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\begin{center}
{\bf Homework Set 1: Solutions}
\end{center}

\vspace*{.3in}
\noindent{\bf 1.} 
Find the operator norm of the linear transformations
$L:\R^2\to\R^2$ with matrices
$$
\left(\begin{matrix} 4 & 0 \\ 0 & -4\end{matrix}\right)\text{ and }
\left(\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}\right).
$$
\vspace*{.2in}

\noindent{\bf Solution:} Let $L$ be the linear transformation
corresponding to the first matrix and 
$\mathbf{v} = (x, y)$ be
a vector.  Then
$$
\norm{L(\mathbf v)} = \norm{(4x, -4y)}
                    = \sqrt{(4x)^2 + (-4y)^2} = 4\sqrt{x^2+y^2} 
		    = 4\norm{\mathbf v}.
$$
Hence $\norm{L(\mathbf v)}/\norm{\mathbf v} = 4$ regardless of $\mathbf v$.
It follows that $\norm{L} = 4$.

Now let $L$ be the linear transformation corresponding to the other matrix.
Note that 
$$
\norm{L(t\mathbf v)}/\norm{t\mathbf v} = \norm{L(\mathbf v)}/\norm{\mathbf v}
$$
for any $t\in\R$.  Hence 
\begin{eqnarray*}
\sup_{\mathbf v\in\R^2} \norm{L(t\mathbf v)}/\norm{t\mathbf v}
& = &
\sup\{ \norm{L(\mathbf v)}/\norm{\mathbf v}: \mathbf v = (x,1), x\in\R\} \\
& = & \sup_{x\in\R} \sqrt{\frac{(x + 1)^2 + x^2}{x^2 + 1}}
\end{eqnarray*}
(OK, so I'm missing a multiple of the vector $(1,0)$, but you can check that 
one yourself, and anyhow I actually do take care of it implicitly below when I 
let $x\to\pm \infty$.).  Call the function inside the square root $f(x)$.
Then $\lim_{x\to\pm\infty} f(x) = 2$.  Moreover, after differentiating,
we see that $f$ has critical points when 
$$
x^2 - x = 0 \Rightarrow x = 1,0.
$$
Since $f(1) = 5/2$ and $f(0) = 1$, we conclude that $\norm{L} = \sqrt{5/2}$.

\vspace*{.3in}
\noindent{\bf 2.} 
Let $V$ be a vector space over the field $\R$ (or $\C$).  A 
\emph{norm} on $V$ is a function $\norm{\cdot}:V\to \R$ such that
for all $\lambda\in\R$ and $\mathbf{v},\mathbf{w}\in V$,
\begin{itemize}
\item $\norm{\mathbf{v}} \geq 0$ with equality if and only if 
$\mathbf{v} = 0$.
\item $\norm{\mathbf{\lambda v}} = |\lambda|\norm{\mathbf{v}}$
\item $\norm{\mathbf{v} + \mathbf{w}} \leq 
  \norm{\mathbf{v}} + \norm{\mathbf{w}}$.
\end{itemize}
Given a norm $\norm{\cdot}$ on $V$, show that 
$$
d(\mathbf{v},\mathbf{w}) = \norm{\mathbf{v} - \mathbf{w}}
$$
defines a metric on $V$.  A set $U$ is said to be \emph{open with respect to
$\norm{\cdot}$} if it is open with respect to the associated metric $d$.
\vspace*{.2in}

\noindent{\bf Solution: }
We first check that $d$ is a metric.  Clearly 
$d(\V v,\V w) = \norm{\V v - \V w} \geq 0$, and 
$$
\norm{\V v - \V w} = 0 \Leftrightarrow 
\V v - \V w = 0 \Leftrightarrow \V v = \V w.
$$ 
Symmetry of $d$ follows from $\norm{\V v - \V w} = |-1|\norm{\V w - \V v}$.
Finally, 
$$
d(\V v,\V w) = 
\norm{\V v - \V w} = 
\norm{(\V v - \V u) - (\V w - \V u)} \leq 
\norm{\V v - \V u} + \norm{\V w - \V u} =
d(\V v, \V u) + d(\V u,\V w),
$$
so the triangle inequality holds.  Thus $d$ is a metric. 

