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Any nonnegative $f\in C_b(X)$ is a pointwise limit of an increasing sequence $f_n\in C_0(X)$ and, therefore $T_tf=\lim\_{n\to\infty}T_tf_n$ is also a limit of an increasing sequence in $C_0(X)$, so is lower semicontinuous. Applying the same statement to $\Vert f\Vert - f$ implies that $f$ is also upper semi-continuous, so is continuous.

2. If $f\in C_0(X)$ is nonnegative and $t_n\to0$ then $(T_{t_n}f-f)\_+\to0$ in the compact-open topology.

(Note: This doesn't use continuity of $Y$. And, in general, it is not necessary that $T_{t_n}f\to f$ in the compact open topology, even for $Y$ right-continuous.) It is enough to consider $f$ with compact support and, by right-continuity of the process $Y$, we have $T_{t_n}f(x)\to f(x)$ as $n\to\infty$ for all $x\in X$. Then, setting $g_m=m\int\_0^{1/m}T_sf\\,ds$, it can be seen that $\Vert T_{t_n}g_m-g_m\Vert\le 2mt_n\Vert f\Vert\to0$ as $n\to\infty$. Now, letting $S_m$ be the space of convex combinations of $\{g_m,g_{m+1},\ldots\}$, $f$ is a pointwise limit of a sequence in $S_m$, so is in its closure under the compact-open topology (this is a consequence of the Hahn-Banach theorem and the Riesz representation theorem). Therefore, there exists $f_m\in S_m$ converging to $f$ in the compact-open topology. For any $\epsilon > 0$ and compact $K\subseteq X$, take $m$ large enough such that $\vert f-f_m\vert\le\epsilon$ on the union of $K$ and the support of $f$. So, $$ T_tf-f \le T_tf_m-f_m+T_t(f-f_m)_++(f_m-f)\le T_tf_m-f_m + 2\epsilon. $$ on $K$. Letting $t$ decrease to zero gives $\limsup\_{t\to 0}(T_tf-f)\le2\epsilon$ uniformly on $K$.

This is the point where continuity is required. It is not too hard to show that the space of $f\in C_b(X)$ satisfying the conclusion forms a $C^*$-algebra and is closed under $T_t$, so it is enough to prove it for nonnegative $f\in C_b(X)$ with compact support $K$. Also, the fact that $T_tC_0(X)\subseteq C_b(X)$ can be used to prove the strong-Markov property. If $Y_0\not\in K$ and, letting $\tau$ be the first time at which $Y$ hits $S$, continuity implies that $f(Y_{\tau})=0$ and $Y_\tau\in K$. So, for $x\not\in K$, $$ \begin{align} T_tf(x)=\mathbb{E}\_x[f(Y_t)]&=\mathbb{E}\_x\left[T_{(t-\tau)\_+}f(Y_{\tau\wedge t})\right]\\\\ &\le\sup\_{y\in K,s\le t}\left(T_sf(y)-f(y)\right)\_+. \end{align} $$ The previous statement says that the right-hand-side tends to zero as $t\to0$ and, as it is independent of $x$, $T_tf\to0$ uniformly on $X\setminus K$.

As $X$ is lccb, $C_0(X)$ has a countable dense subset $S$. The closure of $S$ under products, taking $\mathbb{Q}$-linear combinations and applying $T_t$ for $t\in\mathbb{Q}\_+$ is a countable dense subset of $\mathcal{A}$ (use 3). So $C_0(\hat X)\cong\mathcal{A}$ is separable. This implies that $\hat X$ has a countable base.

Any nonnegative $f\in C_b(X)$ is a pointwise limit of an increasing sequence $f_n\in C_0(X)$ and, therefore $T_tf=\lim_{n\to\infty}T_tf_n$ is also a limit of an increasing sequence in $C_0(X)$, so is lower semicontinuous. Applying the same statement to $\Vert f\Vert - f$ implies that $f$ is also upper semi-continuous, so is continuous.

