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During the next section we shall switch between
various norms on $\RR^n$:
\[ \|\bf{a}\|_\infty = \max|a_j|, \qquad
\|\bf{a}\|_2 = \sqrt{\sum_j |a_j|^2}, \qquad
\|\bf{a}\| = \sum_j |a_j| .\]
There is a general theorem that all norms on
$\RR^n$ are ``equivalent'':
\begin{align*}
  \|\bf{a}\|_\infty
  &\le \|\bf{a}\|_1 \le n\|\bf{a}\|_\infty \\
  \|\bf{a}\|_\infty
  &\le \|\bf{a}\|_2 \le n\|a\|_\infty
\end{align*}

Last time we showed that there exists an $\eps(n)
> 0$ such that if $\bf{a} \in \RR^{n + 1}$ and
$\|\bf{a}\|_\infty \ge 1$ then
\[ -\sup_{t \in [a, b]} \left|\sum_{j = 0}^n a_j
t^j\right| \ge \eps(n) .\]
Thus if we write $T\bf{a} = P$ with $P(t) = \sum_j
a_j t^j$, we have $\|T\bf{a}\|_\infty \to \infty$
as $\|\bf{a}\|_2 \to \infty$.

We now show:

\begin{flashcard}[best-poly-approx-lemma]
\begin{lemma*}
  If $f \in C[a, b]$, then there exists a $P \in
  \mathcal{P}_n$ such that
  \[ \|f - P\|_\infty \le \|f - Q\|_\infty \]
  for all $Q \in \mathcal{P}_n$, i.e. ther exists
  a \emph{best} polynomial approximation (for
  fixed degree).
\end{lemma*}

\begin{proof}
  \cloze{Without loss of generality $[a, b] = [0, 1]$.

  Consider the map $S : \RR^{n + 1} \to \RR$ given
  by
  \[ S(\bf{a})(t) = \left| \sum_{j = 0}^n a_j t^j
  - f(t)\right| .\]
  We claim that $S$ is continuous.
  \begin{align*}
    |S(\bf{a})(t) - S(\bf{b})(t)|
    &= \left| \left| \sum_{j = 0}^n a_j t^j - f(t)
    \right| - \left| \sum_{j = 0}^n b_j t^j - f(t)
    \right| \right| \\
    &\le \left| \sum_j (a_j t^j  - f(t)) - \sum_j
    (b_j t^j - f(t)) \right| \\
    &= \left| \sum_j (a_j - b_j) t^j \right| \\
    &\le \sum_j |a_j - b_j|
  \end{align*}
  so
  \[ \|S(\bf{a}) - S(\bf{b})\| \le \sum_j |a_j -
  b_j| .\]
  So $S$ is continuous as a map $(\RR^{n + 1},
  \|\bullet\|_2) \to \RR$. Now $|S(\bf{a})| \to
  \infty$ as $\|\bf{a}\|_2 \to \infty$ by our
  previous result. So we can find an $R$ such that
  $|S(\bf{a})| \ge S(\bf{0})$ for all
  $\|\bf{a}\|_2 \ge R$. $S$ is continuous on the
  compact set $\ol{B(0, R)}$ (closed ball), so has
  a minimum at some $\bf{c} \in \ol{B(0, R)}$.
  $S(c) \le S(\bf{a}) ~\forall \|\bf{a}\|_2 \le
  R$,
  \[ S(\bf{c}) \le S(\bf{0}) \le S(\bf{a}) \qquad
  \forall \|\bf{a}\|_2 \ge R . \qedhere \]}
\end{proof}
\end{flashcard}

\subsubsection*{* Non-examinable material}

Look at the attained minimum $(S - f) t$. Suppose
that it does not satisfy the equiripple criterion.
Then we can perturb it a little to improve the
approximation, which is a contradiction. This
shows that the equiripple is in fact a
\emph{necessary} condition (not just a sufficient
condition).

