Does Continuity in Weak Operator Topology imply Continuity in Strong Operator Topology? This homework problem has puzzled me for almost a year. As nobody in the class has figured it out, I would like to seek a proof or a disproof here.
Problem (Prove or Disprove) Let $X$ be a Banach space and assume that $T\colon [0,\infty) \to \mathcal{L}(X, X)$ satisfies


*

*$T(0)=I$,

*$T(s+t)=T(s)T(t)$ for all $s,t\in[0,\infty)$,

*For all $x\in X$ and $x^*\in X^*$ we have $x^*(T(t)x)\to x^*(x)$ as $t\to 0^+$.


Show that $T$ is a linear $C_0$-semigroup.
Explanation The third condition states that $T$-is continuous in the weak operator topology, while being a $C_0$-semigroup requires continuity in the strong operator topology.
A Partial Result We can prove for the case when $X$ is reflexive. Again, I want to stress that I am not sure if the statement in the problem is true. The proof for the partial result might be over-complicated. One might want to work on the problem him/herself before looking at our result.
The main idea comes from A Short Course on Operator Semigroups by S. Axler and K.A. Ribet Engel, Klaus-Jochen and Nagel, Rainer, while there are still some original ideas in the proof.
A Technical Lemma Let $f:[0,1]\to\mathbb{R}$ be a bounded right continuous function. Then $f$ has only countably discontinuous points.
We omit the proof the technical lemma. Back to prove the problem assuming $X$ is reflexive.
Proof of partial result
Let $E=\left\{x\in X\mid (T(h)-I)x\to 0\text{ strongly as }h\to 0^+\right\}$. Our goal is to prove $E=X$.
We have convergence of $T$ in weak operator topology at 0 and also have boundedness for sequences that are weakly convergent, so we can still obtain that $||T(t)||\leq Me^{\omega t}$ for all $t\geq 0$ and some $M\geq 1$, $\omega\geq 0$.
Also, we have right continuity in weak operator topology: for all $x\in X^*, x\in X$
$$\langle{x^*,T(t+h)x-T(t)x}\rangle=\langle{x^*,(T(h)-I)T(t)x}\rangle\to 0\text{ as }h\to 0^+.
$$
Now we define $L_r(x),x\in X, 0<r<1$ be such that for all $x^*\in X^*$
$$L_r(x)(x^*)=\frac{1}{r}\int_0^r\langle{x^*,T(s)x}\rangle ds.
$$
This Riemannian integral makes sense because the integrand is right continuous, and is ,by lemma, Riemannian integrable. It is clear that $L_r(x)$ is indeed linear. Note also $|{L_r(x)(x^*)}|\leq||{x^*}||\sup_{0<s<1}||{T(s)x}||$, so $L_r(x)$ is bounded, hence is well-defined.
What's more, since $X$ is reflexive, we can find $y_r(x)\in X$ such that $J_X(y_r(x))=L_r(x)$ and
$$\langle{x^*,y_r(x)}\rangle=\frac{1}{r}\int_0^r\langle{x^*,T(s)x}\rangle ds.(*)
$$
Then we compute that
\begin{eqnarray*}
&&||{(T(h)-I)y_r(x)}||=\sup_{||{x^*}||=1}\langle{x^*,(T(h)-I)y_r(x)}\rangle\\
&=&\sup_{||{x^*}||=1}\frac{1}{r}\left({\int_0^r\langle{x^*,T(s+h)x}\rangle ds-\int_0^r\langle{x^*,T(s)x}\rangle ds}\right)\\
&=&\sup_{||{x^*}||=1}\frac{1}{r}\left({\int_r^{r+h}\langle{x^*,T(s)x}\rangle ds-\int_0^h\langle{x^*,T(s)x}\rangle ds}\right)\\
&\leq&\sup_{||{x^*}||=1}\frac{2h}{r}||{x^*}||\sup_{0<s<1}||{T(s)}||\,||{x}||
\to 0\text{ as }h\to 0^+.
\end{eqnarray*}
Hence $D=\left\{y_r(x)\in X\mid x\in X, 0<r<1\right\}\subset E$. Clearly $y_r(x) \rightharpoonup x$ as $r\to 0^+$, so $D$ is weakly dense in $X$ and so is $E$, where "weakly dense in $X$" means its weak sequential closure is $X$. Note also $E$ is convex, therefore $E$ is strongly dense in $X$.
Finally, fix $\epsilon>0, x\in X$. There is $y\in E$ such that $||{x-y}||<\frac{\epsilon}{3+3Me^\omega}$. Choose $\delta\in(0,1)$ such that for all $0\leq h<\delta$, we have $||{T(h)y-y}||<\frac{\epsilon}{3}$. Then
\begin{eqnarray*}
||{T(h)x-x}||&\leq& ||{T(h)(x-y)}||+||{T(h)y-y}||+||{y-x}||<\epsilon.
\end{eqnarray*}
Hence $E=X$. Q.E.D.
 A: 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 $t > 0$.
