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Last time we introduced Bernstein's version of the
Weierstrass polynomial approximation theorem:

\begin{theorem*}
  If $f : [0, 1] \to \RR$ is continuous then
  writing
  \[ P_n(p) = \sum_r {n \choose r} f \left(
  \frac{r}{n} \right) p^r (1 - p)^{n - r} \]
  then $P_n \to f$ uniformly.
\end{theorem*}

\vspace{-1em}
We use a probabilistic interpretation and
Chebyshev inequality:

\begin{lemma*}
  If $X$ is a bounded random variable, then
  \[ \PP(|X - \EE X| \ge a) \le
  \frac{\Var(X)}{a^2} .\]
\end{lemma*}

\begin{proof}
  See \courseref[Probability]{prob} from IA.
\end{proof}

\begin{proof}[Proof of Bernstein's version of
  Weierstrass polynomial approximation theorem]
  Now consider $X_1, X_2, \ldots, X_n$ independent
  random variables with $\PP(X_j = 1) = p$,
  $\PP(X_j = 0) = 1 - p$. Then
  \[ \Var(X_j) = \EE((X - \EE X)^2) = \EE((X -
  p)^2) = (1 - p) p^2 + p(1 - p)^2 = p(1 - p) \le
  \quarter \]
  by AM-GM. Then
  \[ \Var \left( \frac{X_1 + \cdots + X_n}{n}
  \right) = \frac{1}{n^2} \sum_j \Var(X_j) \le
  \frac{1}{4n} .\]
  Now consider the function $f$. By compactness
  there exists an $M$ such that $|f(p)| \le M$ for
  all $p \in [0, 1]$. Also, $f$ is uniformly
  continuous -- that is to say given $\eps > 0$,
  there exists $\delta(\eps) > 0$ such that
  \[ |s - t| \le \delta \implies |f(s) - f(t)| <
  \eps \]
  We fix $\eps > 0$, $\delta(\eps) > 0$ through
  the proof. Then we say that we can make $\eps$
  as small as we want.
  \begin{align*}
    |P_n(p) - f(p)|
    &= |\EE((\ol{X})) - f(p)| \\
    &= |\EE(f(\ol{X}) - f(p))| \\
    &= \sum_{k = 0}^n \PP \left( \ol{X} =
    \frac{r}{n} \right) \left| f \left(
    \frac{r}{n} \right) - f(p) \right| \\
    &= \sum_{\left| \frac{r}{n} - p \right| \le
    \delta} + \sum_{\left| \frac{r}{n} - p \right|
    > \delta} \\
    &\le \sum_{\left| \frac{r}{n} - p \right| \le
    \delta} \eps \PP\left(\left|\ol{X} -
    \frac{r}{n}\right| \le p\right) + \sum_{\left|
    \frac{r}{n} - p\right| > \delta} M \PP \left(
    \left| \ol{X} - \frac{r}{n} \right| = p\right)
    \\
    &\le \eps + 2M \PP \left( \left| \ol{X} -
    \frac{r}{n} \right| \ge \delta \right) \\
    &\le \eps + 2M \frac{\Var(\ol{X})}{n}
    \frac{1}{\delta^2}
    &&\text{(Chebyshev)} \\
    &\le \eps + \frac{2M}{4} \frac{1}{n} \\
    &< 2\eps
  \end{align*}
  if $n$ is large enough.
\end{proof}

\subsubsection*{Best uniform approximation}

\begin{flashcard}[equiripple-thm]
\begin{theorem*}[Chebyshev equiripple criterion]
  \cloze{If $f : [a, b] \to
  \RR$ is continuous and $P$ is a polynomial of
  degree at most $n$ such that there exists $M \ge
  0$ and $a = a_0 < a_1 < \cdots < a_n < a_{n + 1}
  \le b$ and $P(a_j) - f(a_j) = (-1)^j M$ (or
  $P(a_j) - f(a_j) = (-1)^{j + 1} M$) and such
  that $\|P - f\|_\infty$. Then $\|P - f\|_\infty \le \|Q -
  f\|_\infty$ for all polynomials $Q$ of degree $n$
  or less.}
\end{theorem*}

\fcscrap{
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/b9ef499aec9e4c30.png}
\end{center}
}

\begin{proof}
  \cloze{Without loss of generality $P(a_j) - f(a_j) =
  (-1)^j M$. If $Q \in \mathcal{P}_n$ and $\|P -
  f\|_\infty > \|Q - f\|_\infty$ then $P(a_{2j} -
  f) \ge Q(a_{2j}) - f$, so $P(a_{2j}) >
  Q(a_{2j})$. Similarly for $P(a_{2j + 1}) <
  Q(a_{2j + 1})$. So there exists $c_j$ with $a_0
  < c_0 < a_1 < c_1 < a_2 < \cdots < c_n < a_{n +
  1}$ with $(P - Q)(c_j) = 0$ (intermediate value
  theorem). So $P - Q$ has $n + 1$ zeroes, but $P
  - Q \in \mathcal{P}_n$, so $P - Q = 0$,
  contradiction.}
\end{proof}
\end{flashcard}

Now $\|T_n\|_\infty \le 1$ and $T_n$ is
alternately $\pm 1$ at $n + 1$ points. The
coefficient of $t^n$ in $T_n$ is $2^{-n + 1}$ (for
$n \ge 1$). So $2^{n - 1} T_n(t) = t^n - Q_{n -
1}(t)$ with $Q_{n - 1}$ then best uniform
approximation to $t^n$.

\begin{flashcard}[poly-minima-coeff-coro]
\begin{corollary*}
  There exists an $\eps_n$ such that if $P(T) =
  \sum_{j = 0}^n a_j t^j$ and $\exists k$ such
  that $|a_k| \ge 1$, then $|P(t)| \ge \eps_n
  ~\forall t \in [-1, 1]$.
\end{corollary*}

\begin{proof}
  \cloze{Proof by induction. True for $n = 0, 1$ by
  inspection. Now suppose true for $n = N$.
  \[ P(t) = \sum_{j = 0}^{N + 1} a_j t^j = a_{N +
  1} t^{N + 1} + Q_N(t) .\]
  Say $Q_N(t) = \sum_{j = 0}^N a_j t^j$. If $|a_{N
  + 1}| < \frac{\eps_N}{2}$, then
  \[ \|P\|_\infty \ge \|Q\|_\infty -
  \frac{\eps_N}{2} \ge \frac{\eps_N}{2} .\]
  If $|a_{N + 1}| > \frac{\eps_N}{2}$ and we know
  from Chebyshev that
  \[ |P(t)| \ge \frac{\eps_N}{2} \inf_{Q \in
  \mathcal{P}_n} |Q(t) - t^{N + 1}| = \eps_N' > 0
  . \qedhere \]}
\end{proof}
\end{flashcard}