Suppose $f\in L^2([0,1],\lambda)$. Are there assumptions on the smoothness of $f$ which translate into the particular behavior of Frourier coefficients. Namely, I have arbitrary complete orthonormal basis $(\varphi_j)_{j=1}^\infty$ (not necessary trigonometric) and so $$f = \sum_{j=1}^\infty\langle \varphi_j,f\rangle \varphi_j$$
I would like to asses the order of $$\left\|\sum_{j=m}^\infty \langle \varphi_j,f\rangle \varphi_j\right\|^2 = \sum_{j=m}^\infty|\langle \varphi_j,f\rangle|^2$$
Is it possible to assume that $f$ is smooth in some sense e.g. Holder, or Sobolev with some exponent, which would give me, for instance, $$\sum_{j=m}^\infty|\langle \varphi_j,f\rangle|^2 = O(m^{-\gamma})$$
Suppose $A$ is selfadjoint with a complete orthonormal basis of eigenvectors $\{ e_{n}\}_{n=1}^{\infty}$ and corresponding eigenvalues $\{ \lambda_{n}\}_{n=1}^{\infty}$ that tend to $\infty$ in absolute value. Then $$ f \in \mathcal{D}(A^{N}) \iff \sum _{n=1}^{\infty}\lambda_{n}^{2N}|(f,e_{n})|^{2} < \infty. $$ The behavior of the ordinary Fourier coeffficients of $f$ has to do with how many derivatives the function $f$ has, because the functions $\{ e^{2\pi in\theta}\}_{n=-\infty}^{\infty}$ are the eigenfunctions of $\frac{1}{i}\frac{d}{d\theta}$ with corresponding integer eigenvalues. The more times you can apply the differentiation operator, the better behaved the Fourier coefficients must be. For example, $f \in L^{2}$ is absolutely continuous and periodic on $[0,1]$ with derivative $f'\in L^{2}$ if and only if $$ \sum_{n=0}^{\infty}\left|\frac{1}{2\pi}\int_{0}^{2\pi}f(\theta)e^{-in\theta}d\theta\right|^{2}n^{2} < \infty. $$ If you start with an arbitrary set of orthogonal functions $\{ e_{n} \}$ and arbitrarily assigned eigenvalues $\{ \lambda_{n} \}$, then you can define $$ Af = \sum_{n}\lambda_{n}(f,e_{n})e_{n} $$ Then you definitely get $$ f\in\mathcal{D}(A^{N}) \iff \sum_{n}|\lambda_{n}|^{2N}|(f,e_{n})|^{2} < \infty, $$ but the statement is not of much general use without knowing more about $A$.
The normalized ordinary Legendre polynomials $P_{n}$ are eigenfunctions of the Legendre operator: $$ Lf = -\frac{d}{dx}(1-x^{2})\frac{df}{dx},\\ LP_{n} = n(n+1)P_{n},\;\;\; n=0,1,2,\cdots. $$ The domain of this operator $L$ consists of all bounded twice absolutely continuous functions $f \in L^{2}[-1,1]$ for which $Lf \in L^{2}$. So you can work out what it means for $f \in \mathcal{D}(L^{N})$ in terms of differentiability, but the conditions so required are equivalent to the condition that $$ \sum_{n=0}^{\infty}(n(n+1))^{2N}|(f,P_{n})|^{2} < \infty. $$ (It is critical that $P_{n}$ be normalized so that $(P_{n},P_{n})=1$.) So, the general rule here is also that "the smoother the $f$, the faster the rate of decay of the Legendre Fourier coefficients $(f,P_{n})$." But this is all just related to the fact that $$ L^{N}f = \sum_{n=0}^{\infty}\{n(n+1)\}^{N}(f,P_{n})P_{n}. $$