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\begin{document}
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\begin{center}
\textbf{201B, Winter '11, Professor John Hunter}

\textbf{Homework 7 Solutions}

\end{center}
\
\\
\\
\

\begin{enumerate}

\bigskip
\bigskip\noindent
\textbf{1.} Let $\mathcal{H} = L^2(0,1)$ with the standard inner product
\[
\langle f, g\rangle = \int_0^1 \bar{f}(x) g(x)\, dx.
\]
Define $M : \mathcal{H} \to \mathcal{H}$ by
\[
(M f) (x) = x f(x)
\]
\textit{i.e.} $M$ is multiplication by $x$.\\
\\




(a) Show that $M$ is a bounded self-adjoint linear operator on $\mathcal{H}$ and
find $\|M\|$.
\\
\

\begin{proof}

$\bullet$ The boundedness of the operator $M$ follows from:
\begin{equation*}
\begin{aligned}
\Vert Mf\Vert_{\mathcal{H}}=&\left( \int_{0}^{1}\mid Mf\mid ^2\right)^{\frac{1}{2}} \\ 
=&\left( \int_{0}^{1}\mid xf(x)\mid ^2\right)^{\frac{1}{2}} \\ 
\leq &\sup_{x\in \left[ 0,1\right] }\mid x\mid\left( \int_{0}^{1}\mid f(x)\mid ^2\right)^{\frac{1}{2}} \\
\leq &\sup_{x\in \left[ 0,1\right] }\mid x\mid\left( \int_{0}^{1}\mid f(x)\mid ^2\right)^{\frac{1}{2}} \\
\leq &\left( \int_{0}^{1}\mid f(x)\mid ^2\right)^{\frac{1}{2}} \\
=&\Vert f \Vert_{\mathcal{H}}
\end{aligned}
\end{equation*}
\\
Note, that here we used the fact that $\Vert x \Vert _{\infty}$ is finite since $x$ is a continuous function on a compact set, and therefore it is bounded. It follows that 
\begin{equation*}
\Vert Mf\Vert_{\mathcal{H}}\leq \Vert f \Vert_{\mathcal{H}}
\end{equation*}
i.e., 
\begin{equation*}
\Vert M\Vert_{\mathcal{H}}\leq 1
\end{equation*}

$\bullet $ The linearity of $M$ is trivial to show. \\
\\

$\bullet $ $M$ is self-adjoint because of the following

\begin{equation*}
\begin{aligned}
\left\langle Mf, g \right\rangle =& \int_{0}^{1} x \bar{f(x)}g(x)\, dx\\
=&\int_{0}^{1}  \bar{f(x)\bar{xg(x)}}\, dx\\
=&\left\langle f, Mg \right\rangle .
\end{aligned}
\end{equation*}
Note that here we used the fact that $x=\bar{x}$ since $x$ is a real-valued function.\\
\\

$\bullet$ We want to conclude that the norm of the operator $M$ is indeed $1$.  For this consider the sequence of functions $f_{k}(x)=\sqrt{2k+1}x^{k}$. \\
\\

We have
\begin{equation*}
\int_{0}^{1} (f_{k}(x))^2\, dx=\frac{2k+1}{2k+1} x \mid _{0}^{1}=1,
\end{equation*}
for all $k \in \mathbb{N}$.\\
\\
Computing the norm of $M$ evaluated at $f_{k}$ we get 
\begin{equation*}
\Vert Mf_{k}\Vert =\int_{0}^{1} (2k+1)x^{2k+2}\, dx=\frac{2k+1}{2k+3}.
\end{equation*}
\\

This approaches $1$ as $k$ goes to infinity, and hence we are done showing that the norm of $M$ is indeed equal to $1$.\\

