Hellinger–Toeplitz theorem

We use the closed graph theorem (a standard consequence of the Baire category theorem): a linear operator defined on all of a Hilbert space is bounded if and only if its graph $\{(x,Tx):x\in H\}$ is closed in $H\times H$, i.e. whenever $x_n\to x$ and $Tx_n\to y$, we must have $y=Tx$.

Suppose $x_n\to x$ and $Tx_n\to y$. We must show $y=Tx$. For any fixed $z\in H$, symmetry gives $\langle Tx_n,z\rangle=\langle x_n,Tz\rangle$ for every $n$.

Taking $n\to\infty$ on both sides: the left side $\langle Tx_n,z\rangle\to\langle y,z\rangle$ because $Tx_n\to y$ and the inner product is continuous; the right side $\langle x_n,Tz\rangle\to\langle x,Tz\rangle$ because $x_n\to x$. So $\langle y,z\rangle=\langle x,Tz\rangle=\langle Tx,z\rangle$ (using symmetry once more on the right).

Since $\langle y,z\rangle=\langle Tx,z\rangle$ holds for every $z\in H$, we get $\langle y-Tx,z\rangle=0$ for all $z$; taking $z=y-Tx$ gives $\|y-Tx\|^2=0$, so $y=Tx$. The graph of $T$ is therefore closed, and the closed graph theorem concludes $T$ is bounded.