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...

  • $\begingroup$ I forgot: Normal implies closed and closed implies bounded by the closed graph theorem if the domain was the full space. $\endgroup$ Oct 20 '14 at 7:43

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.

  • $\begingroup$ Thanks for that nice proof!! $\endgroup$ Oct 20 '14 at 7:41
  • $\begingroup$ @Freeze_S You're welcome. $\endgroup$ Oct 20 '14 at 10:50

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