\end{proof}

\
\\
\
(b) What is the kernel of $M$? What is the range of $M$? Is $M$ onto? Is $\ran M$ closed?\\
\\
\
$\bullet$ \textbf{Kernel:} Note for $xf =0$ we must have that $f = 0$ a.e., so $M$ is injective i.e., the kernel of our operator is trivial (means it contains just the $0$ vector).  \\
\\
\
$\bullet$ \textbf{Surjectivity:} $M$ is not onto since for $Mf(x) = xf(x) = g(x) \in L^2(\left[0,1 \right] )$, we must have $f(x) = \frac{g(x)}{x}$. Let's for example pick $g(x)$ to be the constant function $g(x)=1$, then we end up getting that $f(x)=\frac{1}{x}$, which is not an $L^2(0,1)$ function, as we wanted. Therefore, the range  of $M = \left\{ f \in L^2(\left[ 0,1\right) ) : \frac{f}{x} \in L^2(\left[ 0,1\right] ) \right\}$. \\
\\
The fact that range of $M$ is not closed comes from: choose a sequence of functions defined by
\[ f_n(x) = \chi_{\left[ \frac{1}{n},1 \right]} \frac{1}{x} \]
which obviously belongs to $L^2(0,1)$, but it is obvious that  $Mf_n \to 1 \not\in \mbox{Range of }M$.\\
\\
Another way to see that indeed the range of $M$ is not closed, is to use the fact that $M$ is self-adjoint, and by a  theorem studied in class we get
\begin{equation*}
\mathcal{H}=\bar{ran \, M}\oplus ker \, M,
\end{equation*}
but since $ker\, M=\left\lbrace \right 0\rbrace $ and since $ran \, M \neq \mathcal{H}$, we get that $ran\, M\neq \bar{ran\, M}$.

\end{proof}


\bigskip\noindent
\textbf{2.} Let $\mathcal{H}$ be a complex Hilbert space.
\\

(a) If $A, B \in \mathcal{B}(\mathcal{H})$ are bounded linear operators on $\mathcal{H}$ such that
\[
\langle x, A x\rangle = \langle x, B x\rangle\qquad \mbox{for all $x\in \mathcal{H}$}
\]
show that $A=B$.
\begin{proof}
Suppose $\left\langle x,Ax\right\rangle=\left\langle x,Bx\right\rangle$. Let $C=A-B$, then
\begin{equation*}
\left\langle x,Cx\right\rangle=0,\forall x\in\mathcal{H}.
\end{equation*}
Hence for any $x,y\in\mathcal{H}$,
\begin{align*}
0=\left\langle x+y,C(x+y)\right\rangle&=\left\langle x,Cx\right\rangle+\left\langle y,Cy\right\rangle+\langle x,Cy\rangle+\langle y,Cx\rangle\\
&=\langle x,Cy\rangle+\langle y,Cx\rangle.
\end{align*}
Similarly, we can prove $\langle x,Cy\rangle-\langle y,Cx\rangle=0$ by considering $\left\langle x+iy,C(x+iy)\right\rangle$. Then we find
\begin{equation*}
\langle x,Cy\rangle=0,\quad \forall x,y\in\mathcal{H}
\end{equation*}
which implies $C=0$. Therefore $A=B$.

\end{proof}
\smallskip\noindent
(b) Show that an operator $A \in \mathcal{B}(\mathcal{H})$ is self-adjoint if and only if
$\langle x, A x\rangle$ is real for all $x\in \mathcal{H}$.
\begin{proof}
$\bullet$ Suppose that $A\in \mathcal{B}(\mathcal{H})$ is self-adjoint. Then by definition we have:
\begin{equation*}
\left\langle x,Ax\right\rangle =\bar{\left\langle Ax,x\right\rangle}  =\bar{\left\langle x,Ax\right\rangle},
\end{equation*}
for all $x \in \mathcal{H}$. Note that this implies that $\left\langle x, Ax \right\rangle \in \mathbb{R}$ (a number equals its complex conjugate just when that number belongs to $\mathbb{R}$). \\

\\
$\bullet$ Now let's prove the second implication. Suppose that $\left\langle x, Ax \right\rangle \in \mathbb{R}$ then
\begin{equation*}
\left\langle x, Ax\right\rangle =\left\langle A^{*}x, x \right\rangle =\bar{\left\langle  x, A^{*}\right\rangle } =\left\langle x, A^{*}x\right\rangle .
\end{equation*}
>From probelm no $2$, part $a$), we just solved above, we indeed get $A=A^{*}$.