\vspace*{.3in}
\noindent{\bf 3.} 
Different norms $\norm{\cdot}$ and $\norm{\cdot}'$ on the same
vector space are called \emph{comparable} if there are constants $C_1,C_2>0$
such that
$$
C_1\norm{\mathbf{v}} \leq \norm{\mathbf{v}}' \leq C_2\norm{\mathbf{v}}
$$
for all $\mathbf{v}\in V$.  

Supposing that $\norm{\cdot},\norm{\cdot}'$ are comparable, show that
a set $U\subset V$ is open with respect to $\norm{\cdot}$ if and only
if it is open with respect to $\norm{\cdot}'$.  Does the same conclusion
hold if you replace `open' with `closed'? `compact'? `connected'?
Explain.

\vspace*{.2in}
\noindent{\bf Solution: }
Let $U\subset V$ be open with respect to $\norm{\cdot}$ and $\V v\in U$.
Then there exists $r>0$ such that 
$N_r(\V v) = \{\V w\in V:\norm{\V w - \V v}<r\}\subset U$.  But since
$$
\norm{\V w - \V v}' \leq r/C_2 \Rightarrow \norm{\V w - \V v} \leq r, 
$$
we have $N'_{r/C_2}(\V v)\subset N_r(\V v)\subset U$ (where the prime denotes 
`neighborhood with respect to $\norm{\cdot}'$.  That is, any $\V v \in U$
admits a $\norm{\cdot}'$ neighborhood also contained in $U$, so $U$ is open
with respect to $\norm{\cdot}'$.  

The same argument shows that if $U$ is open with respect to $\norm{\cdot}'$,
then $U$ is also open with respect to $\norm{\cdot}$.\qed

The conclusion also works for closed sets, compact sets, and connected sets,
because all of these can be characterized in terms of open sets (e.g.
a set is closed iff it's the complement of an open set, etc, etc.)
\vspace*{.3in}

\noindent{\bf 4.}
Let $n,m\in\Z^+$ be given and $V = L(\R^n,\R^m)$ be the vector space
of linear transformations from $\R^n$ to $\R^m$.  Let $T = (a_{ij})\in V$ be
an arbitrary element.  Show that the following norms on $V$ are all comparable
to the operator norm on $V$.
\begin{itemize}
\item $\norm[\infty]{T} = \max_{i,j}|a_{ij}|$
\item $\norm[1]{T} = \sum_{i,j} |a_{ij}|$
\item $\norm[2]{T} = \sqrt{\sum_{i,j} |a_{ij}|^2}$
\end{itemize}
In fact, it can be shown that pretty much any two norms on a finite dimensional
vector space are comparable (Prove this and you take care of all the
above items at once.  And I'll give you five extra credit points).  

\vspace*{.2in}
\noindent{\bf Solution: }
Let $a = \max|a_{ij}|$.  Then
$$
a = \sqrt{a^2} \leq \sqrt{\sum_{i,j} a_{ij}^2} \leq
                 \sqrt{\left(\sum_{i,j} |a_{ij}|\right)^2} 
		 = \sum_{i,j} |a_{ij}|
		   \leq nm a,
$$
where $nm$ is just the number of entries in $T$.
Since all these inequalities hold regardless of $T$, this shows that 
$\norm[\infty]{\cdot}$, $\norm[2]{\cdot}$ and $\norm[1]{\cdot}$ are
all comparable.  To finish the proof it's enough to show that $\norm{\cdot}$
is comparable to any one of these---say $\norm[\infty]{\cdot}$.  

If $\V v = \V e_j$ is one of the usual basis vectors, then 
$$
\norm{T(\V v)}  = \norm{(a_{1j},a_{2j},\dots,a_{mj})}
                = \sqrt{\sum_i a_{ij}^2} \leq \sqrt{\sum_i a^2}
		= \sqrt{m} a,
$$
and if $\V v = v_1\V e_1 + \dots v_n\V e_n$ is an arbitrary unit vector, 
then 
$$
\norm{T(\V v)} = \norm{\sum_j v_j T(\V e_j)} 
\leq \sum_j |v_j|\norm{T(\V e_j)} \leq n\cdot\sqrt{m}{a}
$$
because $|v_j|\leq 1$ for all $j$.  Hence 
$$
\norm{T(\V v)} = \sup_{\norm{\V v} = 1} \norm{T(\V v)} \leq n\sqrt{m} 
\norm[\infty]{T}.
$$
\qed
\vspace*{.2in}

