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\begin{document}

\centerline{\bf MATH 604 TAKE-HOME EXAM}

\smallskip
\centerline{\bf Due March 15,  2004}

\bigskip

{\bf 1.} Let $l^\infty$ be the collection of bounded sequences in
$\mathbb C$ equipped with the norm $\|x\|_\infty=\sup_{k}|x_k|$
for $x=(x_1, x_2, \cdots)$.

(1) Prove $(l^\infty, \|\cdot\|_\infty)$ is a Banach space.

(2) Let $c_0$ be the collection of sequences in $\mathbb C$
convergent to 0. Prove $c_0$ is a closed subspace of $l^\infty$.

(3) For any $x=(x_1, x_2, \cdots)\in l^\infty$, find dist$(x,
c_0)$. Can this distance be realized by an element in $c_0$?
Justify your answers.

\bigskip
{\bf 2.} Let $T$ be a bounded linear operator in a Hilbert space
$\mathcal{H}$ and satisfy $|(Tx, x)|\ge c\|x\|^2$ for any
$x\in\mathcal{H}$. Then $T$ has a bounded inverse in
$\mathcal{B}(\mathcal{H})$.

\bigskip

{\bf 3.} Let $\mathcal{X}$ be a NVS. A nonempty subset
$O\subset\mathcal{X}$ is  {\it weakly open} if for any $x_0\in O$
there exists a finite collection of $f_1, \cdots, f_n\in
\mathcal{X}^*$ such that
$$x_0+\cap_{k=1}^n\{x\in\mathcal{X}; |f_k(x)|<1\}\subset O.$$
A subset $C$ is weakly closed if $C^c$ is weakly open. Prove the
following results.

(1) $\mathcal{X}$ is weakly open; a countable union of weakly open
subsets is weakly open; a finite intersection of weakly open
subsets is weakly open. Moreover, $\{x\in\mathcal{X}; |f(x)|<1\}$
is weakly open for any $f\in\mathcal{X}^*$.

(2) Any weakly open subset is open (in the norm topology); any
weakly closed subset is closed; the identity map $i: (\mathcal{X},
\|\cdot\|)\to(\mathcal{X}, w)$ is continuous.

(3) If $\mathcal{X}$ is infinite dimensional, any nonempty weakly
open subset is unbounded. (Hint: Prove the following statement
first: If $\cap_{k=1}^n\mathcal{N}(f_k)=\{0\}$ for some $f_1,
\cdots, f_n\in\mathcal{X}^*$, then $\mathcal{X}$ is finite
dimensional.)

(4) Closed subspaces are weakly closed.

(5) Let $x_n$, $n=1, \cdots,$ and $x$ be elements in
$\mathcal{X}$. Then $x_n$ converges to $x$ weakly in $\mathcal{X}$
(i.e., $f(x_n)$ converges to $f(x)$ for any $f\in\mathcal{X}^*$)
if and only if for any weakly open subset $O$ containing $x$ there
exists an $N$ such that $x_n\in O$ for any $n\ge N$.

\bigskip
{\bf 4.} Solve by Riese Representation Theorem the following
boundary value problem for $f\in C[0, 1]$
\begin{align*}&u^{(4)}+u=f\text{ in }(0, 1),\\
&u(0)=u(1)=u'(0)=u'(1)=0.\end{align*} Provide all the necessary
preparations. (You may assume all the results we proved in solving
the Sturm-Liouville system.)

\vfill\eject

\begin{proof} [Solution of Problem 1.] (1) To prove
$\|\cdot\|_\infty$ is a norm, we shall only prove it satisfies the
triangle inequality. There holds for any $k$
$$|x_k+y_k|\le |x_k|+|y_k|\le \|x\|_\infty+\|y\|_\infty.$$
By taking supremum over $k$, we get $\|x+y\|_\infty\le
\|x\|_\infty+\|y\|_\infty$.

Now we prove $(l^\infty, \|\cdot\|_\infty)$ is complete.   Suppose
$\{x^{(n)}=(x_1^{(n)}, x_2^{(n)}, \cdots)\}$ is a Cauchy sequence
in $l^\infty$. By $|x^{(n)}_k-x^{(m)}_k|\le
\|x^{(n)}-x^{(m)}\|_\infty$ for each fixed $k$, the scalar
sequence $\{x^{(n)}_k\}$ is Cauchy and hence is convergent. For
each $k$, we let $x^{(n)}_k\to x_k$ as $n\to\infty$. We shall
prove $x=(x_1, x_2, \cdots)\in l^\infty$ and
$\|x^{(n)}-x\|_\infty\to 0$ as $n\to\infty$. Note $\{x^{(n)}\}$ is
bounded in $l^\infty$ since it is Cauchy. Hence there exists an
$M>0$ such that $\|x^{(n)}\|_\infty\le M$ for any $n$. This
implies in particular for any $k$
$$|x^{(n)}_k|\le M. $$
By letting $n\to\infty$, we get
$$|x_k|\le M. $$
Now we may take the supremum over $k$ to conclude $\|x\|_\infty\le
M$. The proof of the convergence $\|x^{(n)}-x\|_\infty\to 0$ is
similar and is omitted.

