This following question is a question of Second Edition of Fourier Series and Boundary Value Problems by Churchill about Legendre Polynomials $P_n$:
Explain why it's true that
(a) when $-1 < x < 1$, then
$$\lim_{n \rightarrow \infty}P_n(x) = 0 \hspace{3cm} (n = 0, 1, 2, \cdots);$$
(b) when $f$ is sectional continuous on $(0,1)$, then
$$\lim_{n \rightarrow \infty} \sqrt{4n+1} \int_0^1 f(x)P_{2n}(x) dx = 0$$
$\textbf{My attempt:}$
(a)
By the theory developed in Churchill's book, we know that
$$\lim_{n \rightarrow \infty} \left( P_n(x) , P_n(x) \right) = \lim_{n \rightarrow \infty} ||P_n(x)||^2 = \lim_{n \rightarrow \infty} \frac{2}{2n+1} = 0.$$
By the continuity of the norm, we have
$$||\lim_{n \rightarrow \infty} P_n(x)||^2 \lim_{n \rightarrow \infty} ||P_n(x)||^2 = \lim_{n \rightarrow \infty} \frac{2}{2n+1} = 0,$$
which implies, by the definition of the norm, that
$\lim_{n \rightarrow \infty} P_n(x) = 0$.
(b)
By Cauchy-Schwarz Inequality,
$$\int_0^1 f(x) P_{2n}(x) dx \leq \left( \int_0^1 \left[ f(x) \right]^2 dx \right)^{\frac{1}{2}} \left( \int_0^1 \left[ P_{2n}(x) \right]^2 dx \right)^{\frac{1}{2}}$$
Using that $||P_n(x)||^2 = \frac{2}{2n+1}$, we have that $||P_{2n}(x)||^2 = \int_{-1}^1 \left[ P_{2n}(x) \right]^2 dx = 2 \left( \int_0^1 \left[ P_{2n}(x) \right]^2 dx \right) \Longrightarrow \left( \frac{2}{4n+1} \right) = 2 \left( \int_0^1 \left[ P_{2n}(x) \right]^2 dx \right)$, i.e., $\left( \int_0^1 \left[ P_{2n}(x) \right]^2 dx \right)^{\frac{1}{2}} = \frac{1}{\sqrt{4n+1}}$, then
$$\int_0^1 f(x) P_{2n}(x) dx \leq \left( \int_0^1 \left[ f(x) \right]^2 dx \right)^{\frac{1}{2}} \frac{1}{\sqrt{4n+1}}$$
$$\Longrightarrow \lim_{n \rightarrow \infty} \sqrt{4n+1} \int_0^1 f(x)P_{2n}(x) dx \leq \left( \int_0^1 \left[ f(x) \right]^2 dx \right)^{\frac{1}{2}}$$
I'm stuck here and I do not have ideas on what to do in order to prove the item (b).
I would like to know if my attempt for (a) is correct and I would like to receive a hint for solve (b).
Thanks in advance!
$\textbf{EDIT:}$
I saw how to do the exercises and I return to post the solutions if anyone looks for these exercises.
(a)
By the theory developed by Churchil in his book,
$$|P_n(x)| < \sqrt{\frac{\pi}{2n \left( 1 - x^2 \right)}},$$
then
$$\lim_{n \rightarrow \infty} |P_n(x)| \leq \lim_{n \rightarrow \infty} \sqrt{\frac{\pi}{2n \left( 1 - x^2 \right)}} = 0,$$
so that $\lim_{n \rightarrow \infty} |P_n(x)| = 0$, which implies that
$$\lim_{n \rightarrow \infty} P_n(x) = 0.$$
(b)
Considering the pair extension of $f$ and the orthonormal set $\left\{ \frac{P_n}{\sqrt{\frac{2}{2n+1}}} ; n \in \mathbb{N} \right\}$, we define $c_n := \int_{-1}^1 \frac{P_n}{\sqrt{\frac{2}{2n+1}}} f(x) dx$, then $c_n := \int_{-1}^1 \frac{P_n}{\sqrt{\frac{2}{2n+1}}} f(x) dx = \sqrt{\frac{2n+1}{2}} \int_{-1}^1 P_n(x) f(x) dx$
If $n = 2k (k \in \mathbb{N})$:
$$c_{2k} = 2 \sqrt{\frac{4k+1}{2}} \int_0^1 P_{2k}(x) f(x) dx.$$
If $n = 2k+1 (k \in \mathbb{N})$:
$$c_{2k+1} = 0.$$
Observe that $c_n = \left( f, \frac{P_n}{||P_n||} \right)$, i.e., $c_n$ is the coefficient of generalized Fourier Series of $f$, then we can apply the Bessel's inequality:
$\sum_{k=1}^{\infty} \left[ 2 \sqrt{\frac{4k+1}{2}} \int_0^1 P_{2k} (x) f(x) dx \right]^2 = \sum_{k=1}^{\infty} \left[ \sqrt{\frac{4k+1}{2}} \int_{-1}^1 P_{2k} (x) f(x) dx \right]^2 \leq \int_{-1}^1 f(x)^2dx$
$\int_{-1}^1 f(x)^2dx < \infty$, then $\sum_{k=1}^{\infty} \left[ 2 \sqrt{\frac{4k+1}{2}} \int_0^1 P_{2k} (x) f(x) dx \right]^2 < \infty,$ therefore $\lim_{k \rightarrow \infty} \left[ 2 \sqrt{\frac{4k+1}{2}} \int_0^1 P_{2k} (x) f(x) dx \right] = 0$, then $\lim_{k \rightarrow \infty} \left[ \sqrt{4k+1} \int_0^1 P_{2k} (x) f(x) dx \right] = 0.$ $\square$