** This is the end of the non-examinable comments.

\newpage
\section{Gaussian Quadrature}

\textbf{Problem:} Given $f(x_1), \ldots, f(x_n)$,
we would like to estimate
\[ \int_0^b f(t) \dd t .\]

First idea is to use interpolation.

\begin{flashcard}[poly-integral-lemma]
\begin{lemma*}
  If $x_1, \ldots, x_n \in [a, b]$ all distinct.
  Then there exists unique $A_1, \ldots, A_n$ such
  that
  \[ \int_A^b P(t) \dd t = \sum_j A_j P(x_j) \]
  for all $P \in \mathcal{P}_{n - 1}$.
\end{lemma*}

\fcscrap{
So we hope that
\[ \int_a^b f(t) \dd t \stackrel{?}{\approx}
\sum_j A_j f(x_j) \]
}

\begin{proof}
  \cloze{Recall the notation
  \[ e_j(x) = \prod_{i \neq j} \frac{(x -
  x_i)}{(x_j - x_i)} ,\]
  so that $e_i(x_j) = \delta_{ij}$.
  If $P \in \mathcal{P}_{n - 1}$, we know that
  \[ P(t) = \sum_j P(x_j) e_j(t) .\]
  So
  \[ \int_a^ P(t) \dd t \sum_j P(x_j) A_j \]
  with
  \[ A_j = \int_a^b e_j(t) = \dd t .\]

  Conversely, if $\int_a^b P(t) \dd t = \sum_j B_j
  P(x_j)$ for all $P \in \mathcal{P}_n$, then we
  have
  \[ \int_a^b e_i(t) \dd t = \sum_j B_j e_i(x_j) =
  B_i .\]
  So $B_i = A_i$.}
\end{proof}
\end{flashcard}

Without loss of generality $[a, b] = [0, 1]$. If
we choose $x_r = \frac{r}{n}$ and look at $x_0,
\ldots, x_n$ and $n$ is small we get quite good
results. But as $n$ increases, the associated
$A_j$ rapidly take very large values (allowed by
the $A_j$ taking positive and negative values).

Gauss says look at Legendre polynomials.

Recall Gramm-Schmidt:

If $\bf{e}_1, \ldots, \bf{e}_k$ are orthonormal
vectors in an inner product space and $\bf{u}
\notin \Span\{\bf{e}_1, \ldots, \bf{e}_k\}$, we
can find $\bf{e}_{k + 1}$ such that $\bf{e}_1,
\ldots, \bf{e}_{k + 1}$ are orthonormal and
$\bf{u} \in \Span\{\bf{e}_1, \ldots, \bf{e}_{k +
1}\}$.

\begin{flashcard}[leg-poly-defn]
\prompt{Legendre Polynomials? \\}
Recall also that $C([0, 1])$ is an inner product
space if we write
\[ \langle f, g \rangle = \int_0^1 f(t) g(t) \dd t
.\]
Thus we can find $P_0, P_1, P_2, \ldots$ with $P_j
\in \mathcal{P}_j$ and $P_0, P_1, \ldots$
orthonormal, and so we can find unique polynomials
$p_0, p_1, \ldots$ such that $p_n$ has degree $n$
and positive leading coefficient, with $p_0,
p_1, \ldots, p_n$ orthonormal. We call the $p_j$
the \emph{Legendre polynomials}.
\end{flashcard}

\begin{flashcard}[legendre-poly-roots-lemma]
\begin{lemma*}
  The Legendre polynomial $p_n$ has all its roots
  simple, real and lying in $[0, 1]$.
\end{lemma*}

\begin{proof}
  \cloze{Let $\beta_1, \ldots, \beta_m$ be the roots of
  odd order lying in $[0, 1]$ (odd order means
  $p_n$ changes sign as $t$ passes through
  $\beta_j$). Then writing $Q(t) = \prod_j (t -
  \beta_j)$ we have $Q(t) p_n(t)$ is single signed
  and non-zero. So
  \[ \int_0^1 Q(t) p_n(t) \neq 0 \]
  so $\deg Q \ge n$, so $m \ge n$. But $p_n$
  is of degree $n$, so all roots lie in $[0, 1]$
  and are simple.}
\end{proof}
\end{flashcard}
