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\begin{document}

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\section{Approximation by Polynomials}

We know that if $f$ is analytic then it can be approximated by Taylor
polynomials, $P_n(x) = \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$. In
fact $P_n(x) \to f(x)$ uniformly on compact subsets of its interval of
convergence.

Can continuous functions be approximated by polynomials?

\subsection{Lagrange Interpolation}
Let $f$ be continuous on $[a,b]$ and pick $x_0,\dots,x_n\in[a,b]$. Then
there is a polynomial $P_n$ of degree $n$ such that $P_n(x_j) = f(x_j)$ for
$j=0,\dots,n$.

Here is how to construct $P_n$: Define $q_k(x) = \Pi_{j\ne k}(x - x_j)$.
Then $q_k(x_j) = 0$ if $j\ne k$ and $q_k(x_k) \ne 0$. Next define
\[ Q_k(x) = \frac{q_k(x)}{q_k(x_k)} \quad \mbox{so}\quad 
Q_k(x_j) = \left\{\begin{array}{ll} 0 & j\ne k\\ 1 & j = k\end{array}\right.
\]
Finally, letting, $a_k = f(x_k)$, we see that the polynomial
\[ P_n(x) = \sum_{k=0}^n a_k Q_k(x) \]
has the desired properties.

\emph{Example:} If $f(x) = \sin(4x)$ and $x_k = k$, for $k=0,\dots,5$, the
5th degree polynomial obtained by Lagrange interpolation does not
approximate $f(x)$ well.

\[ \includegraphics{Pictures/04.fig1.eps} \]

We could hope that $P_n \to f$ as $n\to\infty$, although, it is not clear
how to prove this. A better idea is to use \emph{convolution with
polynomials}.

\subsection{Convolution}

\begin{definition}
The \emph{support} of a function $f$ is the closure of the set
of points $x$ in the domain of $f$ such that $f(x)\ne 0$.
\end{definition}

\emph{Note:} If the support of a function $f$ is compact then $f$ must
vanish outside a bounded interval: $f(x) = 0$, $\forall\, x\notin[a,b]$.

\begin{definition}
Let $f$ and $g$ be integrable functions and assume either $f$ or $g$
has compact support. The \emph{convolution} of $f$ and $g$ is
\[ f*g(x) = \int_{-\infty}^\infty f(x-t)g(t)\,dt \]
\end{definition}

\emph{Note:} By making a change of variable, $u = x-t$, $t = x-u$,
is it easy to show that $f*g = g*f$.

\begin{theorem}\label{04:POLY}
Let be $f$ integrable and have compact support. If $g$ is a polynomial
of degree $n$, then $f*g$ is a polynomial of degree $\le n$.
\end{theorem}

\proof Let $g(x) = \sum_{k=0}^n a_k x^k$ and suppose $f(x) = 0$, $\forall\,
x\notin[a,b]$. Then
\begin{eqnarray*}
f*g(x) &=& g*f(x) = \int_a^b g(x-t)f(t)\,dt \\
&=&\int_a^b \sum_{k=0}^n a_k (x-t)^k f(t)\,dt\\
&=&\int_a^b \sum_{k=0}^n \sum_{j=0}^k
(-1)^{k-j}a_k{k\choose j}x^jt^{k-j}f(t)\,dt\\
&=&\sum_{j=0}^n b_j x^j
\end{eqnarray*}
where
\[b_j = \sum_{k=j}^n\Big\{(-1)^{k-j}a_k{k\choose j}\int_a^b
t^{k-j}f(t)\,dt\Big\} \in \R \]
\qed

\begin{theorem}\label{04:DERIVATIVE}
Let $f$ be $C^1$ and have compact support. Let $g$ be continuous. Then
$(f*g)' = f'*g$.
\end{theorem}

\proof
We compute the derivative at $x_0$ using an arbitrary sequence $x_n\to
x_0$:
\begin{eqnarray*}
(f*g)'(x_0) &=& \lim_{n\to \infty} \frac{f*g(x_n) - f*g(x_0)}{x_n-x_0}\\
&=&\lim_{n\to \infty}
\frac{1}{x_n-x_0}\Big[\int\!\!f(x_n-t)g(t)\,dt-\int\!\!f(x_0-t)g(t)\,dt\Big]\\
&=&\lim_{n\to \infty} \int\frac{f(x_n-t) - f(x_0-t)}{x_n-x_0}g(t)\,dt\\
&=&\lim_{n\to \infty} \int f'(z_n-t)g(t)\,dt
\end{eqnarray*}
for some $z_n$ between $x_0$ and $x_n$ by the Mean Value Theorem.
We now show that $f'(z_n-t)g(t) \to f(x_0-t)g(t)$ \emph{uniformly} so that
we may interchange the limit and the integral, by Theorem~\ref{02:INTEGRALS},
to get
\begin{eqnarray*}
(f*g)'(x_0) &=& \int\!\!\!\lim_{n\to\infty}f'(z_n-t)g(t)\,dt
= \int f'(x_0-t)g(t)\,dt\\ &=& (f'*g)(x_0)
\end{eqnarray*}