(2) It is obvious that $c_0$ is a subspace. Now we prove $c_0$ is
closed. To this end, we let $\{x^{(n)}\}$ be a sequence in $c_0$
which is convergent to $x$ in $l^\infty$. We need to prove $x_0\in
c$. Let $x^{(n)}=(x_1^{(n)}, x_2^{(n)}, \cdots)$ and $x=(x_1, x_2,
\cdots)$. We need to prove $x_k\to 0$ as $k\to\infty$.  To this
end, we just note
\begin{equation*}|x_k|\le
|x_k^{(n)}-x_k|+|x_k^{(n)}| \le
\|x^{(n)}-x\|_\infty+|x_k^{(n)}|.\end{equation*} We first take a
fixed $n$ sufficiently large so that the first  term is small.
Then we take any $k$ large so that the last term is small. This
proves the convergence of the sequence $\{x_k\}$ to 0.

(3) For a fixed $x\in l^\infty$, we let $D=\limsup|x_k|$. For any
$y=(y_1, y_2, \cdots)\in c_0$, $\lim_{k\to\infty}y_k=0$, we have
$$\|x-y\|_\infty=\sup_{k}|x_k-y_k|\ge
\limsup_{k\to\infty}|x_k-y_k|=\limsup_{k\to\infty}|x_k|=D.$$ Hence
we have
$$dist(x, c_0)\ge D.$$
For each fixed $n$, set $y^{(n)}=(x_1, \cdots, x_n, 0,\cdots)\in
c_0$. Then we have $x-y^{(n)}=(0, \cdots, 0, x_{n+1}, x_{n+2},
\cdots)$ and $$\|x-y^{(n)}\|_\infty=\sup_{k\ge n+1}|x_k|\to
\limsup_k|x_k|=D, \quad\text{as }n\to \infty.$$ Hence
$\{y^{(n)}\}$ is a minimizing sequence and $d(x, c_0)=D$.

Next, for each fixed $x\in l^\infty$ as above, consider $y=(y_1,
y_2, \cdots)$ with
$$y_n=\begin{cases} x_n-D&\text{if }x_n>D\\
0&\text{if }|x_n|\le D\\
x_n+D&\text{if }x_n<-D.\end{cases}$$ Then it is easy to see that
$y\in c_0$ and
$$x_n-y_n=\begin{cases} D&\text{if }x_n>D\\
x_n&\text{if }|x_n|\le D\\
-D&\text{if }x_n<-D.\end{cases}$$ Obviously $\|x-y\|_\infty=D$.

Therefore, $dist(x, c_0)=\limsup_{n\to\infty}|x_n|$ and this
distance is realized by some element in $c_0$.
\end{proof}

\begin{proof} [Solution of Problem 2.] By
$$|(Tx, x)|\le
\|Tx\|\|x\|,\text{ and }|(Tx, x)|=|(x, T^*x)|\le \|T^*x\|\|x\|,$$
we get
$$c\|x\|\le \|Tx\|, c\|x\|\le \|T^*x\|\text{ for any
}x\in\mathcal{H}.$$ This implies both $T$ and $T^*$ are injective,
i.e., $\mathcal{N}(T)=\{0\}$ and $\mathcal{N}(T^*)=0$. Next,
recall $\overline{\mathcal{R}(T)}=\mathcal{N}(T^*)^\perp$. (This
is a homework problem.) We obtain
$\overline{\mathcal{R}(T)}=\mathcal{H}$. Now we prove
$\mathcal{R}(T)$ is closed. To this end, we consider $y_n=Tx_n\in
\mathcal{R}(T)$ such that $y_n\to y$ in $\mathcal{H}$. We shall
prove $y\in \mathcal{R}(T)$. By $c\|x_n-x_m\|\le \|Tx_n-Tx_m\|$,
we note $\{x_n\}$ is a Cauchy sequence in $\mathcal{X}$, and hence
we may assume $x_n\to x$ in $\mathcal{H}$. By the continuity of
$T$, we conclude $y_n=Tx_n\to Tx$, and hence $y=Tx\in
\mathcal{R}(T)$. Therefore $\mathcal{R}(T)=\mathcal{H}$. Hence $T$
is injective and surjective, and then has an inverse. By
$c\|x\|\le \|Tx\|$ for any $x\in\mathcal{H}$, we get
$c\|T^{-1}y\|\le \|y\|$ for any $y\in\mathcal{H}$. Therefore,
$T^{-1}\in\mathcal{B}(\mathcal{H})$. (Note, we did not use the
Banach Inverse Theorem.)
\end{proof}
\bigskip


