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\begin{document}

\title{Math 135-2, Homework 6}
\author{Name: \underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}\quad ID: \underline{\qquad\qquad\qquad\qquad}}
\date{}
\maketitle



\begin{problem}{Problem 37.1}
  One of the important consequences of the orthogonality properties of the
  trigonometric sequence (2) [namely, equations (4) in this section and (2),
    (5), (6), (8) in section 33] is \textit{Bessel's inequality}: If $f(x)$ is
  any function integrable on $[-\pi,\pi]$, its ordinary Fourier coefficients
  satisfy the inequality
  \begin{align}\label{eq:37.1}
    \frac{1}{2} a_0^2 + \sum_{k=1}^\infty (a_k^2 + b_k^2) \le \frac{1}{\pi} \int_{-\pi}^\pi [f(x)]^2\,dx.
  \end{align}
  Prove this by the following steps:
  \qpart{a} For any $n \ge 1$, define 
  \begin{align*}
    s_n(x) = \frac{1}{2} a_0 + \sum_{k=1}^n (a_k \cos kx + b_k \sin kx)
  \end{align*}
  and show that
  \begin{align*}
    \frac{1}{\pi} \int_{-\pi}^\pi f(x) s_n\,dx = \frac{1}{2} a_0^2 + \sum_{k=1}^n (a_k^2 + b_k^2).
  \end{align*}
  \qpart{b} By considering all possible products in the multiplication of
  $s_n(x)$ by itself, show that
  \begin{align*}
    \frac{1}{\pi} \int_{-\pi}^\pi [s_n(x)]^2\,dx = \frac{1}{2} a_0^2 + \sum_{k=1}^n (a_k^2 + b_k^2).
  \end{align*}
  \qpart{c} By writing
  \begin{align*}
    \frac{1}{\pi} \int_{-\pi}^\pi [f(x) - s_n(x)]^2\,dx
    &= \frac{1}{\pi} \int_{-\pi}^\pi [f(x)]^2\,dx - \frac{2}{\pi} \int_{-\pi}^\pi f(x) s_n(x)\,dx + \frac{1}{\pi} \int_{-\pi}^\pi [s_n(x)]^2\,dx \\
    &= \frac{1}{\pi} \int_{-\pi}^\pi [f(x)]^2\,dx - \frac{1}{2} a_0^2 - \sum_{k=1}^n (a_k^2 + b_k^2)
  \end{align*}
  conclude that
  \begin{align*}
    \frac{1}{2} a_0^2 + \sum_{k=1}^n (a_k^2 + b_k^2) \le \frac{1}{\pi} \int_{-\pi}^\pi [f(x)]^2\,dx,
  \end{align*}
  and from this complete the proof.

  Observe that the convergence of the series on the left side of \eqref{eq:37.1}
  implies the following corollary of Bessel's inequality: If $a_n$ and $b_n$ are
  the ordinary Fourier coefficients of $f(x)$, then $a_n \to 0$ and $b_n \to 0$
  as $n \to \infty$.
\end{problem}

\begin{solution}
\end{solution}

\begin{problem}{Problem 37.3}
  Prove the Schwarz inequality (19) [$|(f,g)| \le \|f\| \|g\|$].  \textit{Hint:}
  If $\|g\| \ne 0$, then the function $F(\alpha) = \|f + \alpha g\|^2$ is a
  second degree polynomial in $\alpha$ that has no negative values; examine the
  discriminant.
\end{problem}

\begin{solution}
\end{solution}

\begin{problem}{Problem 34.6}
  Mathematicians prefer the classes of functions they study to be linear space,
  that is, to be closed under the operations of addition and multiplication by
  scalars.  Unfortunately, this is not true for the class of functions defined
  on the interval $-\pi \le x < \pi$ that satisfy the Dirichlet conditions.
  Verify this statement by examining the functions
  \begin{align*}
    f(x) = x^2 \sin \frac{1}{x} + 2x \qquad (x \ne 0), \qquad f(0) = 0
  \end{align*}
  and
  \begin{align*}
    g(x) = -2x.
  \end{align*}
\end{problem}

