Evaluate $\lim_{n \rightarrow \infty}P_n(x)$ and $\lim_{n \rightarrow \infty} \sqrt{4n+1} \int_0^1 f(x)P_{2n}(x) dx$

Question

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 $\lim_{k \rightarrow \infty} \left[ \sqrt{4k+1} \int_0^1 P_{2k} (x) f(x) dx \right] = 0.$ $\square$