2. If $f\in C_0(X)$ is nonnegative and $t_n\to0$ then $(T_{t_n}f-f)_+\to0$ in the compact-open topology.

(Note: This doesn't use continuity of $Y$. And, in general, it is not necessary that $T_{t_n}f\to f$ in the compact open topology, even for $Y$ right-continuous.) It is enough to consider $f$ with compact support and, by right-continuity of the process $Y$, we have $T_{t_n}f(x)\to f(x)$ as $n\to\infty$ for all $x\in X$. Then, setting $g_m=m\int_0^{1/m}T_sf\,ds$, it can be seen that $\Vert T_{t_n}g_m-g_m\Vert\le 2mt_n\Vert f\Vert\to0$ as $n\to\infty$. Now, letting $S_m$ be the space of convex combinations of $\{g_m,g_{m+1},\ldots\}$, $f$ is a pointwise limit of a sequence in $S_m$, so is in its closure under the compact-open topology (this is a consequence of the Hahn-Banach theorem and the Riesz representation theorem). Therefore, there exists $f_m\in S_m$ converging to $f$ in the compact-open topology. For any $\epsilon > 0$ and compact $K\subseteq X$, take $m$ large enough such that $\vert f-f_m\vert\le\epsilon$ on the union of $K$ and the support of $f$. So, $$ T_tf-f \le T_tf_m-f_m+T_t(f-f_m)_++(f_m-f)\le T_tf_m-f_m + 2\epsilon. $$ on $K$. Letting $t$ decrease to zero gives $\limsup_{t\to 0}(T_tf-f)\le2\epsilon$ uniformly on $K$.

This is the point where continuity is required. It is not too hard to show that the space of $f\in C_b(X)$ satisfying the conclusion forms a $C^*$-algebra and is closed under $T_t$, so it is enough to prove it for nonnegative $f\in C_b(X)$ with compact support $K$. Also, the fact that $T_tC_0(X)\subseteq C_b(X)$ can be used to prove the strong-Markov property. If $Y_0\not\in K$ and, letting $\tau$ be the first time at which $Y$ hits $S$, continuity implies that $f(Y_{\tau})=0$ and $Y_\tau\in K$. So, for $x\not\in K$, $$ \begin{align} T_tf(x)=\mathbb{E}_x[f(Y_t)]&=\mathbb{E}_x\left[T_{(t-\tau)_+}f(Y_{\tau\wedge t})\right]\\\\ &\le\sup_{y\in K,s\le t}\left(T_sf(y)-f(y)\right)_+. \end{align} $$ The previous statement says that the right-hand-side tends to zero as $t\to0$ and, as it is independent of $x$, $T_tf\to0$ uniformly on $X\setminus K$.

As $X$ is lccb, $C_0(X)$ has a countable dense subset $S$. The closure of $S$ under products, taking $\mathbb{Q}$-linear combinations and applying $T_t$ for $t\in\mathbb{Q}_+$ is a countable dense subset of $\mathcal{A}$ (use 3). So $C_0(\hat X)\cong\mathcal{A}$ is separable. This implies that $\hat X$ has a countable base.

(Note: This doesn't use continuity of $Y$. And, in general, it is not necessary that $T_{t_n}f\to f$ in the compact open topology, even for $Y$ right-continuous.) It is enough to consider $f$ with compact support and, by right-continuity of the process $Y$, we have $T_{t_n}f(x)\to f(x)$ as $n\to\infty$ for all $x\in X$. Then, setting $g_m=m\int\_0^{1/m}T_sf\\,ds$, it can be seen that $\Vert T_{t_n}g_m-g_m\Vert\le 2mt_n\Vert f\Vert\to0$ as $n\to\infty$. Now, letting $S_m$ be the space of convex combinations of $\{g_m,g_{m+1},\ldots\}$, $f$ is a pointwise limit of a sequence in $S_m$, so is in its closure under the compact-open topology (this is a consequence of the Hahn-Banach theorem and the Riesz representation theorem). Therefore, there exists $f_m\in S_m$ converging to $f$ in the compact-open topology. For any $\epsilon > 0$ and compact $K\subseteq X$, take $m$ large enough such that $\vert f-f_m\vert\le\epsilon$ on the union of $K$ and the support of $f$. So, $$ T_tf-f \le T_tf_m-f_m+T_t(f-f_m)_++(f_m-f)\le T_tf_m-f_m + 2\epsilon. $$ on $K$. Letting $t$ decrease to zero gives $\limsup\_{t\to 0}(T_tf-f)\le2\epsilon$ on $K$.

