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\begin{document}
\vspace*{.3in}
\noindent{\bf Rudin, page 165/23:}

\proof 
I proceed by induction to show that for all $n\geq 0$ and $|x|<1$ that
$$
0\leq P_n(x) \leq P_{n+1}(x) \leq |x|
$$
and 
$$
|x| - P_n(x) \leq |x|\left(1-\frac{|x|}{2}\right)^n.
$$
For the moment, let me suppose that these inequalities are proven.  By
finding roots of the derivative, it is easily shown that the
function $h:[0,1]\to\R$ given by $h(t) = t(1-t/2)^n$ will achieve its
maximum at $t=0,1$ or $2/(n+1)$ (i.e. at endpoints or critical points).
We have $h(0) = 0$, $h(1) = 1/2^n$, and $h(2/n+1) < 2/(n+1)$ (since
$h(x)<x$ when $x\in(0,1]$).  In
any case, $h(t)< 2/(n+1)$ for all $t$.  Thus
$$
||x| - P_n(x)| = |x| - P_n(x) \leq h(x) < 2/(n+1)
$$
for all $x\in[-1,1]$, and it follows that $P_n$ converges uniformly
to $|x|$.

Now I return to the proof of the inequalities asserted earlier.
When $n=0$, we have $P_n(x) = 0$ and $P_{n+1}(x) = x^2/2$, and all
the inequalities are easily verified directly.  So now I assume that 
the inequalities have been verified for $n=k$, and I will prove that
they hold when $n=k+1$.  First of all, we use $0\leq P_k(x) \leq |x|$
to estimate
$$
P_{k+1}(x) = P_k(x) + \frac{x^2-P_k^2(x)}{2} 
           \geq P_k(x) + \frac{x^2 - |x|^2}{2} = P_k(x).
$$
Secondly,
$$
|x| - P_{k+1}(x) = [|x|-P_k(x)]\left[1-\frac{|x|+P_k(x)}{2}\right]    
   \geq [|x|-P_k(x)]\left[1-\frac{|x|+|x|}{2}\right]
   \geq 0
$$
for $|x|\leq 1$.  So $P_{k+1}(x) \leq |x|$.  Finally, in the other direction
\begin{eqnarray*}
|x| - P_{k+1}(x) & = & [|x|-P_k(x)]\left[1-\frac{|x|+P_k(x)}{2}\right] \\
                 & \leq & [|x|-P_k(x)]\left[1-\frac{|x|}{2}\right] \\
		 & \leq & |x|\left(1-\frac{|x|}{2}\right)^k
		         \left[1-\frac{|x|}{2}\right] \\
			 & = & |x|\left(1-\frac{|x|}{2}\right)^{k+1},
\end{eqnarray*}
where the second inequality comes from the induction hypothesis.  This
completes the induction step and the proof.
\qed
\end{document}

