Exercise 7.23

From the definition, it is easy to see that if $P_n$ is an even function, then so is $P_{n+1}$. Hence by induction, since $P_0$ is trivially an even function, all the $P_n$ are even functions. Since $|x|$ is also even, it suffices to show that the $P_n$ converge uniformly to $x$ on $[0,1]$. In what follows, it is always understood that the variable $x$ lies in $[0,1]$.

Following the hint, note that

I want to show that $0\le P_n(x)\le P_{n+1}(x)\le x$, by induction. Since $P_0(x)=0$ and $P_1(x)=x^2∕2$, this is true for the case $n=0$. Suppose it is true for the case $n$. Then the factors

in $\text{(∗)}$ are nonnegative, so $x-P_{n+1}(x)\ge 0$ or $P_{n+1}\le x$. Also, the terms

in the definition of $P_{n+1}$ are nonnegative, so $P_{n+1}(x)\ge 0$. Hence the factor

in $\text{(∗)}$ lies between 0 and 1, so that $x-P_{n+1}(x)\le x-P_n(x)$, or $P_n(x)\le P_{n+1}(x)$. Putting this all together, $\{P_n\}$ is a monotonically increasing sequence of polynomials on $[0,1]$ which lie between $0$ and $x$.

By $\text{(∗)}$ we have

By elementary calculus, the function $x{(1-x∕2)}^n$ has a maximum value at $x=2∕(n+1)$ of

Hence $P_n(x)$ increases monotonically and uniformly to $x$ as $n\to \infty$.