- Hilbert spaces are complex.
- The inner product is anti-linear in its first argument.
- Operators are linear and possibly unbounded.
An everywhere-defined symmetric operator on a Hilbert space is bounded.
A more verbose version of this statement:
If an operator in a Hilbert space is defined everywhere, that is on all of , and symmetric, i.e. satisfies
then is necessarily bounded.
Consequently, a symmetric Hilbert space operator that is (truly) unbounded cannot be defined everywhere.
Assume that is not bounded. Then there exists a sequence of unit vectors in such that . Consider the sequence of linear functionals on , defined by
The second identity is due to the symmetry of . Apply Cauchy-Schwarz to both expressions to obtain the inequalities
for each and . The first inequality shows that the functionals are bounded. The second one shows that, for fixed , the sequence is bounded by , since for all . By the Uniform Boundedness Theorem, is a bounded sequence. One has
Divide by (if nonzero) to obtain for all but finitely many . Thus is a bounded sequence, contradicting .
By the Closed Graph Theorem, it is sufficient to show that the graph of is closed. Let be a convergent sequence of vectors in such that the image sequence converges as well. Naming the limits and , respectively, we have
Continuity of the inner product implies
for all . Since is symmetric, the first assertion can be rewritten as
A sequence of complex numbers has at most one limit, hence for all . By the Riesz representation theorem, .