How to apply continuous functional calculus
Andrew Mclaughlin 
This is the statement I am using.
Theorem 2.17, pg 34: Suppose that $A$ is a unital $C^*$ algebra and that $a$ is a normal element in $A$, then there is a $*$-isometric isomorphism $$C(\sigma(a)) \rightarrow C^*(\{1_A, a \}) \subseteq A$$ such that $id \mapsto a$, where $id$ is the identity function on $\Bbb C$.
I want apply this result to a bounded operator $T$ on Hilbert space $H$.
If $T \in B(H)$ is normal, and $\sigma(a)$ is discrete, then the "spike" functions, $$\delta_\lambda(x):= \begin{cases} 1 \text{ if } \lambda =x \\ 0 \text{ otherwise } \end{cases} $$ Maps to the projection map $$\delta_\lambda(T)= P_\lambda$$ the projection map on to the eigenspaces.
This seems intuitively true. But I don't know how to prove it.
The problem may not be well defined too, since we do not know that the eigenspaces are in fact closed - let us suppose this to be the case.
$\endgroup$1 Answer
$\begingroup$Your functions satisfy $ x\delta_\lambda(x)=\lambda\delta_\lambda(x)$. So you have $$ TP_\lambda=\lambda P_\lambda. $$This shows that any element in the range of $P_\lambda$ is a $\lambda$-eigenvector for $T$.
Conversely, if $Tv=\lambda v$, we get that $p(T)v=p(\lambda)v$ for any polynomial $p$ that has $p(0)=0$. By taking a sequence $\{p_n\}$ of polynomials such that $p_n\to\delta_\lambda$ uniformly (recall that the spectrum of $T$ is discrete), we obtain $$ \delta_\lambda(T)v=\delta_\lambda(\lambda)v, $$which is $P_\lambda v=v$. Thus $P_\lambda$ is the projection onto the $\lambda$-eigenspace of $T$.
The eigenspaces are always closed, they are kernels of bounded operators.
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