216_21_Thomas_Breydo.tex

\documentclass{amsart}
\usepackage{thomas} % my style file, https://git.io/thomas.sty


\newcommand{\pagenum}{216}
\newcommand{\probnum}{21}

\title{\pagenum.\probnum}
\author{Thomas\ Breydo}

\begin{document}

\maketitle

\begin{problem*}
Fix a positive integer $n.$ In the inner product space of
continuous real-valued functions on $[-\pi, \pi]$ with inner
product
\begin{align*}
    \iprod{f,g}=\int_{-\pi}^\pi{\!f(x)g(x)\dx},
\end{align*}
let
\begin{align*}
    V=\vspan(1,\sin x,\sin 2x,\dots,\sin nx,\cos x,\cos 2x,\dots,\cos nx).
\end{align*}
\begin{enumerate}[label=(\alph*),itemsep=1em,topsep=1em]
    \item  Define $D\in\SL(V)$ by $Df=f'.$ Show that $D^*=-D.$
        Conclude that $D$ is normal but not self-adjoint.
    \item Define $T\in\SL(V)$ by $Tf=f''.$ Show that $T$ is
        self-adjoint.
\end{enumerate}
\end{problem*}
\vspace{0.5in}

\begin{claim*}
For all $f,g\in V,$
\begin{align*}
    f(\pi)g(\pi)=f(-\pi)g(-\pi).
\end{align*}
\end{claim*}
\begin{proof}
Suppose $f,g\in V.$ Then,
\begin{align*}
    f(x) &= a_0+a_1\sin x+\dots+a_n\sin nx+a_{n+1}\cos x+\dots+a_{2n}\cos nx \\
    g(x) &= b_0+b_1\sin x+\dots+b_n\sin nx+b_{n+1}\cos x+\dots+b_{2n}\cos nx.
\end{align*}
Since all terms of $f(\pi)g(\pi)$ with $\sin(k\pi)$ will be zero,
\begin{align*}
    f(\pi)g(\pi)= a_0b_0 &+ a_0\sum_{i=1}^n b_{n+i}\cos(i\pi)
    + b_0\sum_{i=1}^n a_{n+i}\cos(i\pi)\\
            &+\sum_{1\le i,j\le n}
            a_{n+i}\cos(i\pi)b_{n+j}\cos(j\pi).
\end{align*} 
Similarly,
\begin{align*}
    f(-\pi)g(-\pi)= a_0b_0 &+ a_0\sum_{i=1}^n b_{n+i}\cos(-i\pi)
    + b_0\sum_{i=1}^n a_{n+i}\cos(-i\pi)\\
            &+\sum_{1\le i,j\le n}
            a_{n+i}\cos(-i\pi)b_{n+j}\cos(-j\pi).
\end{align*}
But since $\cos(-x)=\cos(x),$ we see that
$f(\pi)g(\pi)=f(-\pi)g(-\pi).$
\end{proof}

\begin{claim*}
For all $f,g\in V,$
\begin{align*}
    \int_{-\pi}^{\pi}{\!f'(x)g(x)\dx} = -\int_{-\pi}^{\pi}{\!f(x)g'(x)\dx}
\end{align*}
\end{claim*}
\begin{proof}
Starting with the previous claim,
\begin{align*}
    0 &= f(\pi)g(\pi)-f(-\pi)g(-\pi) \\
      &=(f\cdot g)(x)\bigg\vert_{-\pi}^\pi \\
      &=\int_{-\pi}^{\pi}{\!(f\cdot g)'(x)\dx} \\
      &=\int_{-\pi}^{\pi}{\!f'(x)g(x)\dx}
       +\int_{-\pi}^\pi{\!f(x)g'(x)\dx}.
\end{align*}
Thus,
\begin{align*}
\int_{-\pi}^{\pi}{\!f'(x)g(x)\dx}=-\int_{-\pi}^\pi{\!f(x)g'(x)\dx}
\end{align*}
as desired.\\
\end{proof}

Recall that $D\in\SL(V)$ is defined by $Tf=f'.$
\begin{claim*}
$D^*=-D.$
\end{claim*}
\begin{proof}
Since $D^*$ is unique, all we need to do is show that for all
$f,g\in V,$
\begin{align*}
\iprod{Df,g}=\iprod{f,-Dg}.
\end{align*}
Indeed,
\begin{align*}
    \iprod{Df,g} &= \iprod{f',g}\\
                 &= \int_{-\pi}^{\pi}{\!f'(x)g(x)\dx} \\
                 &=-\int_{-\pi}^\pi{\!f(x)g'(x)\dx} && (\text{previous claim})\\
                 &= -\iprod{f,Dg} \\
                 &= \iprod{f,-Dg}
\end{align*}
as desired.
\end{proof}
\begin{claim*}
    $D$ is normal.
\end{claim*}
\begin{proof}
\begin{align*}
    DD^* &= (D)(-D) \\
         &= (-D)(D) \\
         &= D^*D,
\end{align*}
and thus $D$ is normal.
\end{proof}

\begin{claim*}
    $D$ is not self-adjoint.
\end{claim*}
\begin{proof}
    Clearly $D\ne D^*$ since $D^*=-D.\\$
\end{proof}

Recall that $T\in\SL(V)$ is defined by $Tf=f''.$
\begin{claim*}
    $T$ is self-adjoint.
\end{claim*}
\begin{proof}
Since $T=DD,$
\begin{align*}
    T^* &= (DD)^* \\
        &= D^*D^* \\
        &= (-D)(-D) \\
        &= DD \\
        &= T.
\end{align*}
Thus, $T$ is self-adjoint.
\end{proof}

\vspace{0.5in}

\begin{note*}
You can view the source code for this solution
\href{https://github.com/thomasbreydo/linalg/blob/main/\pagenum_\probnum_Thomas_Breydo.tex}
{here}.
\end{note*}

\end{document}