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 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112  --- title: Hellinger–Toeplitz Theorem parent: Unbounded Operators nav_order: 1 description: > The Hellinger–Toeplitz Theorem states that an everywhere-defined symmetric operator on a Hilbert space is bounded. We give a proof using the Uniform Boundedness Theorem. We give another proof using the Closed Graph Theorem. --- # {{ page.title }} Conventions: {: .mb-0 } - Hilbert spaces are complex. - The inner product is anti-linear in its first argument. - Operators are linear and possibly unbounded. Recall that an operator $T : D(T) \to \hilb{H}$ in a Hilbert space $\hilb{H}$ is called *symmetric*, if it has the property $$\innerp{Tx}{y} = \innerp{x}{Ty} \quad \forall x,y \in D(T).$$ {% theorem * Hellinger–Toeplitz Theorem %} An everywhere-defined symmetric operator on a Hilbert space is bounded. {% endtheorem %} Consequently, a symmetric Hilbert space operator that is (truly) unbounded cannot be defined everywhere. --- ## Proof using the Uniform Boundedness Theorem Assume that $T$ is not bounded. Then there exists a sequence $(x_n)$ of unit vectors in $\hilb{H}$ such that $\norm{Tx_n} \to \infty$. Consider the sequence $(f_n)$ of linear functionals on $\hilb{H}$, defined by $$f_n(y) = \innerp{Tx_n}{y} = \innerp{x_n}{Ty} \quad y \in \hilb{H}.$$ The second identity is due to the symmetry of $T$. Apply Cauchy-Schwarz to both expressions to obtain the inequalities $$\abs{f_n(y)} \le \norm{Tx_n} \norm{y} \quad \text{and} \quad \abs{f_n(y)} \le \norm{x_n} \norm{Ty}$$ for each $n \in \NN$ and $y \in \hilb{H}$. The first inequality shows that the functionals $f_n$ are bounded. The second one shows that, for fixed $y$, the sequence $(\abs{f_n(y)})$ is bounded by $\norm{Ty}$, since $\norm{x_n} = 1$ for all $n$. By the [Uniform Boundedness Theorem]({% link pages/functional-analysis-basics/the-fundamental-four/uniform-boundedness-theorem.md %}), $(\norm{f_n})$ is a bounded sequence. One has $$\norm{Tx_n}^2 = \abs{f_n(Tx_n)} \le \norm{f_n} \norm{Tx_n} \quad n \in \NN.$$ Divide by $\norm{Tx_n}$ (if nonzero) to obtain $\norm{Tx_n} \le \norm{f_n}$ for all but finitely many $n$. Thus $(\norm{Tx_n})$ is a bounded sequence, contradicting $\norm{Tx_n} \to \infty$. --- ## Proof using the Closed Graph Theorem By the [Closed Graph Theorem]({% link pages/functional-analysis-basics/the-fundamental-four/closed-graph-theorem.md %}), it is sufficient to show that the graph of $T$ is closed. Let $(x_n)$ be a convergent sequence of vectors in $\hilb{H}$ such that the image sequence $(Tx_n)$ converges as well. Naming the limits $x$ and $z$, respectively, we have $$x_n \to x \quad \text{and} \quad Tx_n \to z.$$ Continuity of the inner product implies $$\innerp{x_n}{Ty} \to \innerp{x}{Ty} \quad \text{and} \quad \innerp{Tx_n}{y} \to \innerp{z}{y}$$ for all $y \in \hilb{H}$. Since $T$ is symmetric, the first assertion can be rewritten as $$\innerp{Tx_n}{y} \to \innerp{Tx}{y}.$$ A sequence of complex numbers has at most one limit, hence $\innerp{Tx}{y} = \innerp{z}{y}$ for all $y$. By the Riesz representation theorem, $Tx=z$.