\begin{proof} [Solution of Problem 3.] We first note that
$\mathcal{N}(f)$ is a closed subspace of codimension 1 for any
nonzero $f\in\mathcal{X}^*$.

(1) This is straightforward and is omitted.

(2) Let $O$ be weakly open and $x_0\in O$ be an arbitrary point.
Then there exist $f_1, \cdots, f_n\in \mathcal{X}^*$ such that
$$x_0+\cap_{k=1}^n\{x\in\mathcal{X}; |f_k(x)|<1\}\subset O.$$
Take $r=(\sup_{1\le k\le n}\{\|f_k\|\})^{-1}$. Then we have
$$B(0, r)\subset \cap_{k=1}^n\{x\in\mathcal{X}; |f_k(x)|<1\}.$$
This implies $B(x_0, r)\subset O$. Hence $O$ is open (in the norm
topology). The other two statements follow easily.

(3) We shall prove the following first: Let $\mathcal{X}$ be a NVS
and $f_1, \cdots, f_n, f\in \mathcal{X}^*$. Then
$f\in\text{span}\{f_1, \cdots, f_n\}$ if and only if
$\cap_{i=1}^n\mathcal{N}(f_i)\subset\mathcal{N}(f)$.

$\Rightarrow$ Suppose $f=\sum_{i=1}^n c_if_i$. For any $x\in
\cap_{i=1}^n\mathcal{N}(f_i)$, we have $f_i(x)=0$, and then
$f(x)=0$, or $x\in \mathcal{N}(f)$.

$\Leftarrow$. We first consider whether
$$(\cap_{i\neq
n}\mathcal{N}(f_i))\cap  \mathcal{N}(f_n)^c\neq \emptyset.$$ If
$\cap_{i\neq n}\mathcal{N}(f_i)\subset  \mathcal{N}(f_n)$, we
simply drop $f_n$ and consider $\{f_1, \cdots, f_{n-1}\}$.
Otherwise, we consider whether
$$(\cap_{i\neq
n-1}\mathcal{N}(f_i))\cap  \mathcal{N}(f_{n-1})^c\neq \emptyset.$$
We continue this process. Therefore we may assume for any $k=1,
\cdots, n$
$$(\cap_{i\neq
k}\mathcal{N}(f_i))\cap  \mathcal{N}(f_{k})^c\neq \emptyset.$$
Hence there exists an $x_k\in (\cap_{i\neq k}\mathcal{N}(f_i))\cap
\mathcal{N}(f_{k})^c$, or
$$f_i(x_k)=0\text{ for }i\neq k, \quad f_k(x_k)=1.$$
For any $x\in\mathcal{X}$, consider $y=x-\sum_{i=1}^nf_i(x)x_i$.
Then $f_k(y)=f_k(x)-f_k(x)=0$, or
$y\in\cap_{i=1}^n\mathcal{N}(f_i)$. By the assumption, $y\in
\mathcal{N}(f)$, or $f(y)=0$. Hence we have
$f(x)=\sum_{i=1}^nf_i(x)f(x_i)$, or $f=\sum_{i=1}^nf(x_i)f_i$.

Now we comeback to (3). We shall prove that any weakly open set
containing 0 is unbounded. First, by the definition of weakly open
set, there exist $f_1, \cdots, f_n\in\mathcal{X}^*$ such that
$$\{x\in\mathcal{X}; |f(x)|<1\}\subset O.$$
In particular, we have
$$\cap_{i=1}^n\mathcal{N}(f_i)\subset O.$$
If $\cap_{i=1}^n\mathcal{N}(f_i)=\{0\}$, then
$\cap_{i=1}^n\mathcal{N}(f_i)\subset\mathcal{N}(f)$ for any
$f\in\mathcal{X}^*$. This implies $\mathcal{X}^*$ is spanned by
$f_1, \cdots, f_n$, and hence finite dimensional. Then it is easy
to see that $\mathcal{X}$ is finite dimensional, which leads to a
contradiction. Therefore, we conclude that
$\cap_{i=1}^n\mathcal{N}(f_i)$ is a nontrivial subspace, and hence
unbounded. So $O$ is unbounded.


