable at
$t=t_0$, since $g(X)$ is differentiable at $X=A+t_0H$ and so are
$a_r+t_0h_r\ r=1,\ldots,n$.

Furthermore

\begin{eqnarray*}
h'(t)&=&\left\{\frac{d}{dt}A+tH\right\}.\nabla g(A+tH)\textrm{ at
}t=t_0\\ &=&H.\nabla g(A+tH)\ t=t_0
\end{eqnarray*}

And $h(t)$ is continuous for $t$ in $[01]$ hence by MVT
$h(1)-h(0)-h'(\theta$ for some $\theta|0<\theta<1$. Hence

$$g(A+H)-g(A)=H.\nabla g(A+\theta H)$$

\item[Theorem 7 Taylor's Theorem]
Suppose that the function $f(X)$ is such that all its partial
derivatives of (total) order $u-1$ are continuous on the closed
line segment $[A,A+H]$, and differentiable on the open line
segment $(A,A+H)$, then for some $\theta$ with $0<\theta<1$, we
have

\begin{eqnarray*}
f(A+H)&=&\left\{\sum_{r=0}^{u-1}\frac{1}{r!}\Omega^rf(X)\right\}_{X=A}
+\left\{\frac{1}{u!}\Omega^Uf(X)\right\}_{X=A+\theta H}\\
\Omega&=&H.\nabla=h_1\frac{\partial}{\partial
x_1}+\ldots+h_n\frac{\partial}{\partial x_n}
\end{eqnarray*}

\item[Proof]
Write $h(t)=f(A+tH)$

By induction on $r$ we have

$$\frac{d^r}{dt^r}h(t)=\left[\Omega^rf(X)\right]_{X=A+tH}\left\{\begin{array}{c}r=0,1,\ldots
u-1\\ 0\leq t\leq 1\\ r=u\ 0<t<1\end{array}\right.$$

By Taylor's theorem for a function of one variable

$$h(1)=\sum_{r=0}^{u-1}\frac{1}{r!}h^{(r)}(0)+\frac{1}{U!}h^{(n)}\theta\
0<\theta<1$$

Hence the result.

\item[Maxima and Minima]
$f(X)$ has a strict maximum at $X=A$ means $\exists\varepsilon>0|$
in $ 0<(X-A|<\varepsilon$ we have $f(A)>f(X)$.

By a weak minimum we mean that in $0<|X_A|<\varepsilon|\ f(A)\geq
f(X)$.

\item[Theorem 8]
Suppose $f(X)$ has a maximum or a minimum at $X+A$. If $f(X)$ has
first order partial derivatives at $X=A$ then $(\nabla f)_A=0$
i.e. $f_r(A)=0\ r=1,2,\ldots,n$. If $f(X)$ has second order
derivatives continuous in a neighbourhood of $A$, then the
quadratic form $\sum_{ij}h_ih_jf_{ij}(A)$ in $h_1\ldots h_m$ is
negative or positive semi-definite.

\item[Proof]
Consider $f(x_1,x_2\ldots a_n)=\phi(x_1)$. This, as a function of
$x_1$, has a maximum or minimum at $x_1=a_1$ therefore by the
theorem for a function of one variable $\phi'(x_1)=\frac{\partial
f}{\partial; x_1}=0$. Similarly for other variables therefore
$\nabla f_A=0$.

Suppose that the quadratic form is not semi-definite. Then
$\exists U=(u_1\ldots u_n)$ such that

$$\sum_{ij}f_{ij}u_iu_j>0$$

and $V=(v_1\ldots v_n)$ such that

$$\sum_{ij}f_{ij}Av_iv_j<0$$

Let $H^1=(u_1h\ldots u_nh)$.

Using Taylor's Theorem

$$f(A+H@)=f(A)+\frac{h^2}{2!}\sum_{ij}f_{ij}(A+\theta^1H^1)u_iu_j$$

The linear terms vanishing as $\nabla f=0$

Let $H^2=(v_1h\ldots v_nh)$

$$f(A+H^2)=f(A)+\frac{h^2}{2!}\sum_{ij}f_{ij}(A+\theta^1H^2)v_iv_j$$

Since the second derivatives are continuous, $\exists\delta>0|$
for $0<h<\delta$

\begin{eqnarray*}
f(A+H^1)&=&f(A)+(\phi^1)^2\\ f(A+H^2)&=&f(A)-(\phi^2)^2
\end{eqnarray*}

We get neither a maximum nor a minimum since in any sphere of
radius $\varepsilon$ about $A$, we can choose $h$ such that
$|H^1|<\varepsilon\ |H^2|<\varepsilon$.

item[Theorem 9] Suppose that $f(x)$ has second derivatives which
are continuous in the neighbourhood of $A$. Suppose that $\nabla
f_A=0$ and that the quadratic form $\sum_{ij}h_ih_jf_{ij}(A)$ is
negative/positive definite in $h_1\ldots h_n$. Then $f(X)$ has \
maximum/minimum at $X=A$.

\item[Proof]
Given $\sum_{ij}f_{ij}(A)h_ih_j$ positive definite, and the second
derivatives are continuous. Then $\exists\delta>0$ for each $X$
satisfying $|X-A|<\delta$ the quadratic form
$\sum_{ij}f_{ij}(X)h_ih_j$ is also a positive definite form.

[Using the determinant test, as all the determinants are
continuous functions of $X$.]

Using Taylor's Theorem,

$$f(X)=f(A)+\frac{1}{2}\sum_{Ij}f_{ij}(A+\theta H)h_ih_j>f(A)$$

since the quadratic form is positive definite at each point in the
sphere.

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