Let $(H, \langle \cdot, \cdot \rangle)$ be a Hilbert space and let $A : H \to H$ be a linear operator defined everywhere. If $A$ is symmetric i.e. if $$\langle Ax,y \rangle = \langle x, Ay \rangle \quad \text{for all } x, y \in H$$ then $A$ is bounded.


Since $A$ is linear it is enough to prove that it is continuous.

  1. $A^2 = A \circ A$ is symmetric: indeed take $x,y \in H$, then $$ \langle A^2x,y \rangle = \langle A(Ax),y \rangle = \langle Ax,Ay \rangle = \langle x,A(Ay) \rangle = \langle x,A^2y \rangle $$

  2. Let $B: H \to H$ be a symmetric linear operator defined everywhere. Let $\{x_n\}_{n \in \mathbb{N}} \subset H$ and $x \in H$ s.t. $x_n \to x$. Then $Bx_n \rightharpoonup Bx$. Indeed, let $R: H^{*} \to H$ be the Riesz map and take $L \in H^{*}$, then: $$L(Bx_n) = \langle RL,Bx_n \rangle = \langle BRL,x_n \rangle \to \langle BRL,x \rangle = \langle RL,Bx \rangle = L(Bx) $$

  3. Let $\{x_n\}_{n \in \mathbb{N}} \subset H$ and $x \in H$ s.t. $x_n \to x$. Then $\{Ax_n\}_{n \in \mathbb{N}}$ is Cauchy. Indeed:

$$ \|Ax_n -Ax_m\|^2 = \langle (x_n-x_m), A^2(x_n-x_m) \rangle \le \|x_n-x_m\|(\|A^2x_n\|+\|A^2x_m\|) \le \|x_n-x_m\|2M $$

where $\|A^2x_k\| \le M$ is due to the fact that, by 2, this is a weak convergent sequence and then it is bounded.


By 3. there exist $y \in H$ s.t. $Ax_n \to y$ and then it also holds $Ax_n \rightharpoonup y$. But from 2. we have $Ax_n \rightharpoonup Ax$ and then, by uniqueness of weak limit, I can conclude $Ax_n \to Ax$.

Is my proof correct? Thanks in advance!

  • $\begingroup$ This is known as Hellinger-Toeplitz theorem, it is an application of the closed graph theorem. $\endgroup$
    – Evil Witch
    Commented Jun 7 at 23:28


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