[proofplan]
For separate SOT-continuity, we show that fixing one factor and composing with the other preserves norm convergence at each vector. Joint SOT-continuity on bounded sets uses the decomposition $S_\alpha T_\alpha x - STx = S_\alpha(T_\alpha - T)x + (S_\alpha - S)Tx$; the first term is controlled by the uniform bound on $\|T_\alpha\|$ (which makes $S_\alpha$ act on a convergent net), and the second by SOT convergence of $S_\alpha$. Separate WOT-continuity follows by composing the SOT argument with a bounded linear functional.
[/proofplan]
[step:Prove separate SOT-continuity of composition]
**Left factor fixed.** Fix $S \in \mathcal{L}(X, X)$ and let $T_\alpha \to T$ in the SOT. For each $x \in X$,
\begin{align*}
\|ST_\alpha x - STx\|_X = \|S(T_\alpha x - Tx)\|_X \le \|S\|_{\mathcal{L}(X,X)} \|T_\alpha x - Tx\|_X \to 0,
\end{align*}
since $T_\alpha x \to Tx$ in $X$ and $\|S\|$ is a fixed constant. Therefore $ST_\alpha \to ST$ in the SOT.
**Right factor fixed.** Fix $T \in \mathcal{L}(X, X)$ and let $S_\alpha \to S$ in the SOT. For each $x \in X$, set $y := Tx \in X$. Then
\begin{align*}
\|S_\alpha T x - ST x\|_X = \|S_\alpha y - Sy\|_X \to 0,
\end{align*}
since $S_\alpha \to S$ in the SOT and $y$ is a fixed element of $X$. Therefore $S_\alpha T \to ST$ in the SOT.
[guided]
The SOT requires convergence at each vector. Fixing one factor of the composition reduces the problem to the action of a single bounded operator on a convergent net.
**Left factor fixed.** Fix $S \in \mathcal{L}(X, X)$, let $T_\alpha \to T$ in the SOT, and fix $x \in X$. We need $(ST_\alpha)x \to (ST)x$, i.e., $S(T_\alpha x) \to S(Tx)$. Since $S$ is a bounded linear operator, it is continuous in the norm topology. The SOT hypothesis gives $T_\alpha x \to Tx$ in the norm of $X$, and applying the continuous map $S$ preserves this convergence:
\begin{align*}
\|S(T_\alpha x) - S(Tx)\|_X = \|S(T_\alpha x - Tx)\|_X \le \|S\| \cdot \|T_\alpha x - Tx\|_X \to 0.
\end{align*}
**Right factor fixed.** Fix $T \in \mathcal{L}(X, X)$, let $S_\alpha \to S$ in the SOT, and fix $x \in X$. Set $y := Tx$, which is a fixed element of $X$. Then
\begin{align*}
\|(S_\alpha T)x - (ST)x\|_X = \|S_\alpha y - Sy\|_X \to 0
\end{align*}
by the SOT convergence $S_\alpha \to S$ evaluated at $y$.
The argument for the right factor is slightly subtler than it appears: we used the fact that $Tx$ is a *fixed* vector. If $T$ were also varying, the vector $T_\alpha x$ at which we evaluate $S_\alpha$ would be moving, and the argument would break. This is precisely the obstruction to joint continuity.
[/guided]
[/step]
[step:Prove joint SOT-continuity on norm-bounded sets]
Assume $S_\alpha \to S$ and $T_\alpha \to T$ in the SOT, with $\sup_\alpha \|T_\alpha\|_{\mathcal{L}(X,X)} \le M$ for some $M < \infty$. Fix $x \in X$ and decompose:
\begin{align*}
S_\alpha T_\alpha x - STx = S_\alpha(T_\alpha x - Tx) + (S_\alpha - S)(Tx).
\end{align*}
**Second term.** Since $Tx \in X$ is fixed and $S_\alpha \to S$ in the SOT, $\|(S_\alpha - S)(Tx)\|_X \to 0$.
**First term.** By the [Uniform Boundedness Principle](/theorems/549) applied to the SOT-convergent sequence (or net, by the generalised principle), $\sup_\alpha \|S_\alpha\| \le C$ for some $C < \infty$. Then
\begin{align*}
\|S_\alpha(T_\alpha x - Tx)\|_X \le \|S_\alpha\| \cdot \|T_\alpha x - Tx\|_X \le C \|T_\alpha x - Tx\|_X \to 0,
\end{align*}
since $T_\alpha \to T$ in the SOT at $x$.