\end{proof}

\
\\
\
(c) Do these results remain true if $\mathcal{H}$ is a real Hilbert space?
\begin{proof}\emph{Due to Tim.}\\
No, the results do not remain true. To see that $a$) fails in a real Hilbert space, consider the operators $A$ and $B$ on $\mathbb{R}^2$ given by rotation by $\frac{\pi}{2}$ and $\frac{3\pi}{2}$, respectively. Then $\langle x, Ax \rangle = 0 = \langle x, Bx \rangle$, but these operators are clearly not identical.

To see that $b$) fails, consider the left-shift operator on $\ell^2(\mathbb{N})$ over $\mathbb{R}$. Then the inner product will be real, but this operator is not self-adjoint. 


\end{proof}


\
\\
\\
\
\textbf{3.} Suppose that $A : \mathcal{H} \to \mathcal{H}$ is a bounded, self-adjoint linear operator
such that there is a constant $c > 0$ with
\[
c\|x\| \le \|Ax\| \qquad \mbox{for all $x \in \mathcal{H}$}.
\]
Prove that there is a unique solution $x$ of the equation $Ax = y$ for every $y \in \mathcal{H}$.
\begin{proof}
Note that the problem is asking you to prove that in the conditions mentioned above, $A$ has an inverse, which is a bounded operator. This problem can be solved in at least two ways, but here is one of the proofs.\\

If $x\in\ker(A)$, then $Ax=0$. Since $c\|x\|\leq \|Ax\|=0$, we find $x=0$. Therefore $A$ is an injection. \\
Since $A$ is self-adjoint, we find
\begin{equation}
\ran (A)^{\perp}=\ker(A)=\{0\}.
\end{equation*}
Hence $\bar{\ran(A)}=\hs$. If $y\in\hs$ and $(x_{n})$ is a sequence in $\hs$ so that
\begin{equation}
\lim_{n\to\infty}\|Ax_{n}-y\|=0.
\end{equation*}
\
\\
We find $(Ax_{n})$ is a Cauchy sequence in $\hs$. Since
\begin{equation}
c\|x_{n}-x_{m}\|\leq\|A(x_{n}-x_{m})\|=\|Ax_{n}-Ax_{m}\|,
\end{equation}
we find $(x_{n})$ is also a Cauchy sequence in $\hs$. Since $\hs$ is a Hilbert space, we can find $x\in\hs$ so that $\lim_{n\to\infty}\|x_{n}-x\|=0$. Since $A$ is a bounded operator, we find $y=\lim_{n\to\infty}Ax_{n}=Ax$. \\
Thus $y\in\bar{\ran(A)}$. Hence $\ran(A)=\hs$. This proves that $A$ is a bounded open bijection. \\
The open mapping theorem implies that $A$ is invertible. Therefore for each $y\in\hs$, $Ax=y$ has a unique solution.
\end{proof}

