Context
(Approximation to the identity) Let $\epsilon >0$ and $0 < \delta <1$. A function $f: \mathbb{R} \to \mathbb{R}$ is said to be an $(\epsilon, \delta)$- approximation to the identity if it obeys the following three properties:
(1) $f$ is supported on $[-1, 1]$, and $f(x) \ge 0$ for all $-1 \le x \le 1$.
(2) $f$ is continuous, and $\int_{-\infty}^\infty f =1$.
(3) $|f(x)| \le \epsilon$ for all $\delta \le |x| \le 1$.
Question
(a) Prove that for any real number $0 \le y \le 1$ and any natural number $n \ge 0$, that $(1-y)^n \ge 1-ny$.
(b) Show that $\int_{-1}^1 (1-x^2)^n \ge \frac1{\sqrt{n}}$. (Hint: for $|x| \le 1/\sqrt{n}$, use part (a); for $|x| \ge 1/\sqrt{n}$, just use the fact that $(1-x^2)$ is positive. It is also possible to proceed via trigonometric substitution, but I would not recommend this unless you know what you are doing.)
(c) Prove Lemma 3.8.8: For every $\epsilon >0$ and $0 < \delta <1$ there exists an $(\epsilon, \delta)$- approximation to the identity which is a polynomial $P$ on $[-1, 1]$ .
(Hint: choose $f(x)$ to equal $c(1-x^2)^N$ for $x \in [-1, 1]$ and to equal zero for $x \not\in [-1, 1]$, where $N$ is a large number $N$, where $c$ is chosen so that $f$ has integral $1$, and use (b).) .
Attempt
a. Done using induction.
b. as suggested in the hint, $$\int_{-1}^1 (1-x^2)^n = \int_{-1/\sqrt{n}}^{1/\sqrt{n}} (1-x^2)^n +\int_{-1}^{-1/\sqrt{n}} (1-x^2)^n+\int_{1/\sqrt{n}}^1 (1-x^2)^n.$$
I get $\int_{-1/\sqrt{n}}^{1/\sqrt{n}} f\ge \frac4{3\sqrt{n}}$. But, I am not sure how to compute the second and third term. How can I use the fact that $(1-x^2)$ is positive?
c. I understand that $f$ satisfies (1). But, I think $c$ must depends on $N$ to make the integral 1. Doesn't it? Then, does it still satisfy property (2)? I am also not sure about property (3). How the result in (b) helps to show (3)?
I appreciate if you give some help!
For (b) you only need the fact that the second and third terms are $\geq 0$. So $\int_{-1}^{1}(1-x^{2})^{n}dx \geq \frac 4 {3\sqrt n} \geq \frac 1 {\sqrt n}$. For (c) we choose $c$ so that $c\int_{-1}^{1}(1-x^{2})^{N}dx =1$. By (b) this gives $c \leq \sqrt N$. For $\delta \leq |x|\leq 1$ we get $|f(x)| \leq \sqrt N (1-\delta^{2})^{N}$. Can you show that $\sqrt N t^{N} \to 0$ as $ N \to \infty$ for any $t \in (0,1)$? Once you show this we can choose $N$ large enough to make $f$ an $(\epsilon, \delta)$ approximation to the identity.