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REAL ANALYSIS

INTERCHANGE THEOREMS

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\item[Theorem 1]
Suppose $f_{12}(xy)$ is continuous at $(ab)$ and $\exists\delta&gt;0$
such that $f_2(xb)$ exists for $|x-a|&lt;\delta$. Then $f_{21}(ab)$
exists and $=f_{12}(ab)$.

\item[Proof]
Without loss of generality we may take $(ab)=(0,0)$.

Let $\varepsilon$ be given. Choose $\delta$, such that
$0&lt;\delta_1&lt;\delta$ and $|f_{12}(xy)-f_{12}(00)|&lt;\varepsilon$
whenever $|x|&lt;\delta_1$ and $|y|&lt;\delta_1$. Suppose
$0&lt;|h|&lt;\delta_1$.

Consider

$$\frac{f_2(ho)-f_2(00)}{h}=\lim_{k\to0}\frac{\Delta_{hk}}P{hk}$$

$$\Delta_{hk}=\{f(hk)-f(h0)\}-\{f(ok)-f(00)\}$$

We regard $k$ as being temporarily fixed with $|k|$ sufficiently
small, and write $F(h)=f(hk)-f(h0)$ so that

\begin{eqnarray*}
\frac{\Delta_{hk}}{hk}&amp;=&amp;\frac{F(h)-F(0)}{hk}\\ &amp;=&amp;\frac{F'(\theta
h)}{k}\textrm{by MVT} 0&lt;\theta&lt;1\\ &amp;=&amp;\frac{f_1(\theta
h,k)-f_1(\theta h,0)}{k}\\ &amp;=&amp;f_{12}(\theta h \theta'k)\textrm{ my
MVT} 0&lt;\theta'&lt;1
\end{eqnarray*}

Hence

$$\left|\frac{\Delta_{hk}}{hk}-f_{12}(00)\right|&lt;\varepsilon.$$

Letting $k\to0$ we have by (1)

$$\left|\frac{f_2(h0)-f_2(00)}{h}-f_{12}(00)\right|\leq\varepsilon.$$

Hence $f_{21}(00)$ exists and is equal to $f_{12}(00)$

In the following results $R$ de3notes the closed rectangle $a\leq
x\leq b\ c\leq y\leq d$.

\item[Lemma]
Let $f(xy)$ be continuous on $R$. Then we have
$\phi(x)=\sum_c^df(xy)\,dy$ is continuous on $[ab]$.

\item[Proof]
$f(xy)$ is uniformly continuous on $R$. Hence, given
$\varepsilon&gt;0,\
\exists\delta&gt;0||f(P)-f(Q)|&lt;\frac{\varepsilon}{d-c}$ whenever
$P\in R\ Q\in R$ and $|PQ|&lt;\delta$.

Now if $x_1,x_2$ are each in $[ab]$ and $|x_1-x_2|&lt;\delta$:

$$|\phi(x_1)-\phi(x_2)|\leq\int_c^d|f(x_1y)-f(x_2y)|\,dy&lt;d-c\frac{\varepsilon}{d-c}=\varepsilon$$

\item[Theorem 2]
Let $f(xy)$ be continuous as a function of $y$ for $c\leq y\leq d$
relative to this interval, for each $x$ with $a\leq x\leq b$.
Suppose that $f_1(xy)$ is continuous with respect to $(xy)$ on the
region $a&lt;x&lt;b\ c\leq y\leq d$. Then

$$\frac{\partial}{\partial
x}\int_c^df(xy)\,dy=\int_c^d\frac{\partial}{\partial
x}f(xy)\,dy\textrm{ for }a&lt;x&lt;b$$.

\item[Proof]
Let $x_0$ satisfy $a&lt;x_0&lt;b$.

Choose $\eta$ to satisfy $a&lt;a+\eta&lt;x_0&lt;b-\eta&lt;b$. Let $R_\eta$ be
the closed rectangle $a+\eta\leq x\leq b-\eta, c\leq y\leq d$, and
$I_\eta$ the interval $a+\eta\leq x\leq b-\eta$.

$f_1(xy)$ is uniformly continuous on $R_\eta$. Given
$\varepsilon&gt;0\exists\delta||f_1(P)-f_1(Q)|&lt;\frac{\varepsilon}{d-c}$
whenever $|PQ|&lt;\delta$ and $P,Q\in R_\eta$.