\\
\
\newpage\noindent
\textbf{4.} A Laurent operator (or discrete convolution) is a bounded linear
operator $A$ on $\ell^2(\mathbb{Z})$ whose matrix with respect to the standard basis
is
\[
\left[A\right] = \left[\begin{array}{ccccc}
\cdot &\cdot &\cdot &\cdot &\cdot
\\
\cdot & a_0 & a_{-1} & a_{-2} &\cdot
\\
\cdot & a_1 & a_0 & a_{-1} & \cdot
\\
\cdot & a_2 & a_1 & a_{0} & \cdot
\\
\cdot &\cdot &\cdot &\cdot &\cdot
\end{array}\right]
\]
where $a_n \in \mathbb{C}$, meaning that
\[
\left(A x\right)_m = \sum_{n=-\infty}^\infty a_{m-n} x_n.
\]
(a) Let $S : \ell^2(\mathbb{Z}) \to \ell^2(\mathbb{Z})$ denote the right shift operator, defined by
\[
(S x)_m = x_{m-1}.
\]
Show that a bounded linear operator on $\ell^2(\mathbb{Z})$
is a Laurent operator if and only if it commutes with $S$.

\begin{proof}
$\bullet$ Suppose $A$ is a Laurent operator. First implication I am going to prove is to assume $A$ is a Laurent operator and I want to show that it commutes with $S$. \\
Then
\[ A(Sx)_m = (Ax)_{m-1} = \sum_{n \in \mathbb{Z}} a_{m - n -1}x_n. \]
Also,
\begin{equation*}
\begin{aligned}
S(Ax)_m =& S\left( \sum_{n \in \mathbb{Z}} a_{m - n}x_n \right) \\
=& \sum_{n \in \mathbb{Z}} S(a_{m-n}x_n)\\
=& \sum_{n \in \mathbb{Z}} a_{m - n}x_{n-1} \\
=& \sum_{k \in \mathbb{Z}} a_{m - k -1}x_k. 
\end{aligned}
\end{equation*}
Hence, $S$ and $A$ commute.
\\

$\bullet$ Now, let $A \in \mathcal{B}(\ell^2(\mathbb{Z}))$. The goal now is to show that $A$ is a Laurent operator, knowing that it commutes with $S$.\\
 We can represent $A$ as a matrix of the form as follows
\begin{equation*}
 A\mathbf{x} = (a_{m,n})(x_n) = \sum_{m \in \mathbb{Z}}\sum_{n \in \mathbb{Z}}a_{m,n}x_n. 
 \end{equation*}
 
As I said, we are in the case where we suppose that $A$ and $S$ commute. \\

Then,
\begin{equation*}
\begin{aligned}
  AS\mathbf{x} =&\sum_{m \in \mathbb{Z}} \sum_{n \in \mathbb{Z}} a_{m,n}x_{n-1} \\
 =& SA\mathbf{x} \\
 =& S\left( \sum_{m \in \mathbb{Z}}\sum_{n \in \mathbb{Z}} a_{m,n}x_n  \right)\\
  = &\sum_{m \in \mathbb{Z}}\sum_{n \in \mathbb{Z}} a_{m-1,n-1}x_{n-1}.
 \end{aligned}
 \end{equation*}
\
\\

Hence, we got that $a_{m,n} = a_{m-1,n-1}$. Therefore, $A$ is a Laurent operator.
\end{proof}
\
\\
\\
\


\smallskip\noindent
(b) Let $\mathcal{F} : L^2(\mathbb{T}) \to \ell^2(\mathbb{Z})$ denote the unitary Fourier transform
\[
\mathcal{F} f = \hat{f}\qquad \hat{f}(n) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{T}} f(x) e ^{-inx}\, dx.
\]
Suppose that $M : L^2(\mathbb{T}) \to L^2(\mathbb{T})$ is the bounded multiplication operator
\[
(M f)(x) = a(x) f(x)
\]
corresponding to multiplication by a function $a \in L^\infty(\mathbb{T})$. Show that
\[
A = \mathcal{F} M \mathcal{F}^{-1}
\]
is a Laurent operator whose matrix entries are the Fourier coefficients of $a$.
What function $s(x)$ corresponds to $S$?
\begin{proof}
The idea is that when you want to determine an operator,  it is enough to determine how the operator, which in our case is $A$, acts on the basis of the space on which was defined; in our case on the basis $\{e_n\}$ of $\ell^2(\mathbb{Z})$. \\
To determine the entries of the matrix, we need to see how the matrix acts on the basis mentioned above. Note that the $n$th basis vector $e_{n}$, is in fact the Fourier coefficient of $e^{inx}$.\\
\\
Then
\begin{equation*}
 A(e_n) = \mathcal{F}M\mathcal{F}^{-1}(e_n) = \mathcal{F}M(e^{inx}) = \mathcal{F}(a(x)e^{inx}), 
 \end{equation*}
which implies that
\begin{equation*}
 \widehat{a(x)e^{inx}}_k = \int_\mathbb{T} a(x)e^{inx}e^{-ikx} \, dx = \int_\mathbb{T} a(x)e^{i(n-k)x} \, dx = \hat{a}_{k-n}. 
 \end{equation*}
 \
 \\
Since any sequence $b \in l^2(\mathbb{Z})$ can be written as a linear combination of the elements of the basis, i.e., $\mathbf{b} = \sum\limits_{n \in \mathbb{Z}} b_ne_n$, we have:

