Does Continuity in Weak Operator Topology imply Continuity in Strong Operator Topology?
Solution 1:
Here's a proof that $T$ is continuous in the strong operator topology. As noted in the question, for each $x\in X$, the map $t\mapsto T_tx$ is right-continuous under the weak topology, and norm-bounded on each bounded interval. To complete the proof given in the question for an arbitrary Banach space, we just require the following lemma.
Lemma: Let $X$ be a Banach space, $a < b\in\mathbb{R}$, and $f\colon[a,b]\to X$ be weakly right-continuous and norm-bounded. Then, there is a (unique) $y\in X$ such that $$ x^*(y)=\int_a^b x^*(f(t))\,dt $$ for all $x^*\in X$.
We can write $y=\int_a^bf(t)\,dt$, which is the Bochner integral. Rather than defining $L_r$, we can define $y_r(x)\in X$ directly as $y_r(x)=\frac1r\int_0^rT_tx\,dt$, which satisfies $$ x^*(y_r(x))=\frac1r\int_0^r x^*(T_tx)\,dt $$ for all $x^*\in X^*$. So, the remainder of the proof given in the question follows through unchanged for arbitrary Banach spaces.
The main step in proving the lemma above is the Pettis measurability theorem. For a measurable space $(E,\mathcal{E})$ and Banach space $X$, a function $f\colon E\to X$ is called simple if it is of the form $f(t)=\sum_{k=1}^n1_{t\in E_k}x_k$ for $E_k\in\mathcal{E}$ and $x_k\in X$, and $f$ is said to be strongly $\mathcal{E}$-measurable if there is a sequence $\{f_n\colon n=1,2,\ldots\}$ of simple functions converging pointwise to $f$. A function $f\colon E\to X$ is separably valued if its image is contained in a separable subspace of $X$.
Theorem (Pettis): Let $(E,\mathcal{E})$ be a measurable space, $X$ a Banach space, and $f\colon E\to X$. The following are equivalent
- $f$ is strongly $\mathcal{E}$-measurable.
- $f$ is separably valued and $t\mapsto x^*(f(t))$ is $\mathcal{E}$-measurable for each $x^*\in X^*$.
A statement and proof of the Pettis measurability theorem in this form is given in these notes. Then, for any finite measure $\mu$ on $(E,\mathcal{E})$, any $f\colon E\to X$ which is norm-bounded and strongly measurable is also Bochner integrable, so the Bochner integral $y=\int f\,d\mu$ is well-defined and satisfies $x^*(y)=\int x^*(f)\,d\mu$ for all $x^*\in X^*$. Constructing the Bochner integral is straightforward. For a simple $f(t)=\sum_{k=1}^n1_{t\in E_k}x_k$ the Bochner integral is $\int f\,d\mu=\sum_{k=1}^n\mu(E_k)x_k$. For a strongly measurable and norm-bounded $f$, we choose a sequence of simple functions $f_n$ converging pointwise to $f$. If $f$ is bounded in norm by $K$ then we can assume that $f_n$ are norm-bounded by $K+1$ (replace $f_n(t)$ by $1_{\{\lVert f_n(t)\rVert\le K+1\}}f_n(t)$). Then it can be seen that $\int f_n\,d\mu$ is Cauchy in $X$, and we set $\int f\,d\mu$ to be its limit.
To prove the lemma then, applying the Pettis integrability theorem, it is enough to show that if $f\colon[a,b]\to X$ is weakly right-continuous then it is separably valued and weakly measurable with respect to the Borel sigma-algebra on $[0,1]$.
Measurability of $t\mapsto x^*(f(t))$ follows from the fact that it is right-continuous and that right-continuous real-valued functions are Borel measurabile. To show that it is separably valued, let $S$ be the union of $\{b\}$ and the rational numbers in $[a,b]$. Let $Y_0$ be the subspace spanned by $\{f(t)\colon t\in S\}$, and $Y$ be the norm-closure of $Y_0$ in $X$. Then, $Y$ is separable. Also, as $Y_0$ is a subspace (in particular, it is convex), $Y$ is also the weak closure of $Y_0$. Now, for any $t\in[a,b]$ there exists $t_n\in S$ with $t_n\downarrow t$ and, by weak right-continuity, we have $f(t_n)\to f(t)$ weakly, so $f(t)\in Y$. Hence, the image of $f$ is contained in $Y$, and $f$ is separably valued.
Update: A proof of the result is also given in these notes, and uses the Bochner integral in essentially the same way as in my answer. The paper On semigroups of operators in locally convex spaces relaxes the conditions further. We only need to assume that $X$ is a metrizable locally convex space and, then, if $T_t$ is strongly measurable over $t > 0$ (or, by Pettis' theorem, almost separably valued and weakly measurable) then it is strongly continuous over $t > 0$.