(4) Let $\mathcal{M}$ be a closed subspace in $\mathcal{X}$. We
shall prove that $\mathcal{M}^c$ is weakly open. For any
$x_0\notin \mathcal{M}$, there holds dist$(x_0, \mathcal{M})>0$.
By Hahn-Banach Theorem, there exists an $f\in\mathcal{X}^*$ such
that $f|\mathcal{M}=0$ and $f(x_0)=1$. This implies
$x_0+\{x\in\mathcal{X}; |f(x)|<1\}\subset \mathcal{M}^c$, since
$f(x_0+x)=f(x_0)+f(x)=1+f(x)\neq 0$ if $|f(x)|<1$. Hence
$\mathcal{M}^c$ is weakly open.

(5) We shall only prove the case $x=0$.

$\Rightarrow$ Let $\{x_m\}$ be a sequence convergent to 0 weakly.
Consider any weakly open subset $O$ of 0. Then there exist $f_1,
\cdots, f_n\in \mathcal{X}^*$ such that
$$\cap_{k=1}^n\{x\in\mathcal{X}; |f_k(x)|<1\}\subset O.$$
By the weak convergence of $\{x_m\}$ to 0, there holds
$f_k(x_m)\to 0$ as $m\to\infty$ for any $k=1, \cdots, n$. Then
there exists an $N$ such that $|f_k(x_m)|<1$ for any $m\ge N$ and
any $k=1, \cdots, n$. This implies $x_m\in O$ for any $m\ge N$.

$\Leftarrow$ Let $\{x_m\}$ be a sequence such that for any weakly
open set $O$ containing 0 there exists an $N$ such that $x_n\in O$
for any $m\ge N$. For any $f\in\mathcal{X}^*$, we shall prove
$f(x_m)\to 0$. To this end, we consider an arbitrary
$\varepsilon>0$ and note that $O_\varepsilon=\{x\in\mathcal{X};
|f(x)|<\varepsilon\}$ is weakly open by (1). Hence there exists an
$N$ such that for any $m\ge N$ there holds $x_m\in O_\varepsilon$,
or $|f(x_m)|<\varepsilon$.
\end{proof}

\bigskip


\begin{proof} [Solution of Problem 4.] Set
$$X=\{v\in C^2[0, 1]; v(0)=v(1)=v'(0)=v'(1)=0\}.$$
For any $v\in X$, we multiply $\bar v$ to the equation and
integrate by parts to get
$$\int_0^1[u''\bar v''+u\bar v]dx=\int_0^1f\bar vdx.$$

For any $u, v\in X$, set
$$(u, v)_{\mathcal{H}^2}=\int_0^1[u''\bar v''+u\bar v]dx.$$
This is an inner product in $X$, and it induces a norm
$$\|u\|_{\mathcal{H}^2}=\left(\int_0^1[|u''|^2+|u|^2]dx\right)^{\frac12}.$$
In the following, we  also set
\begin{align*}\|u\|_{\mathcal{L}^2}&=
\left(\int_0^1|u|^2dx\right)^{\frac12},\\
\|u\|_{\mathcal{H}^1}&=
\left(\int_0^1[|u'|^2+|u|^2]dx\right)^{\frac12},\end{align*} and
$$\mathcal{H}^1_0(0, 1)=\{u\in AC[0, 1]; u'\in\mathcal{L}^2(0, 1),
u(0)=u(1)=0\}.$$ We proved that $(\mathcal{H}^1_0(0, 1), (\cdot,
\cdot)_{\mathcal{H}^1})$ is a Hilbert space.

{\it Step 1.} We set \begin{align*}\mathcal{H}^2_0(0, 1)=\{u\in&
\mathcal{H}^1_0(0, 1); u'\in \mathcal{H}^1_0(0, 1)\}\\ =\{u\in
&AC[0, 1]; u'\in AC[0, 1], u''\in L^2(0, 1),\\ &
u(0)=u(1)=u'(0)=u'(1)=0\}.\end{align*} We claim
$(\mathcal{H}^2_0(0, 1), (\cdot, \cdot)_{\mathcal{H}^2})$ is a
Hilbert space. We only need to prove $\mathcal{H}^2_0(0, 1)$ is
complete with respect to the norm $\|\cdot\|_{\mathcal{H}^2}$.

