I'm trying to solve the next:
Find the spectral family of the operator $T:l^{2}\rightarrow l^{2}$ defined by $$(\xi_{1},\xi_{2},\xi_{3},\ldots)\mapsto(\xi_{1}/1,\xi_{2}/2,\xi_{3}/3,\ldots).$$ Find an orthonormal set of eigenvectors.
What form does $T=\displaystyle\int_{m-0}^{M}\lambda dE_{\lambda}$ take in this place?
So, I've proved that $T$ is linear, bounded and self-adjoint operator, then,for spectrum theorem $T$ has the integral representation $\displaystyle\int_{m-0}^{M}\lambda dE_{\lambda},$ where $m=\displaystyle\inf_{||x||=1}\langle Tx,x\rangle$ and $M=\displaystyle\sup_{||x||=1}\langle Tx,x\rangle.$
An orthonormal set can be given by $\{e_{k}\}_{k\in\mathbb{N}},$ where $e_{k}=(0,0,\ldots,1,0,0,\ldots)$ and the number $1$ is on the place $k.$Such set is a set of eigenvectors because if $\lambda_{k}\in\mathbb{R}$ we have that $Te_{k}=\lambda_{k}e_{k}$ if and only if $\lambda_{k}=k$ for each $k\in\mathbb{N}.$
But I can't see how is the form of the integral representation of $T.$ I'd like to know how is $E_{\lambda}$ for each $\lambda$ and computing such integral. My feeling is that such integral is an infinite sum because of the eigenvectors, but I'm not sure.
Any kind of help is thanked in advanced.
This operator is selfadjoint and has eigenvalues $1,1/2,1/3,\cdots$ with $Te_n = \frac{1}{n}e_n$ where $e_n$ is the standard basis function with $1$ in the n-th place and $0$ elsewhere, for $n=1,2,3,\cdots$. The spectral measure is $$ E(S)f = \sum_{\{ n \in \mathbb{N} : \frac{1}{n} \in S \}}\langle f,e_n\rangle e_n. $$ If you are using a version where $E_{\lambda}f$ is $E_{\lambda} = E(-\infty,\lambda]$, then $$ E_{\lambda}f = \sum_{\{ n\in\mathbb{N} : 1/n \le \lambda \}}\langle f,e_n\rangle e_n. $$ So, for $M > 0$, $$ \int_{0}^{M}\lambda d_{\lambda}E_{\lambda}f = \sum_{n \ge 1/M}\frac{1}{n}\langle f,e_n\rangle e_n. $$