Then if $|H|&lt;\delta$ and $x_0+h\in I_\eta$ we have

\begin{eqnarray*}
&amp;&amp;\left|\frac{\phi(x_0+h)-\phi(x_0)}{h}-\int_c^df_1(x_0y)\,dy\right|\\
&amp;=&amp;\left|\int_c^d\left\{\frac{f(x_0+h,y)-f(x_0y)}{h}-f_1(x_0y)\right\}\,dy\right|\\
&amp;=&amp;\left|\int_c^d\{f_1(x_0+\theta yh,y)-f_1(x_0y)\}\,dy\right|\
0&lt;\theta_y&lt;1\\ &amp;&lt;&amp;d-c.\frac{\varepsilon}{d-c}=\varepsilon
\end{eqnarray*}

\item[Theorem 3]
Let $f(xy)$ be continuous on $R$. Then

$$I_1=\int_a^b\,dx\int_c^af(xy)\,dy=\int_c^d\,dy\int_a^bf(xy)\,dx=I_2$$

\item[Note]
This can be justified under more general conditions.

\item[Proof]
Subdivide $R$ into $n^2$ small rectangles. If $R_{ij}$ is one of
these then $A=A_{ij}n^2$.

Let $\displaystyle M_{ij}=\overline{bd}_{P\in R_{ij}}f(P)\ \
m_{ij}=\underline{bd}_{p\in R_{ij}}f(P)\ \ i=1,2,\ldots,n,\\
j=1,2,\ldots,n.$

Since $f$ is uniformly continuous on $R$ we may choose $n$
sufficiently large so that $M_{ij}-m_{ij}&lt;\varepsilon$. Thus for
each $ij$,

\begin{eqnarray*}
m_{ij}\frac{A}{n^2}&amp;\leq&amp;\int_{a_{i-1}}^{a_i}\,dx\int_{b_{j-1}}^{b_j}f(xy)\,dy\leq
M_{ij}\frac{A}{n^2}\\
\int_a^b\,dx\int_c^d\,dy&amp;=&amp;\sum_{i=1}^n\sum_{j=1}^n\int_{a_{i-1}}^{a_i}\,dx\int_{b_{j-1}}^{b_j}f(xy)\,dy\textrm{
therefore}\\
\sum_{i=1}^n\sum_{j=1}^nm_{ij}\frac{A}{n^2}\leq&amp;I_1&amp;\leq\sum_{i=1}^n\sum_{j=1}^nM_{ij}\frac{A}{n^2}\textrm{
similarly}\\
\sum_{i=1}^n\sum_{j=1}^nm_{ij}\frac{A}{n^2}\leq&amp;I_2&amp;\leq\sum_{i=1}^n\sum_{j=1}^nM_{ij}\frac{A}{n^2}\\
&amp;\Rightarrow&amp;|I_1-I_2|&lt;A\varepsilon\Rightarrow I_1=I_2
\end{eqnarray*}

\item[Theorem 4]
Let $T$b be the triangular region $a\leq y\leq x\leq b$.

Let $f(xy)$ be continuous on $T$. Then

$$\int_a^b\,dx\int_a^xf(xy)\,dy=\int_a^b\,dy\int_y^bf(xy)\,dx.$$

\item[Proof]
We first establish the existence of these integrals.

Let $\displaystyle M=\overline{bd}_{P\in T}|f(P)|$. Given
$\varepsilon&gt;)$ choose $\delta$ such that

\begin{enumerate}

\item
$M\delta&lt;\frac{1}{2}\varepsilon$

\item
$|f(P)-f(Q)|&lt;\frac{1}{2}\frac{\varepsilon}{b-a}$ whenever
$|PQ|&lt;\delta$ and $P,Q\in T$.

\end{enumerate}

If $a\leq x_1\leq x_2\leq b$ and $x_2-x_1&lt;\delta$ we have

\begin{eqnarray*}
&amp;&amp;\left|\int_a^{x_2}f(x_2y)\,dy-\int_a^{x_1}f(x_1y)\,dy\right|\\
&amp;\leq&amp;\left|\int_a^{x_1}f(x_2y)\,dy-f(x_1y)\,dy+\int_{x_1}^{x_2}f(x_2y)\,dy\right|\\
&amp;\leq&amp;(x_2-x_1)M+\int_a^{x_1}|f(x_2y)-f(x_1y)|\,dy\\ &amp;\leq&amp;\delta
M+b-a\frac{1}{2}\frac{\varepsilon}{b-a}=\varepsilon
\end{eqnarray*}

This proves continuity of $\int_a^xf(xy)\,dy$, and hence the
existence of the LHS.

