We place ourselves in the particular case where $E = \mathbb{R}_{2m}[X]$, with $m \geq 2$ a fixed natural integer. This vector space is equipped with the scalar product $$\forall (P,Q) \in E^2, \quad (P \mid Q) = \int_{-1}^{1} P(t)Q(t)\,dt$$ The endomorphisms $T(P) = P'$ and $M(P) = P^*$ are defined as before. We set $$\mathbb{R}_{2m-1}^0[X] = \{P \in \mathbb{R}_{2m-1}[X] \mid P(-1) = 0 \text{ and } P(1) = 0\}$$ The polynomials $L_n$ are defined by $L_n(X) = \frac{1}{2^n n!} R_n^{(n)}(X)$ where $R_n(X) = (X^2-1)^n$. Let $\lambda \in \mathbb{R}$. We consider the problem: find $P \in \mathbb{R}_{2m-1}^0[X]$ such that $$\forall Q \in \mathbb{R}_{2m-1}^0[X], \int_{-1}^{1} P(t)Q(t)\,dt - \lambda \int_{-1}^{1} P'(t)Q'(t)\,dt = 0$$ Show that this problem admits a non-identically zero solution $P$ if and only if $\lambda$ is a root of the polynomial $$K(X) = \left(\sum_{k=0}^{m-1} (-1)^k L_{2m-1}^{(2k+1)}(1) X^k\right) \cdot \left(\sum_{k=0}^{m-1} (-1)^k L_{2m}^{(2k+1)}(1) X^k\right)$$
We place ourselves in the particular case where $E = \mathbb{R}_{2m}[X]$, with $m \geq 2$ a fixed natural integer. This vector space is equipped with the scalar product
$$\forall (P,Q) \in E^2, \quad (P \mid Q) = \int_{-1}^{1} P(t)Q(t)\,dt$$
The endomorphisms $T(P) = P'$ and $M(P) = P^*$ are defined as before. We set
$$\mathbb{R}_{2m-1}^0[X] = \{P \in \mathbb{R}_{2m-1}[X] \mid P(-1) = 0 \text{ and } P(1) = 0\}$$
The polynomials $L_n$ are defined by $L_n(X) = \frac{1}{2^n n!} R_n^{(n)}(X)$ where $R_n(X) = (X^2-1)^n$.
Let $\lambda \in \mathbb{R}$. We consider the problem: find $P \in \mathbb{R}_{2m-1}^0[X]$ such that
$$\forall Q \in \mathbb{R}_{2m-1}^0[X], \int_{-1}^{1} P(t)Q(t)\,dt - \lambda \int_{-1}^{1} P'(t)Q'(t)\,dt = 0$$
Show that this problem admits a non-identically zero solution $P$ if and only if $\lambda$ is a root of the polynomial
$$K(X) = \left(\sum_{k=0}^{m-1} (-1)^k L_{2m-1}^{(2k+1)}(1) X^k\right) \cdot \left(\sum_{k=0}^{m-1} (-1)^k L_{2m}^{(2k+1)}(1) X^k\right)$$