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\noindent Solutions for Homework 8
\vskip .2in

  
\vskip .2in

\noindent 
{\bf Problem 1:}  Compute the operator norm of the matrix
$$
\left(\begin{array}{cc} 3 & 5 \\ 5 & 3 \end{array}\right).
$$
(Hint: a vector in $\mathbf{R}^2$ can be written in polar coordinates
$(x,y) = (r\cos\theta,r\sin\theta)$).
\bigskip

\noindent{\bf Solution:}  Consider a vector 
$(x,y) = (r\cos\theta, r\sin\theta)$ in $\R^2$.  Note that $||(x,y)|| = r$.
Define a new vector $(x',y')$ by 
$$
\left(\begin{array}{c} x' \\ y' \end{array}\right) =
\left(\begin{array}{cc} 3 & 5 \\ 5 & 3 \end{array}\right)
\left(\begin{array}{c} x \\ y \end{array}\right) = 
\left(\begin{array}{c} 3r\cos\theta + 5r\sin\theta \\ 
                       5r\cos\theta + 3r\sin\theta \end{array}\right)
$$
The operator norm of the matrix is the maximum value of the quantity
$$
\frac{||(x',y')||}{||(x,y)||} = \sqrt{34 + 60\cos\theta\sin\theta}
                              = \sqrt{34 + 30\sin2\theta}
$$
The expression under the square root is largest when $\sin 2\theta = 1$
so the operator norm of the matrix is $\sqrt{64} = 8$.
\bigskip

\noindent
{\bf Problem 2:} Show {\it using the definition of limit}
that the function $L:\mathbf{R^2} \to \mathbf{R^3}$
given by 
$$
L(x,y) = \left(\begin{array}{cc} 1 & -2 \\ e & 1 \\ 1 & \pi \end{array}\right)
         \left(\begin{array}{c} x \\ y \end{array}\right) + 
         \left(\begin{array}{c} 2 \\ 5 \\ 1 \end{array}\right)
$$
is continuous.
\bigskip

\noindent{\bf Solution:}
Let $A$ denote the $3\times 2$ matrix in the definition of $L$.  From
class we know that the operator norm of $A$ is no greater than $Cmn = 6\pi$
where $C$ is the maximum of the absolute values of the entries of $A$ and
$m$ and $n$ are the number of rows and columns, respectively, of $A$.

Let $\mathbf{a}\in\mathbf{R^n}$ and  $\epsilon > 0$ be given and choose 
$\delta = \epsilon/6\pi$.  Then if $||\mathbf{x}-\mathbf{a}|| < \delta$,
we have
$$
||L(\mathbf{x}) - L(\mathbf{a})|| = ||A\cdot(\mathbf{x} - \mathbf{a})||
     \leq 6\pi||\mathbf{x} - \mathbf{a}|| < 6\pi\delta = \epsilon.
$$
Therefore $L$ is continuous at $\mathbf{a}$.  Since $\mathbf{a}$ was
arbitrary, $L$ is continuous on all of $\mathbf{R}^n$.
\qed

\bigskip
\noindent
{\bf Problem 3:} Show that the intersection of finitely many open sets
is open.  Give an example of a countable collection of open sets 
$U_j\subset\mathbf{R}^2$, 
$j\in\mathbf{N}$ whose intersection $\bigcap_{j=1}^\infty U_j$ is not open.
\bigskip

\noindent{\bf Solution:}
Suppose that $U_1,\dots,U_k\subset\mathbf{R}^n$ are open sets 
and let $U = U_1\cap\dots\cap U_k$ be their intersection.  Then if
$\mathbf{a}\in U$, we must show that $\mathbf{a}$ is an interior point
of $U$.  To do this, we point out that $\mathbf{a}\in U$ implies that
$\mathbf{a}\in U_j$ for $j=1,\dots,k$.  But each $U_j$ is open, so there
exists a number $r_j>0$ such that $B_{r_j}(\mathbf{a})\subset U_j$.
It follows that the number $r = \min\{r_1,\dots,r_k\}$ is positive and
that $B_r(\mathbf{a})\subset U_j$ for every $j=1,\dots,k$.  Therefore
$B_r(\mathbf{a})\subset U$, too.  We conclude that $\mathbf{a}$ {\it is}
an interior point of $U$ and that $U$ is open. \qed
\bigskip


\bigskip
\noindent
{\bf Problem 4:} Let $f:\mathbf{R}^n\to\mathbf{R}^m$ be a continuous function
and let $U\subset\mathbf{R^m}$ be an open set.  Show that the
set
$$
f^{-1}(U) \eqdef \{\mathbf{x}\in\mathbf{R}^n: f(\mathbf{x}) \in U\}
$$
is open. 
\bigskip

