Normal Operator: Everywhere defined implies bounded? Given a Hilbert space $\mathcal{H}$.
Consider a normal operator $N:\mathcal{H}\to\mathcal{H}$.
If its domain is the whole Hilbert space then is it necessarily bounded?
The point is that I'm trying to imagine how a Borel function can induce a normal operator via a spectral measure that is unbounded w.r.t. the spectral measure but still being everywhere defined...
 A: Suppose $E$ is a spectral measure on $\mathbb{C}$. Choose any unit vector $x \in \mathcal{H}$, and look at the probability measure $\mu_{x}(S)=\|E(S)x\|^{2}$. If there are countably many disjoint Borel subsets $\{ S_{j}\}_{j=1}^{\infty}$ with $\mu_{x}(S_{j})\ne 0$, then you can construct a Borel function $f$ such that $\int |f|^{2}d\mu_{x} =\infty$. Then $\int fdE$ is a closed densely-defined normal operator whose domain does not include $x$. If you choose $f$ to be real, then $\int f dE$ is selfadjoint.
If $f$ is a Borel function for which $\int |f|^{2}d\mu_{x} < \infty$ for all $x$, then $\int f dE$ is a closed normal operator defined on the full space, which makes it bounded (being closed is enough to make it bounded.)
Now suppose $f$ is Borel function which is unbounded on the support of $E$. Then there are infinitely many of the following orthogonal projections which are non-zero:
$$
                            E_{n} = E\{ n \le |f| \lt n+1 \},\;\;\; n=0,1,2,3,\cdots.
$$
These are all mutually orthogonal as well, meaning that $E_{n}E_{m}=0$ for $n \ne m$. For any $n$ for which $E_{n}\ne 0$, there exists a unit vector $e_{n}$ in the range of $E_{n}$. Then $\{ e_{n_{k}}\}_{k=1}^{\infty}$ is an infinite orthonormal subset of $\mathcal{H}$ for which
$$
    \left\|\int fdE e_{n_{k}}\right\|^{2} =\int |f|^{2} d\|Ee_{n_{k}}\|^{2}\ge n_{k}^{2}\|e_{n_{k}}\|^{2}=n_{k}^{2}.
$$
If $\{ \alpha_{k}\}_{k=1}^{\infty}$ is a square summable sequence, then $y=\sum_{k=1}^{\infty}\alpha_{k}e_{n_{k}}$ converges in $\mathcal{H}$, and
$$
   \int |f|^{2}d\|Ey\|^{2}=\sum_{k}\int_{n_{k}\le |f|< n_{k}+1}|f|^{2}d\|Ey\|^{2} \\
   = \sum_{k}\int|f|^{2}d\|E\alpha_{k}e_{n_{k}}\|^{2} \\
   \ge \sum_{k}\int |\alpha_{k}|^{2}n_{k}^{2}.
$$
Because $\sum_{k}n_{k}^{2}=\infty$, then there is a square summable sequence $\{ \alpha_{k}\}$ such that $\sum_{k}|\alpha_{k}|^{2}n_{k}^{2}=\infty$. For any such sequence, it follows that $y=\sum_{k}\alpha_{k}e_{n_{k}}$ is not in the domain of $\int fdE$. In other words, if $f$ is unbounded on the support of $E$, then $\int fdE$ cannot be defined everywhere.
