Monday:
- Work:
- Show that $X$ is a Baire space if and only if whenever a
countable union $\bigcup F_n$ of closed sets in $X$ has
interior in $X$ at least one of the sets $F_n$ has interior
in $X$.
- (In this problem, we will assume that if $(X,\rho)$ and
$(Y,\sigma)$ are metric spaces then so is $(X\times
Y,\delta)$ where
$\delta((x,y),(x',y'))=\rho(x,x')+\sigma(y,y')$. You can
also assume that with respect to this product metric,
$(x_n,y_n)\to (x,y)$ if and only if $x_n\to x$ and $y_n\to
y$. In particular, if $(X,\rho)$ and $(Y,\sigma)$ are
complete, so is $(X\times Y,\delta)$.) Let $U$ be a nonempty
open subset of a complete metric space $(X,\rho)$. Show that
$U$ admits a complete metric which is equivalent to that
inherited from $X$. I suggest the following.
- It suffices to find a homeomorphism
$\phi:(U,\rho)\to (Y,\sigma)$ where $(Y,\sigma)$ is
complete.
- Let $A=X\setminus U$ and define $f:U\to \mathbf R$
by $f(x)=\rho(x,A)^{-1}$. Then the map $\phi(x)=(x,f(x))$ is
continuous from $(U,\rho)$ to $(X\times\mathbf
R,\delta)$ where $\delta$ is the obvious complete product
metric. It suffices to see that that the range of
$\phi$ is closed.
- OP: The ruler
function is an example of a function $f:\mathbf R\to
\mathbf R$ which continuous at every irrational and
discontinuous at each rational. In this problem, we want
to see that it is impossible to construct a function which
is continous exactly on the rationals. In fact, we are to
prove that if $D$ is a
countable dense subset of $\mathbf
R$, then there is no function $f:\mathbf R\to \mathbf R$
such that the set of points $C$ where $f$ is continuous is
equal to $D$. I suggest the following.
- Let $U_n$ be the union of all open sets $U\subset
\mathbf R$ such that $\operatorname{diam}(f(U))<\frac1n$.
Show that $C=\bigcap_n U_n$. (A subset of $\mathbf R$, such
as $C$, which is the countable intersection of open sets is
called a $G_\delta$ subset).
- Show that $D$ can't be a $G_\delta$ subset. (Consider: if
$D=\bigcap W_n$ and $V_d:=\mathbf R\setminus \{d\}$ for each
$d\in D$, then $W_n$ and $V_d$ are dense open subsets of
$\mathbf R$.
- OP:
Every vector space $V$ has a basis --- that is, a
linearly independent subset $B$ such that every element in
$V$ is a finite linear combination of elements of $B$. The
dimension of $V$, $\operatorname{dim} V$, is the cardinality
of any such basis. (In analysis, such a basis is sometimes
called a Hamel basis to stress that it is a bonifide
vector space basis.) Show that if $V$ is a Banach space,
then its dimension is either finite or uncountable. (Use
problem #26.)
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Wednesday:
- Turn in: Please turn in problems 13, 16, 17, 20, 22,
23, 24, 25, 26, and 27 by lecture on Monday the 21st. Let me
know if you need more time.
- Work:
- Suppose that $X$ and $Y$ are normed vector spaces.
-
Show
that $\mathcal L(X,Y)$ is a normed vector space with respect to
the "operator norm" defined in lecture such that \[ \|T(x)\|\le
\|T\|\|x\|. \]
- Also show that if $S\in \mathcal L(Y,Z)$ then
$\|ST\|\le \|S\|\|T\|$.
- Show that
\[ \|T\|=\inf\{ \alpha\ge0: \|T(x)\|\le
\alpha\|x\|\quad\text{for all $x\in X$}\}.\]
- Suppose $X$ and $Y$ are Banach spaces with $T\in \mathcal
L(X,Y)$. Suppose that $E$ is a closed proper subspace of
$X$ such that $E\subset \ker T$. Show that there is a unique
operator $\overline{T}\in\mathcal L(X/E,Y)$ such that
$\overline{T}(q(x))=T(x)$ for all $x\in X$ where $q:X\to X/E$ is
the quotient map. Moreover, $\|\overline T\|=\|T\|$.
- OP:
Suppose that $X$ and $Y$ are Banach spaces, that $D$ is a dense
subspace of $X$ and that $T_0\in\mathcal L(D,Y)$. Show that there
is a unique $T\in \mathcal L(X,Y)$ such that $T(x)=T_0(x)$ for all
$x\in D$. (Let $(x_n)$ and $(y_n)$ be sequences in $D$ converging
to $x\in X$. Show that $(T(x_n))$ and $T(y_n))$ must converge to
the same element of $y$.)
- For Fun Only: The existence of continuous functions that
fail to have a derivative at any point (aka nowhere
differentiable) was greeted with sckepticism when Wierestrass
first proved such things existed. He was forced to produce an
example. (Spivak produces a simpler version of Wierestrass's
example in his Calculus book (see Chapter 23, Theorem 5).)
Using the Baire Category Theorem, we can easily see that the set of
continuous nowhere differentiable functions is dense in
$C[0,1]$. My proof of this is attached
for your amusement.
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