(Note: This doesn't use continuity of $Y$. And, in general, it is not necessary that $T_{t_n}f\to f$ in the compact open topology, even for $Y$ right-continuous.) It is enough to consider $f$ with compact support and, by right-continuity of the process $Y$, we have $T_{t_n}f(x)\to f(x)$ as $n\to\infty$ for all $x\in X$. Then, setting $g_m=m\int\_0^{1/m}T_sf\\,ds$, it can be seen that $\Vert T_{t_n}g_m-g_m\Vert\le 2mt_n\Vert f\Vert\to0$ as $n\to\infty$. Now, letting $S_m$ be the space of convex combinations of $\{g_m,g_{m+1},\ldots\}$, $f$ is a pointwise limit of a sequence in $S_m$, so is in its closure under the compact-open topology (this is a consequence of the Hahn-Banach theorem and the Riesz representation theorem). Therefore, there exists $f_m\in S_m$ converging to $f$ in the compact-open topology. For any $\epsilon > 0$ and compact $K\subseteq X$, take $m$ large enough such that $\vert f-f_m\vert\le\epsilon$ on the union of $K$ and the support of $f$. So, $$ T_tf-f \le T_tf_m-f_m+T_t(f-f_m)_++(f_m-f)\le T_tf_m-f_m + 2\epsilon. $$ on $K$. Letting $t$ decrease to zero gives $\limsup\_{t\to 0}(T_tf-f)\le2\epsilon$ uniformly on $K$.

$\hat X$ is an lccb space.

As $X$ is lccb, $C_0(X)$ has a countable dense subset $S$. The closure of $S$ under products, taking $\mathbb{Q}$-linear combinations and applying $T_t$ for $t\in\mathbb{Q}\_+$ is a countable dense subset of $\mathcal{A}$. So $C_0(\hat X)\cong\mathcal{A}$ is separable. This implies that $\hat X$ has a countable base.

$T_tC_b(X)\subseteq C_b(X)$.

For any $f\in C_b(X)$ and $t_n\to0$, $T_{t_n}f\to f$ in the compact-open topology.

This is a fairly standard kind of argument showing that weak continuity of semigroups implies strong continuity. By right-continuity of the process $Y$, we have $T_{t_n}f(x)\to f(x)$ as $n\to\infty$ for all $x\in X$. Then, setting $g_m=m\int\_0^{1/m}T_sf\\,ds$, it can be seen that $\Vert T_{t_n}g_m-g_m\Vert\le 2mt_n\Vert f\Vert\to0$ as $n\to\infty$. Now, letting $S_m$ be the space of convex combinations of $\{g_m,g_{m+1},\ldots\}$, $f$ is a pointwise limit of a sequence in $S_m$, so is in its closure under the compact-open topology (this is a consequence of the Hahn-Banach theorem and the Riesz representation theorem). Therefore, there exists $f_m\in S_m$ converging to $f$ in the compact-open topology. However, by construction, $T_{t_n}f_m\to f_m$ uniformly as $n\to\infty$ and, taking the limit $m\to\infty$, we can conclude that $T_{t_n}f\to f$ under the compact-open topology (you do need to show that you can commute the limits like this, but it's not too hard an exercise).

For any $f\in\mathcal{A}$ and $t_n\to0$, $T_{t_n}f\to f$ uniformly as $n\to\infty$.

This is the point where continuity is required. It is not too hard to show that the space of $f\in C_b(X)$ satisfying the conclusion forms a $C^*$-algebra and is closed under $T_t$, so it is enough to prove it for $f\in C_b(X)$ with compact support $S$. Also, the fact that $T_tC_0(X)\subseteq C_b(X)$ can be used to prove the strong-Markov property. So, letting $\tau$ be the first time at which $Y$ hits $S$, continuity implies that $f(Y_{\tau})=f(Y_0)$ and, $$ \begin{align} T_tf(x)-f(x)=\mathbb{E}\_x[f(Y_t)-f(Y_0)]&=\mathbb{E}\_x\left[T_{(t-\tau)\_+}f(Y_{\tau\wedge t})-f(Y_{\tau\wedge t})\right]\\\\ &\le\sup\_{y\in S,s\le t}\left\vert T_sf(y)-f(y)\right\vert. \end{align} $$ The previous statement says that the right-hand-side tends to zero as $t\to0$ and, as it is independent of $x$, it bounds $\Vert T_tf-f\Vert$.