\begin{solution}
\end{solution}

\begin{problem}{Problem 38.1}
  Consider the sequence of functions $f_n(x)$, $n=1, 2, 3, \ldots$, defined on
  the interval $[0,1]$ by
  \begin{align*}
    f_n(x) = \begin{cases} 0, & 0 \le x \le 1/n,\\ \sqrt{n}, & 1/n < x < 2/n, \\ 0, & 2/n \le x \le 1. \end{cases}
  \end{align*}
  \qpart{a} Show that the sequence $\{f_n(x)\}$ converges to the zero function
  on the interval $[0,1]$.
  \qpart{b} Show that the sequence $\{f_n(x)\}$ does \textit{not} converge in
  the mean to the zero function on the interval $[0,1]$.
\end{problem}

\begin{solution}
\end{solution}

\begin{problem}{Problem 38.2}
  Consider the following sequence of closed subintervals of $[0,1]$:
  $[0,\frac{1}{2}]$, $[\frac{1}{2},1]$, $[0,\frac{1}{4}]$,
  $[\frac{1}{4},\frac{1}{2}]$, $[\frac{1}{2},\frac{3}{4}]$, $[\frac{1}{4},1]$,
  $[0,\frac{1}{8}]$, $[\frac{1}{8},\frac{1}{4}]$, $\ldots$, and denote the $n$th
  subinterval by $I_n$.  Now define the sequence of functions $f_n(x)$ on
  $[0,1]$ by
  \begin{align*}
    f_n(x) = \begin{cases} 1 & \mbox{for $x$ in $I_n$}, \\ 0 & \mbox{for $x$ not in $I_n$}. \end{cases}
  \end{align*}
  \qpart{a} Show that the sequence $\{f_n(x)\}$ converges in the mean to the
  zero function in the interval $[0,1]$.
  \qpart{b} Show that the sequence $\{f_n(x)\}$ does \textit{not} converge
  pointwise at any point in the interval $[0,1]$.
\end{problem}

\begin{solution}
\end{solution}

\begin{problem}{Problem 38.5}
  The function $f(x) = x$ to be approximated on $[0,\pi]$ by
  \begin{align*}
    p(x) = b_1 \sin x + b_2 \sin 2x + b_3 \sin 3x
  \end{align*}
  in such a way that $\int_0^\pi [x-p(x)]^2\,dx$ is minimized.  What values
  should the coefficients of $b_k$ have?
\end{problem}

\begin{solution}
\end{solution}

\begin{problem}{Problem A}
  Operations like differentiation and integration are called \textit{operators}.
  They take one function and give you another one.  For example, let $D$ be the
  derivative operator, so that $D f = f'$.  Then, $L f = D(Df) = f''$ is the
  Laplacian operator, which in the case of real functions is just the second
  derivative.  An operator $A$ is called \textit{symmetric} if $\ip{Af,g} =
  \ip{f,Ag}$ for all $f$ and $g$, where $\ip{u,v}$ represents an inner product.
  \qpart{a} In the case where $f$ and $g$ are ordinary vectors in
  $\mathbbm{R}^n$ with the usual inner product $\ip{\uu,\vv} = \uu \cdot \vv$,
  operators $\AA$ are just matrices.  Show that in this case, the above
  definition of symmetric corresponds to the usual definition of a symmetric
  matrix.
  \qpart{b} For what follows, we will use the inner product defined on $[a,b]$
  by
  \begin{align*}
    \ip{u,v} = \int_a^b u(x) v(x)\,dx.
  \end{align*}
  Show that the Laplacian operator is symmetric on $[a,b]$, provided we limit
  ourselves to the space of function $f$ defined on $[a,b]$, where $f(a) = f(b)
  = 0$ and $f$ has continuous second derivatives on $[a,b]$.  (The definition of
  symmetric requires a property be satisfied for all $f$ and $g$; only $f$ and
  $g$ with the above properties are to be tested.)

  \qpart{c} A symmetric operator $A$ is called \textit{positive definite} if for
  all $f$ we have $\ip{Af,f} \ge 0$, with $\ip{Af,f} = 0$ only when $f = 0$.
  Show that $-L$, defined by $(-L)f = -f''$, is positive definite, under the
  same restrictions on $f$.
\end{problem}

\begin{solution}
\end{solution}

\end{document}
