Let $\mathcal H$ be a infinite dimensional Hilbert space and $T \in \mathcal L (\mathcal H)$ be a compact normal operator. Let $\sigma (T) = \{\lambda_1,\lambda_2, \cdots\} \cup \{0\}$ be the spectrum of $T.$ Let $P_j$ be the projection onto $\text {ker} (T - \lambda_j I).$ If $\{\alpha_n \}_{n \geq 1} \in c_0$ (sequences converging to $0\ $) then show that $\sum\limits_{i=1}^{\infty}\alpha_i P_i$ converges in norm.
If $\lambda_j$'s are all distinct then $P_j \mathcal H \perp P_k \mathcal H,$ for $j \neq k.$ But then I find that $$\left \| \sum\limits_{i=1}^{\infty} \alpha_i P_i \right \|^2 \leq \sum\limits_{i=1}^{\infty} \left \lvert \alpha_i \right \rvert^2.$$ Now I got stuck. Also I don't have any clear idea about how to tackle this question if $\lambda_j$ are not all distinct. Any help is very much required at this stage.
Thanks in advance.
Essentially what the operator $\sum_i\alpha_iP_i$ is doing is to replace the $\lambda_i$ of $T$ with $\alpha_i$, on each of its eigenspaces. For example, if $H=\ell^2$,
$$T(a_j)=(\lambda_1a_1,\lambda_2a_2,\lambda_3a_3,\ldots)$$ $$\sum_i\alpha_iP_i(a_j)=(\alpha_1a_1,\alpha_2a_2,\ldots)$$ Clearly, the new operator is also compact if $\alpha_i\to0$.
Take an orthonormal basis for each finite-dimensional eigenspace $\ker(T-\lambda_i)$, and denote them all together by $e_i$, $i\in\mathbb{N}$. Extend this basis by vectors $f_k$ to cover the complementary perpendicular space (so $Tf_j=0$), and thus $H$. To avoid too many different subscripts, relabel the $\lambda_i$ by $\lambda_j$, $\alpha_i$ by $\alpha_j$, and $P_i$ by $P_j$, with repetitions whenever $e_j$ belong to the same eigenspace.
If $x=\sum_j\langle e_j,x\rangle e_j$, then define $Sx:=\sum_j\alpha_j\langle e_j,x\rangle e_j$. Since $\alpha_j$ are bounded and $(\langle e_j,x\rangle)\in\ell^2$, it follows that $Sx$ is defined, and in fact a linear bounded operator.
\begin{align} \left\|\sum_{i=1}^n\alpha_iP_ix-Sx\right\|^2&=\left\|\sum_j^{n'}\langle e_j,x\rangle\alpha_j e_j-\sum_j\alpha_j\langle e_j,x\rangle e_j\right\|^2\\ &=\sum_{j=n'+1}^\infty|\alpha_j|^2|\langle e_j,x\rangle|^2\le\sup_{j\ge n'}|\alpha_j|^2\|x\|^2 \end{align} Here $n'\ge n$ since there may be several indices $j$ that correspond to the same $i$, but what's important is that $n'\to\infty$ as $n\to\infty$.
Hence $\|\sum_{i=1}^n\alpha_iP_i-S\|\le\sup_{j\ge n'}|\alpha_j|\to0$ as $n\to\infty$.