Recall the estimate for any $u\in \mathcal{H}^1_0(0, 1)$
$$\|u\|_{\mathcal{L}^2}\le \|u'\|_{\mathcal{L}^2}.$$
This implies for $u\in \mathcal{H}^2_0(0, 1)$
$$\|u\|_{\mathcal{H}^1}+ \|u'\|_{\mathcal{H}^1}\le
4\|u\|_{\mathcal{H}^2}.$$

Let $\{u_n\}$ be a Cauchy sequence in $\mathcal{H}^2_0(0, 1)$.
Then $\{u_n\}$ and $\{u_n'\}$ are Cauchy in $\mathcal{H}^1_0(0,
1)$. There exist $u, v\in \mathcal{H}^1_0(0, 1)$ such that $u_n
\to u$ and $u'_n\to v$ in $\mathcal{H}^1_0(0, 1)$. In particular,
$u'_n\to u'$ and $u_n'\to v$ in $\mathcal{L}^2(0, 1)$. This
implies $u'=v$. Hence we have $u\in \mathcal{H}^2_0(0, 1)$, and we
can check easily $u_n\to u$ in $\mathcal{H}^2_0(0, 1)$.

{\it Remark.} We may  prove directly that the completion of $X$
under the norm $\|\cdot\|_{\mathcal{H}^2}$ is exactly
$\mathcal{H}^2_0(0, 1)$.

{\it Step 2.} Define
$$F(v)=\int_0^1 v\bar fdx\text{ for any }v\in\mathcal{H}^2_0(0, 1).$$
Then obviously $F$ is a linear functional on $\mathcal{H}^2_0(0,
1)$. Next, we have by Schwartz inequality
$$|F(v)|\le \|f\|_{\mathcal{L}^2}\|v\|_{\mathcal{L}^2}\le
\|f\|_{\mathcal{L}^2}\|v\|_{\mathcal{H}^2_0(0, 1)}.$$ Hence $F$ is
a bounded linear functional on $\mathcal{H}^2_0(0, 1)$. By Riesz
Representation Theorem, there exists a unique $u\in
\mathcal{H}^2_0(0, 1)$ such that $(u, \varphi)_{\mathcal{H}^2_0(0,
1)}=F(\varphi)$ for any $\varphi\in \mathcal{H}^2_0(0, 1)$, i.e.,
$$\int_0^1[u''\bar \varphi''+u\bar \varphi]dx=\int_0^1f\bar \varphi dx
\text{ for any } \varphi\in\mathcal{H}^2_0(0, 1).$$

{\it Step 3.} By setting $v=f-u$, we obtain
$$\int_0^1u''\bar \varphi''dx=\int_0^1v\bar \varphi dx
\text{ for any } \varphi\in\mathcal{H}^2_0(0, 1).\leqno{(1)}$$ We
shall prove $u''\in AC$, $u^{(3)}\in AC$, $u^{(4)}\in L^2$ and
$u^{(4)}=v$. The crucial step is the following. We shall prove,
under the assumption (1), there exists a $w\in \mathcal{L}^2(0,
1)$ such that
$$\int_0^1 u''\bar \psi'dx=\int_0^1 w\psi dx
\text{ for any } \psi\in\mathcal{H}^1_0(0, 1).\leqno{(2)}$$

Suppose this is already done. We conclude from (2) that $u''\in
AC(0, 1)$ and $u^{(3)}=-w\in \mathcal{L}^2$. Now we may integrate
by parts the left side of (1) to get
$$\int_0^1u^{(3)}\bar \varphi'dx=-\int_0^1v\bar \varphi dx
\text{ for any } \varphi\in\mathcal{H}^2_0(0, 1).$$ Then we get
$u^{(3)}\in AC(0, 1)$ and $u^{(4)}=v$, or $u^{(4)}+u=f$.

To prove (2), we set $g(x)=\int_{1/2}^x v(t)dt$. Then (1) implies
$$\int_0^1u''\bar \varphi''dx=-\int_0^1g\bar \varphi' dx
\text{ for any } \varphi\in\mathcal{H}^2_0(0, 1).\leqno{(3)}$$ Fix
a $\phi\in\mathcal{H}^2_0(0, 1)$ such that $\int_0^1\phi =1$. For
any $\psi \in \mathcal{H}^1_0(0, 1)$, consider
$$\varphi(x)=\int_0^x\psi-\int_0^1\psi\int_0^x\phi.$$
It is easy to check that $\varphi\in\mathcal{H}^2_0(0, 1)$.
Substituting such a $\varphi$ in (3), we get
$$\int_0^1 u''\bar \psi'=\int_0^1 \left(-g+\int_0^1(u''\bar
\phi')+\int_0^1(g\bar\phi)\right)\bar \psi.$$ This finishes the
proof of (2) with $w=-g+\int_0^1(u''\bar
\phi')+\int_0^1(g\bar\phi)$.
\end{proof}



\end{document}
