Let $Lu = (p(x)u'(x))' + q(x)u(x), p \in C^1(I), p \geq 0, q \in C^0(I), I = [a, b]$ be a Sturm-Liouville operator with regular boundary conditions $R_1u = u(a) = 0$ and $R_2u = u(b) = 0$.
We know that there exists $(u_j)_{j\geq1} \subset C^2(I)$, $R_1u_j = R_2u_j = 0$, an orthonormal sequence of eigenvectors with eigenvalues $(\lambda_j)_{j\geq1} \subset \Bbb R \setminus \{0\}$, $\lim\limits_{j\to\infty} \lambda_j = -\infty$, $Lu_j = \lambda_j u_j$, such that for all $v \in C^2(I)$ with $R_1v = R_2v = 0$, the sequence of partial sum of the Fourier series converges uniformly, absolutely, to $v$, i.e. $$\lim\limits_{n,m \to\infty} \sum_{j=n}^m \lvert \lvert \langle v, u_j \rangle u_j \rvert \rvert_{\infty} = 0,$$ $$\lim\limits_{n \to\infty}\lvert \lvert v - \sum_{j=1}^n \langle v, u_j \rangle u_j \rvert \rvert_{\infty} = 0,$$
where $\langle v, u_j \rangle = \int_a^b v(x)u_j(x)dx.$
Furthermore, we can deduce from Bessel inequality that $\lim\limits_{j \to\infty} \lvert \langle v, u_j \rangle \rvert = 0$.
In standard Fourier analysis, if one assumes $v \in C^k$, then the Fourier coefficients $c_n$ decrease as $\frac{1}{n^k}.$
If we assume $v \in C^k, k \geq 2$, $v(a) = v(b) = 0$, can we say anything about the decay of $\lvert \langle v, u_j \rangle \rvert$ in terms of $\lambda_j$ or about the convergence of the sum $\sum_{j=1}^n \lvert \lambda_j \rvert^s \lvert \langle v, u_j \rangle \rvert$ for some $s \gt 1$ ?
For $s = 1$, one can prove that $\lvert \lambda_j \rvert \lvert \langle v, u_j \rangle \rvert = \lvert \langle Lv, u_j \rangle \rvert$, so we deduce the convergence of the sum by Bessel inequality. And for $s \in [0, 1]$, it's about the same.