We have the space $H = \Big \{ f \in L^2\big([0,2\pi]\big) \colon \int_{0}^{2\pi} f(x) dx = 0\Big\}$ and the operator $T$ on $H$ which is defined as
\begin{equation} (Tf)(x) = i \int_0^x f(y)dy + \frac{i}{2\pi} \int_0^{2\pi} y f(y)dy \end{equation}
I have already shown that this operator is is compact and self-adjoint on $H$ and that its eigenvalues are given by $\lambda = \frac{1}{n}$ with $n \in \mathbb{Z}_0$, with eigenvectors $f_n(x) = e^{inx}$. I now have to show that $(e_n)_{n \in \mathbb{Z}_0}$ given by $e_n(x) = \frac{1}{\sqrt{2\pi}} e^{inx}$ is an orthonormal basis for H if we assume that $\ker T = 0$. Then I need to conclude that $(e_n)_{n \in \mathbb{Z}}$ is an orthonormal basis for $L^2([0,2\pi])$.
We know that $T$ admits a basis of eigenvalues because it is compact and self-adjoint but my eigenvectors don't agree with the given basis and I have no clue how I could conclude the statement about $L^2$.