Since $f$ has compact support, say contained in $[a,b]$, the
support of $f'$ must necessarily fall in that same interval
and therefore is compact (if $f\equiv 0$ then $f'\equiv 0$).
By Theorem~\ref{01:UNIFORMCONTINUITY}, $f'$ is uniformly continuous
on $[a,b]$, and hence uniformly continuous on $\R$ since $f'\equiv 0$
outside $[a,b]$.  Since $z_n \to x_0$, there is
some closed interval, $[c,d]$, that contains all $z_n$ and $x_0$.
Then $z_n-t, x_0-t\in[a,b] \Rightarrow t\in[c-b,d-a]$.
Let $M$ be the maximum of $g$ on $[c-b,d-a]$.

Since $f'$ is uniformly continuous, $\forall\, 1/m$, $\exists\, 1/k$ such that
\[ | f'(z_n - t) - f'(x_0 - t) | < \frac{1}{m\cdot M} \]
whenever
\[ |(z_n-t)-(x_0-t)| = |z_n - x_0| < \frac{1}{k} \]
However, $\exists\, N$ such that
\[ |z_n - x_0| < \frac{1}{k}\quad \forall\, n\ge N \]
Therefore, if $n \ge N$, then
\[ | f'(z_n - t)g(t) - f'(x_0 - t)g(t) | \le
   | f'(z_n - t) - f'(x_0 - t) | M < \frac{1}{m}\quad \forall\, t \]
and hence $f'(z_n-t)g(t) \to f(x_0-t)g(t)$ uniformly.
\qed

\subsection{Weierstrass Approximation Theorem}

\begin{theorem}\label{04:WAT}
Let $f$ be continuous on $[a,b]$. Then there exists a sequence of polynomials
converging uniformly to $f$ on $[a,b]$.
\end{theorem}

\proof We may assume $[a,b] = [0,1]$, for if the theorem is true for
\[ g(t) = f(a + (b-a)t)\quad t\in[0,1] \]
then there exists polynomials $Q_m(t)\to g(t)$ uniformly on $[0,1]$
and this implies that the polynomials $P_m(x) = Q_m((x-a)/(b-a)) \to
f(x)$ uniformly on $[a,b]$. We may also assume $f(0) = f(1) = 0$, for
if the theorem is true for
\[ g(x) = f(x) - f(0) - x(f(1)-f(0)) \]
then there exist polynomials $Q_m\to g$ uniformly on $[0,1]$ and this
implies that the polynomials $P_m(x) = Q_m(x) + f(0) - x(f(1)-f(0))\to
f(x)$ uniformly on $[0,1]$.  We extend $f(x)$ to a continuous function
on $\R$ by $f(x) = 0$, $\forall\, x\notin[0,1]$.