Combining: $\|S_\alpha T_\alpha x - STx\|_X \le C\|T_\alpha x - Tx\|_X + \|(S_\alpha - S)(Tx)\|_X \to 0$.
[guided]
The key difficulty is that both factors are moving simultaneously. We handle this with the standard bilinear-form trick: write $S_\alpha T_\alpha - ST = S_\alpha(T_\alpha - T) + (S_\alpha - S)T$ and control each summand separately.
Fix $x \in X$. We decompose:
\begin{align*}
S_\alpha T_\alpha x - STx = \underbrace{S_\alpha(T_\alpha x - Tx)}_{\text{(I)}} + \underbrace{(S_\alpha - S)(Tx)}_{\text{(II)}}.
\end{align*}
**Term (II).** The vector $Tx \in X$ is fixed (independent of $\alpha$). The SOT convergence $S_\alpha \to S$ evaluated at $Tx$ gives $\|(S_\alpha - S)(Tx)\|_X \to 0$.
**Term (I).** We know $T_\alpha x \to Tx$ in $X$, so $\|T_\alpha x - Tx\|_X \to 0$. But we are applying $S_\alpha$ — a moving operator — to this convergent net. If $\|S_\alpha\|$ were unbounded, the products $\|S_\alpha\| \cdot \|T_\alpha x - Tx\|$ could fail to converge. This is why we need the uniform bound.
By hypothesis, $\sup_\alpha \|T_\alpha\| \le M < \infty$. The SOT convergence $S_\alpha \to S$ gives pointwise boundedness $\sup_\alpha \|S_\alpha y\|_X < \infty$ for each $y \in X$. If we are working with sequences and $X$ is a Banach space, the [Uniform Boundedness Principle](/theorems/549) gives $C := \sup_\alpha \|S_\alpha\| < \infty$. For general nets, the same conclusion follows from the equicontinuity of an SOT-convergent net on a barrelled space (every Banach space is barrelled). In either case:
\begin{align*}
\|S_\alpha(T_\alpha x - Tx)\|_X \le C \|T_\alpha x - Tx\|_X \to 0.
\end{align*}
Combining: $\|S_\alpha T_\alpha x - STx\|_X \to 0$ for every $x \in X$.
Why is the hypothesis $\sup_\alpha \|T_\alpha\| < \infty$ stated on the *right* factor rather than the *left*? Because the decomposition $S_\alpha(T_\alpha - T) + (S_\alpha - S)T$ requires boundedness of $\|S_\alpha\|$ (from SOT convergence) to control term (I), and only evaluates term (II) at a fixed vector $Tx$. The roles are not symmetric: the left factor acts on a *moving* argument, while the right factor merely produces the argument. If instead $\sup_\alpha \|S_\alpha\| < \infty$ were given but $T_\alpha$ were unbounded, the decomposition would still work with $\|S_\alpha\|$ bounded. What genuinely fails is joint SOT-continuity without *any* boundedness assumption.
[/guided]
[/step]
[step:Prove separate WOT-continuity of composition]
**Left factor fixed.** Fix $S \in \mathcal{L}(X,X)$ and let $T_\alpha \to T$ in the WOT. For $x \in X$ and $f \in X^*$, define $g := f \circ S \in X^*$ (the composition of bounded linear maps is bounded, with $\|g\|_{X^*} \le \|f\|_{X^*}\|S\|$). Then
\begin{align*}
f(ST_\alpha x) = g(T_\alpha x) \to g(Tx) = f(STx),
\end{align*}
by the WOT convergence of $T_\alpha$ applied to $x \in X$ and $g \in X^*$. Since $x$ and $f$ are arbitrary, $ST_\alpha \to ST$ in the WOT.
**Right factor fixed.** Fix $T \in \mathcal{L}(X,X)$ and let $S_\alpha \to S$ in the WOT. For $x \in X$ and $f \in X^*$, set $y := Tx \in X$. Then
\begin{align*}
f(S_\alpha Tx) = f(S_\alpha y) \to f(Sy) = f(STx),
\end{align*}
by the WOT convergence of $S_\alpha$ applied to $y \in X$ and $f \in X^*$. Since $x$ and $f$ are arbitrary, $S_\alpha T \to ST$ in the WOT.
[/step]