1. $T_tC_b(X)\subseteq C_b(X)$.

2. If $f\in C_0(X)$ is nonnegative and $t_n\to0$ then $(T_{t_n}f-f)\_+\to0$ in the compact-open topology.

(Note: This doesn't use continuity of $Y$. And, in general, it is not necessary that $T_{t_n}f\to f$ in the compact open topology, even for $Y$ right-continuous.) It is enough to consider $f$ with compact support and, by right-continuity of the process $Y$, we have $T_{t_n}f(x)\to f(x)$ as $n\to\infty$ for all $x\in X$. Then, setting $g_m=m\int\_0^{1/m}T_sf\\,ds$, it can be seen that $\Vert T_{t_n}g_m-g_m\Vert\le 2mt_n\Vert f\Vert\to0$ as $n\to\infty$. Now, letting $S_m$ be the space of convex combinations of $\{g_m,g_{m+1},\ldots\}$, $f$ is a pointwise limit of a sequence in $S_m$, so is in its closure under the compact-open topology (this is a consequence of the Hahn-Banach theorem and the Riesz representation theorem). Therefore, there exists $f_m\in S_m$ converging to $f$ in the compact-open topology. For any $\epsilon > 0$ and compact $K\subseteq X$, take $m$ large enough such that $\vert f-f_m\vert\le\epsilon$ on the union of $K$ and the support of $f$. So, $$ T_tf-f \le T_tf_m-f_m+T_t(f-f_m)_++(f_m-f)\le T_tf_m-f_m + 2\epsilon. $$ on $K$. Letting $t$ decrease to zero gives $\limsup\_{t\to 0}(T_tf-f)\le2\epsilon$ on $K$.

3. For any $f\in\mathcal{A}$ and $t_n\to0$, $T_{t_n}f\to f$ uniformly as $n\to\infty$.

This is the point where continuity is required. It is not too hard to show that the space of $f\in C_b(X)$ satisfying the conclusion forms a $C^*$-algebra and is closed under $T_t$, so it is enough to prove it for nonnegative $f\in C_b(X)$ with compact support $K$. Also, the fact that $T_tC_0(X)\subseteq C_b(X)$ can be used to prove the strong-Markov property. If $Y_0\not\in K$ and, letting $\tau$ be the first time at which $Y$ hits $S$, continuity implies that $f(Y_{\tau})=0$ and $Y_\tau\in K$. So, for $x\not\in K$, $$ \begin{align} T_tf(x)=\mathbb{E}\_x[f(Y_t)]&=\mathbb{E}\_x\left[T_{(t-\tau)\_+}f(Y_{\tau\wedge t})\right]\\\\ &\le\sup\_{y\in K,s\le t}\left(T_sf(y)-f(y)\right)\_+. \end{align} $$ The previous statement says that the right-hand-side tends to zero as $t\to0$ and, as it is independent of $x$, $T_tf\to0$ uniformly on $X\setminus K$.

Now let $f_n$ be the sequence converging uniformly on compacts to $f$ constructed in the proof of 2. Then, we have $f_n\to f$ uniformly on $X\setminus K$ as $n\to\infty$ so, in fact, $f_n$ tends uniformly to $f$. Taking the limits $t\to0$ and $n\to\infty$ in $$ \Vert T_tf-f\Vert\le\Vert T_tf_n-f_n\Vert+2\Vert f_n-f\Vert $$ shows that $\Vert T_tf-f\Vert\to0$.

This almost shows that $\hat T$ is Feller, just the following technical lemma is left.

4. $\hat X$ is an lccb space.

As $X$ is lccb, $C_0(X)$ has a countable dense subset $S$. The closure of $S$ under products, taking $\mathbb{Q}$-linear combinations and applying $T_t$ for $t\in\mathbb{Q}\_+$ is a countable dense subset of $\mathcal{A}$ (use 3). So $C_0(\hat X)\cong\mathcal{A}$ is separable. This implies that $\hat X$ has a countable base.