Let $\{e_n\}$ be an orthonormal basis of $L^2([0,1],\mathbb R)$. Let us investigate the convergence of the series $$ \sum_n|\langle f,e_n\rangle|, $$ where $f\in L^2([0,1],\mathbb R)$. Do there exist conditions on $f$ which ensure the convergence of the series? If yes, what are these conditions?
This series reminds the series of Parseval's identity which states that $\sum_n|\langle f,e_n\rangle|^2=\|f\|^2$. So the series $\sum_n|\langle f,e_n\rangle|^2$ converges for each $f\in L^2([0,1],\mathbb R)$ without any additional assumptions. However, the term of the series is squared in Parseval's identity and it seems that additional assumptions are required for the series without $2$ in the exponent to converge.
Using the triangle and the Cauchy-Schwarz inequalities, we can obtain the following inequality \begin{align*} \int_0^1|f(x)|\mathrm dx &=\int_0^1\biggl|\sum_n\langle f,e_n\rangle e_n(x)\biggr|\mathrm dx\\ &\le\sum_n|\langle f,e_n\rangle|\int_0^1|e_n(x)|\mathrm dx\\ &\le\sum_n|\langle f,e_n\rangle|. \end{align*} Hence, if the function $f$ is not integrable, the series diverges.
Any help is much appreciated!
The condition $\langle f,f\rangle <\infty$ is obviously not enough: The sequence $x = (1, \frac{1}{2},\frac{1}{3},\dots)$ is obviously in $\ell^2$, thus it corresponds to a function in $L^2[-\pi,\pi]$ which we will denote by $f$. We have $\langle f,f \rangle = \sum_{k=1}^\infty \frac{1}{k^2} <\infty$ due to parseval equality, but the harmonic series diverge.
However, if we have a $2\pi$-peridioc function $f$ with the condition $f' \in L^2$ then $\sum_{k=1}^\infty |\langle f, e_n \rangle | <\infty$.
To see this, first show that the fourier coefficients of $f,f'$ are connected with $$ \hat{f'}(n) = in\hat{f}(n)$$ Then we have $$\sum_{k=1}^\infty | \hat{f}(n) | =\sum_{k=1}^\infty \frac{|\hat{f'}(n)|}{n} $$ which we bound using Holder's inequality.