Similarly $\int_y^bf(xy)\,dx$ is continuous and hence the RHS
exists.

We write $\int\!\!\!\int_Tf(xy)\,dxdy$ for
$\int_a^b\,dx\int_a^xf(xy)\,dx$.

We write $\int\!\!\!\int_Tf(xy)\,dydx$ for
$\int_a^b\,dy\int_y^bf(xy)\,dx$.

Also let $A(T)$ denote the area of $T$.

We have

\begin{equation}
\left|\int\!\!\!\int_Tf(xy)\,dxdy-\int\!\!\!\int_Tf(xy)\,dydx\right|\leq2NA(T)
\end{equation}

\item[Population $P_r$]
Let $T$ be the region $a\leq y\leq x\leq b$. Let $f(xy)$ be
continuous on $T$. Let $\displaystyle M=\overline{bd}_{p\in
T}|f(P)|$ and let $A(T)$ be the area of $T$.

Then for $r=0,1,2,\ldots,n$ we have

$$\left|\int\!\!\!\int_Tf(xy)\,dxdy-\int\!\!\!\int_Tf(xy)\,dydx\right|\leq2MA(T)\left(\frac{1}{2}\right)^r$$

$P_o$ is true from (1). Let $r\geq0$ and suppose $P_r$ is true. It
remains to prove that $P_{r+1}$ is true.

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$$\left|\int\!\!\!\int_Tf\,dxdy-\int\!\!\!\int_Tf\,dydx\right|=$$
$$\left|\int\!\!\!\int_{TR_1}f\,dxdy-\int\!\!\!\int_{T_1}f\,dydx
+\int\!\!\int_{T_2}f\,dxdy-\int\!\!\!\int_{T_2}f\,dydx+\int\!\!\!\int_Cf\,dxdy-\int\!\!\!\int_Cf\,dydx\right|$$

By Theorem 3 $\int\!\!\!\int_Cf\,dxdy=\int\!\!\!\int_Cf\,dydx$
therefore

\begin{eqnarray*}
\left|\int\!\!\!\int_Tf\,dxdy-\int\!\!\!\int_Tf\,dydx\right|&amp;\leq&amp;
\left|\int\!\!\!\int_{T_1}f\,dxdy-\int\!\!\!\int_{T_1}f\,dydx\right|+
\left|\int\!\!\!\int_{T_2}f\,dxdy-\int\!\!\!\int_{T_2}f\,dydx\right|\\
&amp;\leq&amp;2\left(\frac{1}{2}\right)^rM(A(T_1)+A(T_2))\\
&amp;=&amp;2\left(\frac{1}{2}\right)^rM\left(2.\frac{1}{4}A(T)\right)\\
&amp;=&amp; 2\left(\frac{1}{2}\right)^{r+1}MA(T)
\end{eqnarray*}

Hence the theorem follows.

The following two theorems are generalisations.

\item[Theorem 5]
Suppose $c(x)\ d(x)$ are continuous on $I=[ab]$ and that $c(x)\leq
d(x)$, for $x$ in $I$. Let $S$ be the region of points $(x,y)$
satisfying $a\leq x\leq b\ c(x)\leq y\leq d(x)$. Suppose that
$f(xy)$ is continuous on $S$.

Then $\displaystyle\int_{c(x)}^{d(x)}f(xy)\,dy$ is continuous on
$I$.

\item[Theorem 6]
Let $c(x)\ d(x)$ be continuous on $I=[ab]$ Suppose $c(x)&lt;d(x)$ and
that $c(x)\ d(x)$ are differentiable for $a&lt;x&lt;b$. Let $S$ be the
set of points with $a\leq x\leq b\ c(x)\leq y\leq d(x)$. Let $S_0$
be the set of points with $a&lt;x&lt;b\ c(x)&lt;y&lt;d(x)$. Suppose that
$f(xy)$£ is continuous on $S$ and that $f_1(xy)$ is uniformly
continuous in $S_0$. Then we have

$$\frac{d}{dx}\int_{c(x)}^{d(x)}f(xy)\,dy=d'(x)f(x,d(x))-c'(x)f(x,c(x))+\int_{c(x)}^{d(x)}f_1(x,y)\,dy$$

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