I want to prove that a sequence $(\lambda_n)_n$ of eigenvalues of a specific bounded linear operator $T:H \to H$ defined on a Hilbert Space $H$ is the set of all eigenvalues of $T$ in $H$. I know that there is a sequence of orthonormal eigenfunctions $(u_n)_n$, with $Tu_n=\lambda_nu_n$, which is complete in $H$, I mean $(u_n)_n$ is orthonormal basis of Hilbert for $H$. Can I say with that conditions that $(\lambda_n)_n$ are all eigenvalues of $T$?. Can I say that if $\sigma(T)$ is the spectrum of $T$ then $\sigma(T)=(\lambda_n)_n$?.
Thanks in advance!
Suppose that $T(x)=\lambda x$, write $x=\sum_n x_ne_n$, since $T$ is continuous, $T(x)=\sum_n x_n\lambda_ne_n=\sum_n x_n\lambda e_n$. This implies that $\langle T(x),e_m\rangle=\langle \sum_nx_n\lambda_ne_n,e_m\rangle=\langle \sum_nx_n\lambda e_n,e_m\rangle$. We deduce that $x_m\lambda_m=x_m\lambda$. Since $x\neq 0$, there exists $m$ such that $x_m\neq 0$, we deduce that $x_m\lambda =x_m\lambda_m$ and $\lambda=\lambda_m$.