[proofplan] Both implications are direct consequences of the definitions of the three [operator topologies](/page/Operator%20Topologies). Operator norm convergence controls $\|(T_i-T)\xi\|_H$ uniformly through the operator norm estimate, so it implies pointwise norm convergence on $H$. Strong convergence already gives $\|(T_i-T)\xi\|_H\to 0$ for each fixed vector $\xi$, and the Hilbert-space [Cauchy-Schwarz inequality](/theorems/432) converts this vector convergence into convergence of every matrix coefficient. [/proofplan] [step:Use the operator norm estimate to obtain strong convergence] Assume first that $T_i\to T$ in operator norm. Let $\xi\in H$ be fixed. For each $i\in I$, the definition of the operator norm gives \begin{align*} \|(T_i-T)\xi\|_H \leq \|T_i-T\|_{\mathcal{L}(H)}\|\xi\|_H. \end{align*} Since $\|T_i-T\|_{\mathcal{L}(H)}\to 0$ as a net in $\mathbb{R}$ and $\|\xi\|_H$ is a fixed finite number, the right-hand side converges to $0$. Hence \begin{align*} \|T_i\xi-T\xi\|_H=\|(T_i-T)\xi\|_H\to 0. \end{align*} Because $\xi\in H$ was arbitrary, $T_i\to T$ strongly. [/step] [step:Use Cauchy-Schwarz to turn strong convergence into weak operator convergence] Assume now that $T_i\to T$ strongly. Let $\xi,\eta\in H$ be fixed. Define the scalar net $a:I\to\mathbb{C}$ by \begin{align*} a(i)=((T_i-T)\xi,\eta)_H. \end{align*} For every $i\in I$, the Hilbert-space Cauchy-Schwarz inequality applied to the vectors $(T_i-T)\xi$ and $\eta$ gives \begin{align*} |a(i)|=|((T_i-T)\xi,\eta)_H|\leq \|(T_i-T)\xi\|_H\|\eta\|_H. \end{align*} Strong convergence gives $\|(T_i-T)\xi\|_H\to 0$, and $\|\eta\|_H$ is fixed. Therefore $a(i)\to 0$, that is, \begin{align*} (T_i\xi,\eta)_H\to (T\xi,\eta)_H. \end{align*} Since $\xi,\eta\in H$ were arbitrary, this is exactly convergence of $T_i$ to $T$ in the weak operator topology. [guided] We assume $T_i\to T$ strongly, which means that for every fixed vector $\xi\in H$ one has \begin{align*} \|(T_i-T)\xi\|_H\to 0. \end{align*} To prove weak operator convergence, we must prove convergence of scalar matrix coefficients. So fix two vectors $\xi,\eta\in H$ and define the scalar net $a:I\to\mathbb{C}$ by \begin{align*} a(i)=((T_i-T)\xi,\eta)_H. \end{align*} The goal is to prove $a(i)\to 0$. The bridge from vector norm convergence to scalar convergence is the Hilbert-space Cauchy-Schwarz inequality. Applying it to the vector $(T_i-T)\xi\in H$ and the vector $\eta\in H$ gives \begin{align*} |((T_i-T)\xi,\eta)_H|\leq \|(T_i-T)\xi\|_H\|\eta\|_H. \end{align*} The first factor on the right tends to $0$ by strong convergence, because $\xi$ was fixed before taking the limit. The second factor is the fixed finite number $\|\eta\|_H$. Hence the product tends to $0$, so \begin{align*} ((T_i-T)\xi,\eta)_H\to 0. \end{align*} Equivalently, \begin{align*} (T_i\xi,\eta)_H\to (T\xi,\eta)_H. \end{align*} Because this argument works for every pair $\xi,\eta\in H$, the defining condition for weak operator convergence is satisfied. [/guided] [/step]