\begin{equation*}
(A\mathbf{b})_k = A\left( \sum_{n \in \mathbb{Z}} b_ne_n \right) = \sum_{n \in \mathbb{Z}} b_nA(e_n) = \sum_{n \in \mathbb{Z}}b_n\hat{a}_{k-n},  
\end{equation*}
which is of the form of a Laurent operator. 

\\
Note that the matrix representation for $S$ is

\[
\left[ S\right] = \left[\begin{array}{ccccc}
\cdot &\cdot &\cdot &\cdot &\cdot
\\
\cdot & 0 & 0 & 0 &\cdot
\\
\cdot & 1 & 0 & 0 & \cdot
\\
\cdot & 0 & 1 & 0 & \cdot
\\
\cdot &\cdot &\cdot &\cdot &\cdot
\end{array}\right]
\]
\\
\

Therefore we need a function $f$ that has the only nonzero Fourier coefficient is $\hat{f}(1)=1$. Hence $S$
corresponds to multiplication by $s(x) = e^{ix}$.
\end{proof}


\smallskip\noindent
(c) Deduce that if $a$ is nonzero, except possibly on a set of measure zero, and $1/a \in L^\infty(\mathbb{T})$, then the corresponding
Laurent operator $A$ is invertible. If
\[
\left[A^{-1}\right] = \left[\begin{array}{ccccc} \cdot &\cdot &\cdot &\cdot &\cdot
\\
\cdot & b_0 & b_{-1} & b_{-2} &\cdot
\\
\cdot & b_1 & b_0 & b_{-1} & \cdot
\\
\cdot & b_2 & b_1 & b_{0} & \cdot
\\
\cdot &\cdot &\cdot &\cdot &\cdot
\end{array}\right]
\]
give an expression for the coefficients $b_n$ in terms of $a$.
\begin{proof}{From Tim's .tex file}
Note that if $a(x)$ is non-zero off of a set of measure zero and $\frac{1}{a(x)} \in L^\infty(\mathbb{T})$, then the operator $N: L^2(\mathbb{T}) \to L^2(\mathbb{T})$ define by $Nf = \frac{f(x)}{a(x)}$ is bounded and linear.\\

 Now, consider the operator $B = \mathcal{F}N\mathcal{F}^{-1}$. \\
 
 \\
 
 Then,
\begin{equation*}
\begin{aligned}
 AB = &\mathcal{F}M\mathcal{F}^{-1}\mathcal{F}N\mathcal{F}^{-1}\\
  =& \mathcal{F}MN\mathcal{F}^{-1} \\
  =& \mathcal{F}\mathcal{F}^{-1} \\
  =& I. 
 \end{aligned}
 \end{equation*}
 \\
 
So, by part (b), the matrix entries of $A^{-1}$ will be given by the Fourier coefficients of $\dfrac{1}{a(x)}$.
\\
\\
\
\end{proof}

\end{document}