Define $h_m(x) = c_m^{-1}(1-x^2)^m$ where $c_m = \int_{-1}^1(1-x^2)^mdx$,
so
\begin{equation}\label{04:ESTIMATE1}
\int_{-1}^1 h_m(x)\,dx = 1
\end{equation}
In order to estimate $h_m$ we need a lower bound for $c_m$:
\[ c_m = \int_{-1}^1(1-x^2)^mdx = 2\int_0^1(1-x^2)^mdx
\ge 2\int_0^{1/\sqrt{m}}(1-x^2)^mdx \]
It is easy to verify the inequality $1 - mx^2 \le (1-x^2)^m$ for
$x\in[0,1]$: If $q(x) = (1-x^2)^m - (1 - mx^2)$ then $q(x) \ge 0$ because
$q(0) = 0$ and $q'(x) \ge 0$ for $x\in[0,1]$. Inserting this inequality
above gives
\[ c_m \ge 2\int_0^{1/\sqrt{m}}(1-m x^2)dx = \frac{4}{3\sqrt{m}} >
\frac{1}{\sqrt{m}} \]
\[ \resizebox{!}{1.5in}{\includegraphics{Pictures/04.fig2.eps}} \]
Since, $c_m^{-1} < \sqrt{m}$ we see that $h_m \to 0$ uniformly for $1/n
\le |x| \le 1$:
\begin{equation}\label{04:ESTIMATE2}
h_m(x) = c_m^{-1}(1-x^2)^m \le \sqrt{m}\Big(1-\frac{1}{n^2}\Big)^m \to 0,
\quad m\to\infty
\end{equation}
On the other hand $\int_{-1}^1h_m(x)\,dx = 1$, so the graph of $h_m(x)$
is more and more concentrated at 0 as $m\to\infty$. In fact, we may
think of the limit as the Dirac delta function, $\lim_{m\to\infty}h_m(x)
= \delta_0(x)$.
\[ \resizebox{!}{1.5in}{\includegraphics{Pictures/04.fig3.eps}} \]
Define $P_m(x) = \int_{-1}^1 f(x-1)h_m(t)\,dt = f*h_m(x)$ for
$x\in[0,1]$. By Theorem~\ref{04:POLY}, $P_m$ is a polynomial of degree
$\le 2m$. We now show that $P_m\to f$ uniformly.

Given $\epsilon>0$, choose $1/n$ such that if $|y-x| < 1/n$, then
$|f(y)-f(x)|<\epsilon/2$ (uniform continuity). Using the fact that
$\int_{-1}^1h_m(t)\,dt = 1$, we get
\begin{eqnarray*}
|P_m(x) - f(x)| &=&\Big|\int_{-1}^1 f(x-t)h_m(t)\,dt -
f(x)\int_{-1}^1h_m(t)\,dt\Big|\\
&=&\Big|\int_{-1}^1 \big(f(x-t) - f(x)\big)h_m(t)\,dt\Big|\\
&\le&\int_{-1}^1 \big|f(x-t) - f(x)\big|h_m(t)\,dt\\
\end{eqnarray*}
Now, $|f(x-t)-f(t)| < 2M$ where $M$ is the maximum of $f$ on $[0,1]$
and $|f(x-t)-f(x)|<\epsilon/2$ if $|(x-t)-x| = |t| < 1/n$. So, breaking
$[-1,1]$ up into $[-1,-1/n]\cup[-1/n,1/n]\cup[1/n,1]$, we get
\[ |P_m(x) - f(x)| \le2M\!\!\int_{-1}^{-1/n}\hspace{-.25in} h_m(t)\,dt
+ \frac{\epsilon}{2} \int_{-1/n}^{1/n}\hspace{-.15in} h_m(t)\,dt
+ 2M\!\!\int_{1/n}^1\hspace{-.1in} h_m(t)\,dt \]
The first and third integrals are $\le 2M\sqrt{m}(1-1/n^2)^m(1-1/n) <
2M\sqrt{m}(1-1/n^2)^m$ by estimate (\ref{04:ESTIMATE2}). The middle
integral is $< \epsilon/2$ by property~(\ref{04:ESTIMATE1}). So
\[ |P_m(x)-f(x)| \le 4M\sqrt{m}\Big(1-\frac{1}{n^2}\Big)^m +
\frac{\epsilon}{2} < \epsilon \]
for $m$ large enough and for all $x\in[0,1]$. Therefore, $P_m \to f$
uniformly on $[0,1]$. \qed

One of the advantages of using convolution for polynomial approximations is
that it gives more information about derivatives.

\begin{corollary} If $f\in C^1[a,b]$, then there exists a sequence
of polynomials $P_m$ such that $P_m\to f$ and $P_m'\to f'$ uniformly
on $[a,b]$.
\end{corollary}

\proof As in the previous proof, we may assume that $[a,b] = [0,1]$.
Furthermore, by subtracting the cubic polynomial
\begin{eqnarray*}
\lefteqn{[2f(0)-2f(1)+f'(0)+f'(1)]x^3}&&\\
&&{}+[3f(1)-3f(0)+2f'(0)-2f'(0)-f'(1)]x^2+f'(0)x + f(0)
\end{eqnarray*}
from $f$ we may assume $f$ and $f'$ can be extended continuously by 0
to $\R$. Let $P_m = f*h_m$ as above.  By Theorem~\ref{04:DERIVATIVE},
$P_m' = f'*h_m$. Therefore both $P_m\to f$ and $P_m'\to f'$ uniformly
on $[a,b]$ by the previous proof.
 \end{document}
