Hellinger–Toeplitz Theorem


  • Hilbert spaces are complex.
  • The inner product is anti-linear in its first argument.
  • Operators are linear and possibly unbounded.

Recall that an operator in a Hilbert space is called symmetric, if it has the property

Hellinger–Toeplitz Theorem

An everywhere-defined symmetric operator on a Hilbert space is bounded.

Consequently, a symmetric Hilbert space operator that is (truly) unbounded cannot be defined everywhere.

Proof using the Uniform Boundedness Theorem

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 .

Proof using the Closed Graph Theorem

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, .