Hellinger–Toeplitz Theorem


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

Hellinger–Toeplitz theorem

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

A more verbose version of this statement:

Hellinger–Toeplitz theorem

If an operator TT in a Hilbert space H\hilb{H} is defined everywhere, that is on all of H\hilb{H}, and symmetric, i.e. satisfies

Tx,y=x,Tyx,yH,\innerp{Tx}{y} = \innerp{x}{Ty} \quad \forall x,y \in \hilb{H},

then TT is necessarily bounded.

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

Proof using the Uniform Boundedness Theorem

Assume that TT is not bounded. Then there exists a sequence (xn)(x_n) of unit vectors in H\hilb{H} such that Txn\norm{Tx_n} \to \infty. Consider the sequence (fn)(f_n) of linear functionals on H\hilb{H}, defined by

fn(y)=Txn,y=xn,TyyH.f_n(y) = \innerp{Tx_n}{y} = \innerp{x_n}{Ty} \quad y \in \hilb{H}.

The second identity is due to the symmetry of TT. Apply Cauchy-Schwarz to both expressions to obtain the inequalities

fn(y)Txnyandfn(y)xnTy\abs{f_n(y)} \le \norm{Tx_n} \norm{y} \quad \text{and} \quad \abs{f_n(y)} \le \norm{x_n} \norm{Ty}

for each nNn \in \NN and yHy \in \hilb{H}. The first inequality shows that the functionals fnf_n are bounded. The second one shows that, for fixed yy, the sequence (fn(y))(\abs{f_n(y)}) is bounded by Ty\norm{Ty}, since xn=1\norm{x_n} = 1 for all nn. By the Uniform Boundedness Theorem, (fn)(\norm{f_n}) is a bounded sequence. One has

Txn2=fn(Txn)fnTxnnN.\norm{Tx_n}^2 = \abs{f_n(Tx_n)} \le \norm{f_n} \norm{Tx_n} \quad n \in \NN.

Divide by Txn\norm{Tx_n} (if nonzero) to obtain Txnfn\norm{Tx_n} \le \norm{f_n} for all but finitely many nn. Thus (Txn)(\norm{Tx_n}) is a bounded sequence, contradicting Txn\norm{Tx_n} \to \infty. \square\enspace

Proof using the Closed Graph Theorem

By the Closed Graph Theorem, it is sufficient to show that the graph of TT is closed. Let (xn)(x_n) be a convergent sequence of vectors in H\hilb{H} such that the image sequence (Txn)(Tx_n) converges as well. Naming the limits xx and zz, respectively, we have

xnxandTxnz.x_n \to x \quad \text{and} \quad Tx_n \to z.

Continuity of the inner product implies

xn,Tyx,TyandTxn,yz,y\innerp{x_n}{Ty} \to \innerp{x}{Ty} \quad \text{and} \quad \innerp{Tx_n}{y} \to \innerp{z}{y}

for all yHy \in \hilb{H}. Since TT is symmetric, the first assertion can be rewritten as

Txn,yTx,y.\innerp{Tx_n}{y} \to \innerp{Tx}{y}.

A sequence of complex numbers has at most one limit, hence Tx,y=z,y\innerp{Tx}{y} = \innerp{z}{y} for all yy. By the Riesz representation theorem, Tx=zTx=z. \square\enspace