\noindent{\bf Solution:}
Let $U\subset\R^n$ be open and $W = f^{-1}(U)$.  We must show that
any point $\mathbf{a}\in W$ is an interior point.  To do this, we note
that by definition of $W$, $f(\mathbf{a})\in U$. Since $U$ is open, 
we know that there exists an $\epsilon>0$ such that 
$B_\epsilon(f(\mathbf{a})) \subset U$.  Since $f$ is continuous, we
know that there exists $\delta >0$ such that 
$||\mathbf{x} - \mathbf{a}|| < \delta$ implies that 
$||f(\mathbf{x})-f(\mathbf{a})|| < \epsilon$.  In other words
$x\in B_\delta(\mathbf{a})$ implies that 
$f(\mathbf{x})\in B_\epsilon(f(\mathbf{a}))$--in particular, that
$f(\mathbf{x})\in U$.  Therefore 
$B_\delta(\mathbf{a})\subset f^{-1}(U) = W$.  This shows that $\mathbf{a}$
is an interior point of $W$ and, since $\mathbf{a}$ was chosen arbitrarily,
that $W$ is open.
\qed



\bigskip
\noindent
{\bf Problem 5:}
Let $f,g:\mathbf{R}^n\to\mathbf{R}^m$ be functions and $\mathbf{a}\in\R^n$,
be a point at which both 
$\lim_{\mathbf{x}\to\mathbf{a}} f(\mathbf{x})$ and
$\lim_{\mathbf{x}\to\mathbf{a}} g(\mathbf{x})$ 
exist.  Show that 
$
\lim_{\mathbf{x}\to\mathbf{a}} f(\mathbf{x}) + g(\mathbf{x})
$
exists and is equal to 
$
\lim_{\mathbf{x}\to\mathbf{a}} f(\mathbf{x}) + 
\lim_{\mathbf{x}\to\mathbf{a}} g(\mathbf{x}).
$
\bigskip

\noindent{\bf Solution:}
Let $\mathbf{a}\in\R^n$ and $\epsilon > 0$ be given.  If we set
$\epsilon' = \epsilon/2 > 0$, then (since $f$ and $g$ are continuous)
there exist numbers $\delta_1,\delta_2 > 0$ such that 
$||\mathbf{x}-\mathbf{a}||<\delta_1$ implies 
$||f(\mathbf{x})-f(\mathbf{a})||<\epsilon'$ and
$||\mathbf{x}-\mathbf{a}||<\delta_2$ implies 
$||g(\mathbf{x})-g(\mathbf{a})||<\epsilon'$.  So if we set 
$\delta = \min\{\delta_1,\delta_2\}>0$, we see that 
$||\mathbf{x} - \mathbf{a}||<\delta$ implies that 
$$
||(f+g)(\mathbf{x}) + (f+g)(\mathbf{a})|| \leq ||f(\mathbf{x})-f(\mathbf{a})||
         + ||g(\mathbf{x})-g(\mathbf{a})|| 
     \leq \epsilon' + \epsilon' = \epsilon.
$$
Therefore $f+g$ is continuous at $\mathbf{a}$.
\qed



\bigskip
\noindent
{\bf Problem 6:}
\# 2 on page 244 of Wade's book---no proofs necessary.

\begin{description}
\item[a] $\lim_{x\to 0}\lim_{y\to 0} f(x,y) 
     = \lim_{y\to 0}\lim_{x\to 0} f(x,y) = 0.$  However, 
    $\lim_{(x,y)\to(0,0)} f(x,y)$ does
    not exist, because if $(x_n,y_n) = (1/n,1/n)$ then 
$$
\lim_{n\to\infty} f(x_n,y_n) = \lim_{n\to\infty} \frac{1}{2}n^2(\sin(1/n))^2
                             = \frac12,
$$
which is different than either of the iterated limits.
\item[b] $\lim_{x\to 0}\lim_{y\to 0} f(x,y) = 1$ and 
$ \lim_{y\to 0}\lim_{x\to 0} f(x,y) = 0 = 1/2$.  Again,
$\lim_{(x,y)\to(0,0)} f(x,y)$ does not exist.
\item[c] $\lim_{x\to 0}\lim_{y\to 0} f(x,y) 
     = \lim_{y\to 0}\lim_{x\to 0} f(x,y) = 0.$  Moreover, 
$\lim_{(x,y)\to(0,0)} f(x,y) = 0$ because 
\begin{eqnarray*}
|f(x,y)| & = & \left|\frac{x-y}{(x^2+y^2)^\alpha}\right| \\
         & \leq & \frac{|x|}{(x^2+y^2)^\alpha}
             + \frac{|y|}{(x^2+y^2)^\alpha} \\
         & \leq & \frac{|x|}{|x|^{2\alpha}}
                + \frac{|y|}{|y|^{2\alpha}} 
            = |x|^{1-2\alpha} + |y|^{1-2\alpha},
\end{eqnarray*}
and since $\alpha<1/2$, the last expression tends to zero as $(x,y)\to(0,0)$.
\end{description}





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