[proofplan] For part (1), WOT-continuity of the adjoint follows from the Riesz identification $H^* \cong H$: the WOT convergence $f(T_\alpha x) \to f(Tx)$ for all $x, f$ translates via the inner product to $(T_\alpha x, y)_H \to (Tx, y)_H$, and taking adjoints gives $(x, T_\alpha^* y)_H \to (x, T^* y)_H$, which is WOT convergence of $T_\alpha^*$. For part (2), we exhibit an explicit counterexample using the unilateral right shift on $\ell^2$. For part (3), SOT convergence $T_\alpha x \to Tx$ combined with the inner product formula for the adjoint gives WOT convergence of $T_\alpha^*$. [/proofplan]

[step:Show that $T \mapsto T^*$ is WOT-to-WOT continuous]Let $T_\alpha \to T$ in the WOT on $\mathcal{L}(H, H)$. Fix $x, y \in H$. By the definition of the WOT, we need $(T_\alpha^* x, y)_H \to (T^* x, y)_H$. Using the defining property of the adjoint, \begin{align*} (T_\alpha^* x, y)_H = (x, T_\alpha y)_H = \overline{(T_\alpha y, x)_H}. \end{align*} The WOT convergence $T_\alpha \to T$ with the pair $(y, x)$ gives $(T_\alpha y, x)_H \to (Ty, x)_H$, and complex conjugation is continuous, so \begin{align*} (T_\alpha^* x, y)_H = \overline{(T_\alpha y, x)_H} \to \overline{(Ty, x)_H} = (x, Ty)_H = (T^* x, y)_H. \end{align*} Since $x$ and $y$ are arbitrary, $T_\alpha^* \to T^*$ in the WOT.[/step]

[guided]We use the inner product characterisation of the WOT on a Hilbert space. On $\mathcal{L}(H, H)$, the WOT is generated by the seminorms $T \mapsto |(Tx, y)_H|$ for $x, y \in H$. This is because the [Riesz Representation Theorem](/theorems/218) identifies every $f \in H^*$ with a unique $y_f \in H$ via $f(\cdot) = (\cdot, y_f)_H$, so $f(Tx) = (Tx, y_f)_H$. Fix $x, y \in H$. We need $(T_\alpha^* x, y)_H \to (T^* x, y)_H$. The key is the symmetry of the inner product via the adjoint: \begin{align*} (T_\alpha^* x, y)_H = (x, T_\alpha y)_H = \overline{(T_\alpha y, x)_H}. \end{align*} The WOT convergence $T_\alpha \to T$ evaluated at the vector $y \in H$ and the functional $(\cdot, x)_H \in H^*$ gives \begin{align*} (T_\alpha y, x)_H \to (Ty, x)_H. \end{align*} Since complex conjugation is a continuous operation on $\mathbb{C}$ (or the identity on $\mathbb{R}$), \begin{align*} (T_\alpha^* x, y)_H = \overline{(T_\alpha y, x)_H} \to \overline{(Ty, x)_H} = (x, Ty)_H = (T^* x, y)_H. \end{align*} Since $x, y \in H$ were arbitrary, $T_\alpha^* \to T^*$ in the WOT. The essential feature of Hilbert spaces used here is that the WOT is determined by the inner product, and the adjoint simply swaps the two arguments of the inner product. This swap is invisible in the WOT because WOT convergence quantifies over *all* pairs $(x, y)$.[/guided]

