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\begin{document}


\centerline{\textbf{Problem Set 7: Math 201B}}
\centerline{Due: Friday, February 18}

\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 $x$.

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

\smallskip\noindent
(b) What is the kernel of $M$? What is the range of $M$? Is $M$ onto? Is $\ran M$ closed?

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

\smallskip\noindent
(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 $A=B$.

\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 $x\in \mathcal{H}$.

\smallskip\noindent
(c) Do these results remain true if $\mathcal{H}$ is a real Hilbert space?


\bigskip\noindent
\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 $y \in \mathcal{H}$.



\newpage\noindent
\textbf{4.} A Laurent operator (or discrete convolution) is a bounded linear
operator $A$ on $\ell^2(\Ir)$ 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 $a_n \in \Cx$, meaning that
\[
\left(A x\right)_m = \sum_{n=-\infty}^\infty a_{m-n} x_n.
\]
(a) Let $S : \ell^2(\Ir) \to \ell^2(\Ir)$ denote the right shift operator, defined by
\[
(S x)_m = x_{m-1}.
\]
Show that a bounded linear operator on $\ell^2(\Ir)$
is a Laurent operator if and only if it commutes with $S$.

\smallskip\noindent
(b) Let $\mathcal{F} : L^2(\Ts) \to \ell^2(\Ir)$ denote the unitary Fourier transform
\[
\mathcal{F} f = \hat{f}\qquad \hat{f}(n) = \frac{1}{\sqrt{2\pi}} \int_{\Ts} f(x) e ^{-inx}\, dx.
\]
Suppose that $M : L^2(\Ts) \to L^2(\Ts)$ is the bounded multiplication operator
\[
(M f)(x) = a(x) f(x)
\]
corresponding to multiplication by a function $a \in L^\infty(\Ts)$. 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 $S$?

\smallskip\noindent
(c) Deduce that if $a$ is nonzero, except possibly on a set of measure zero, and $1/a \in L^\infty(\Ts)$, 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 $a$.
\end{document} 