I am stuck trying to solve this problem. It's given a linear operator $T:C^{\infty}[-1,1]\rightarrow C^{\infty}[-1,1]\,$ $Tf(x)=\left((x^2-1)f'(x)\right)' $ and I have to show that there exists $\lambda \in \mathbb R $ with $$f(x)=\frac{1}{2^nn!}\left((x^2-1)^n\right)^{(n)}$$ so that $Tf(x)=\lambda f(x)$ (the notation $f=a^{(n)}$ is equivalent to $\frac{d^nf}{da^n}$)
I tried to replace $f(x)$ in the relation from above to get: $$\left((x^2-1)\frac{1}{2^nn!}\left((x^2-1)^n\right)^{(n+1)}\right)'=\lambda \frac{1}{2^nn!}\left((x^2-1)^n\right)^{(n)}$$ Now I multiply both sides by $2^n n!$ and derivate the LHS:$$2x\left((x^2-1)^n\right)^{(n+1)}+(x^2-1)\left((x^2-1)^n\right)^{(n+2)}=\lambda \left((x^2-1)^n\right)^{(n)}$$ And now integrate both sides then use IBP $$2x\left((x^2-1)^n\right)^{(n)}-\int 2\left((x^2-1)^n\right)^{(n)}dx+(x^2-1)\left((x^2-1)^n\right)^{(n+1)}-\int 2x\left((x^2-1)^n\right)^{(n+1)}dx=\lambda \int \left((x^2-1)^n\right)^{(n)}dx$$ Now there is a nice simplification and I am left with:$$(x^2-1)\left((x^2-1)^n\right)^{(n+1)}=\lambda \left((x^2-1)^n\right)^{(n-1)}$$ After this no attempt was successful, I tried to derivate two times $(x^2-1)^n$ on the LHS or maybe to write $(x^2-1)^n=(x-1)^n(x+1)^n\,$ but I didnt found anything relevant.
I would appreciate if you can help me finish this problem. I suspect it has a nice answer.
Let $\mathbb{R}_n[X]$ be the vector space of real polynomials of degree $\leq n$, endowed with the scalar product: $$\langle f,g \rangle=\int_{-1}^1 f(x)g(x)dx.$$
Legendre polynomials $$f_m(x)=\frac{1}{2^m m!} \frac{d^m}{dx^m}\left((x^2-1)^m \right),$$ are orthogonal to each others for this scalar product. In particular, the orthogonal to $\mathbb{R}_{n-1}[X]$ in $\mathbb{R}_n[X]$ is $\mathbb{R}f_n$.
From degree consideration, it is clear that the restriction of $T$ to $\mathbb{R}_n[X]$ is an endomorphism.
Notice that $T$ is self-adjoint. Indeed,
$$\langle Tf,g\rangle=\langle ((x^2-1)f')',g\rangle=-\langle (x^2-1)f',g'\rangle= -\langle f',(x^2-1)g'\rangle=\langle f,Tg\rangle,$$ using integration by part twice.
Since $T$ is self-adjoint, by induction, it maps the orthogonal of $\mathbb{R}_{n-1}[X]$ in $\mathbb{R}_n[X]$ to itself, namely $\mathbb{R}f_n$ into $\mathbb{R}f_n$.
This doesn't give us the value of $\lambda$, but at least show that $f_n$ is an eigenvector.
EDIT : to obtain the value of $\lambda_n$ we may proceed as follows. First, we compute, for $k\geq 2$, $$T(X^k)=(k+1)k X^k-k(k-1)X^{k-2},$$ and $T(X)=2X$. So the matrix for $T$ in the basis $(1,X,X^2,...,X^n)$ is upper-triangular, and the trace is simply $\sum_{k=1}^n k(k+1)$. The eigenvalue for $f_n$ is the difference between the trace of $T$ on $\mathbb{R}_n[X]$ and the trace of $T$ on $\mathbb{R}_{n-1}[X]$. In other words, if $Tf_n=\lambda_nf_n$, we have $$\lambda_n=n(n+1).$$