[step:Exhibit a counterexample showing $T \mapsto T^*$ is not SOT-to-SOT continuous]Let $H = \ell^2(\mathbb{N})$ with the standard orthonormal basis $(e_k)_{k \in \mathbb{N}}$. Define the unilateral right shift \begin{align*} S: \ell^2(\mathbb{N}) &\to \ell^2(\mathbb{N}) \\ (x_1, x_2, x_3, \ldots) &\mapsto (0, x_1, x_2, \ldots). \end{align*} The adjoint is the left shift \begin{align*} S^*: \ell^2(\mathbb{N}) &\to \ell^2(\mathbb{N}) \\ (x_1, x_2, x_3, \ldots) &\mapsto (x_2, x_3, x_4, \ldots). \end{align*} Consider the sequence $T_k := S^k$ (the $k$-fold composition of $S$). For any $x = (x_1, x_2, \ldots) \in \ell^2$, \begin{align*} \|S^k x\|^2 = \sum_{j=1}^{\infty} |x_j|^2 = \|x\|^2, \end{align*} since $S^k$ shifts all entries $k$ places to the right. The adjoint satisfies $(S^k)^* = (S^*)^k$, and for any $x \in \ell^2$, \begin{align*} \|(S^*)^k x\|^2 = \sum_{j=k+1}^{\infty} |x_j|^2 \to 0 \quad \text{as } k \to \infty, \end{align*} since $\sum_{j=1}^{\infty} |x_j|^2 < \infty$ and the tail of a convergent series tends to zero. Therefore $(S^*)^k \to 0$ in the SOT. If the adjoint map were SOT-continuous, then $(S^*)^k \to 0$ in the SOT would imply $((S^*)^k)^* = S^k \to 0$ in the SOT. But $\|S^k e_1\| = \|e_{k+1}\| = 1$ for every $k$, so $S^k \not\to 0$ in the SOT. This is a contradiction, proving that $T \mapsto T^*$ is not SOT-continuous.[/step]

[guided]The WOT-to-WOT continuity of the adjoint map does not upgrade to SOT-to-SOT continuity. The obstruction is that the SOT breaks the symmetry between the two slots of the inner product: SOT convergence tests $\|T_\alpha x - Tx\|_H$ for each $x$, which involves *both* slots through the norm $\|v\|^2 = (v, v)_H$, whereas the adjoint swaps slots asymmetrically. To produce an explicit counterexample, we use the unilateral right shift $S$ on $\ell^2(\mathbb{N})$. Define $S(x_1, x_2, \ldots) = (0, x_1, x_2, \ldots)$. This is an isometry: $\|Sx\| = \|x\|$ for all $x$. Its adjoint is the left shift $S^*(x_1, x_2, \ldots) = (x_2, x_3, \ldots)$. Consider the sequence $T_k = (S^*)^k$. For any $x = (x_1, x_2, \ldots) \in \ell^2$, \begin{align*} \|(S^*)^k x\|^2 = \sum_{j=k+1}^{\infty} |x_j|^2. \end{align*} This is the tail of the convergent series $\sum |x_j|^2 = \|x\|^2 < \infty$, so it tends to zero as $k \to \infty$. Therefore $(S^*)^k \to 0$ in the SOT. If the adjoint map $\Phi: T \mapsto T^*$ were SOT-continuous, then $\Phi((S^*)^k) = ((S^*)^k)^* = S^k$ would also converge to $\Phi(0) = 0$ in the SOT. But $S$ is an isometry, so $S^k$ is also an isometry for every $k$: \begin{align*} \|S^k e_1\|_{\ell^2} = \|e_{k+1}\|_{\ell^2} = 1 \quad \text{for all } k \in \mathbb{N}. \end{align*} Therefore $S^k e_1 \not\to 0$, and $S^k \not\to 0$ in the SOT. This contradicts the assumed SOT-continuity of $\Phi$.[/guided]

[step:Show that $T \mapsto T^*$ is SOT-to-WOT continuous] Let $T_\alpha \to T$ in the SOT. Fix $x, y \in H$. We compute: \begin{align*} (T_\alpha^* x, y)_H = (x, T_\alpha y)_H. \end{align*} Since $T_\alpha \to T$ in the SOT, $T_\alpha y \to Ty$ in the norm of $H$. The inner product is jointly continuous in the norm topology (by the Cauchy-Schwarz inequality), so with $x$ fixed: \begin{align*} |(x, T_\alpha y)_H - (x, Ty)_H| = |(x, T_\alpha y - Ty)_H| \le \|x\|_H \|T_\alpha y - Ty\|_H \to 0. \end{align*} Therefore $(T_\alpha^* x, y)_H = (x, T_\alpha y)_H \to (x, Ty)_H = (T^* x, y)_H$, and since $x, y$ are arbitrary, $T_\alpha^* \to T^*$ in the WOT. [/step]