By the way, 

\noindent{\bf Theorem. } Any two norms on a finite dimensional real
(or complex) vector space $V$ are comparable.
\vspace*{.1in}

\proof
Let $\{\V e_1,\dots,\V e_n\}$ be a basis for $V$ and $\norm[\infty]{\cdot}$
be the norm on $V$ given by
$$
\norm[\infty]{\V v} = \max_{1\leq j\leq n}|c_j|
$$
where the numbers $c_j$ come from writing $\V v = c_1\V e_1\dots c_n\V e_n$
as a linear combination of basis vectors.  It is enough to show that any
other norm $\norm{\cdot}$ on $V$ is comparable to $\norm[\infty]{\cdot}$.
Now on the one hand, we have
$$
\norm{\V v} \leq |c_1|\norm{\V e_1} + \dots + \norm{\V e_n} 
\leq n(\max\norm{\V e_j})\norm[\infty]{\V v},
$$
which gives comparability in one direction.

To get comparability in the other direction, I suppose for the sake of
obtaining a contradiction that for any $C>0$ there exists $\V v\in V$
such that $\norm[\infty]{\V v} > C\norm{\V v}$.  Then in particular,
by choosing a sequence of $C$'s tending to $\infty$, we can find
a sequence of vectors $\{\V v_j\}\subset V$ such that 
$\norm[\infty]{\V v_j} = 1$ whereas $\lim_{j\to\infty}\norm{\V v_j} = 0$.

Given this, I claim that after passing to a subsequence, we can further assume
that $\{\V v_j\}$ converges to some vector $\V v \in V$.  And I never claim
anything that I can't prove.  Never.  If we write
$$
\V v_j = c_{1j} \V e_1 + \dots c_{nj} \V e_n,
$$
then the `coordinate vectors' $(c_{1j},\dots,c_{nj})\in\R^n$ all lie in the 
compact (because closed and bounded) set 
$\{(x_1,\dots,x_n)\in\R^n: \max|x_k|=1\}$, so after passing to a subsequence,
we can assume that $c_{1j}\to c_1$, \dots $c_{nj}\to c_n$ where
$\max|c_k| = 1$.  But, from the definition of $\norm[\infty]{\cdot}$, this
is the same as saying that 
$$
\lim_{n\to\infty}\norm[\infty]{\V v_j - \V v} = 0
$$
where $\V v = c_1 \V e_1 + \dots + c_n\V e_n$.  So the claim is true.

We get our contradiction as follows.  By the triangle inequality
$$
\norm{\V v_j} - \norm{\V v}, \norm{\V v} - \norm{\V v_j} \leq
\norm{\V v_j - \V v}.
$$
That is,
$$
|\norm{\V v_j} - \norm{\V v}| \leq \norm{\V v_j - \V v} \leq 
C\norm[\infty]{\V v_j - \V v} \to 0
$$
as $j\to\infty$.  So $\norm{\V v} = 0$.  On the other hand $\V v$ is certainly
non-zero, because the basis vectors $\V e_j$ are linearly independent and
at least one of the coefficients $c_j$ used to define $\V v$ has magnitude
$1$.  Since non-zero vectors must have non-zero norm, we have found our
impasse and conclude that there really does exist $C>0$ such that
$$
\norm[\infty]{\V v}\leq C\norm{\V v}
$$
for every $v\in V$.
\qed
 


\vspace*{.3in}
\noindent{\bf 5.} Give
an example of two \emph{incomparable} norms on the (infinite dimensional) 
vector space $C([0,1],\R)$ of continuous functions from $[0,1]$ to $\R$.

\vspace*{.2in}
\noindent{\bf Solution:}
The norms 
$$
\norm[\infty]{f} \eqdef \max_{x\in [0,1]} |f(x)| 
\text{ and }
\norm[1]{f} \eqdef \int_0^1 |f(x)|\,dx
$$
are incomparable.  Consider for instance the functions
$f_n(x) = x^n$.  We have
$$
\norm[\infty]{f_n} = |f_n(1)| = 1
$$
for every $n\in\N$, but 
$$
\norm[1]{f_n} = \frac{1}{n+1} \to 0.
$$
Hence, there is no constant $C>0$ such that
$$
\norm[\infty]{f} \leq C\norm[1]{f}
$$
for all $f\in C([0,1],\R)$.
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