This is an exercise (1.4.2) from Applebaum's Levy Processes and Stochstic Calculus.
I want to show that a convolution semigroup is weakly continuous if and only if \begin{equation} \lim_{s\searrow t} \int_{\mathbb{R}^d} f(y) p_s(dy) = \int_{\mathbb{R}^d} f(y) p_t(dy), \end{equation} for all bounded and continuous $f$ ($f \in C_b(\mathbb{R}^d)$), for all $t\geq 0$.
Definition: A convolution semigroup is a family of probability measures on $\mathbb{R}^d$, $(p_t; t\geq 0)$ with $p_0 = \delta_0$ and $p_s *p_t = p_{s+t}$. It is weakly continuous if \begin{equation} \lim_{t\searrow 0} \int_{\mathbb{R}^d} f(y) p_t(dy) = f(0), \end{equation} for all $f \in C_b(\mathbb{R}^d)$.
My attempt: The if statement is obvious, just take $t=0$. For the only if: assume $(p_t)$ is a weakly continuous convolution semigroup and let $f \in C_b(\mathbb{R}^d)$ and $t\geq 0$, then \begin{align} \lim_{s\searrow t} \int_{\mathbb{R}^d} f(y) p_s(dy) &= \lim_{s\searrow t} \int_{\mathbb{R}^d} f(y) (p_{t} * p_{s-t})(dy) \\ &= \lim_{s\searrow t} \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} f(x+z) p_{t}(dx) p_{s-t}(dz) \\ &= \lim_{u\searrow 0} \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} f(x+z) p_{t}(dx) p_{u}(dz), \end{align} If I can show that the integrand of the outer integral, $g(z):=\int_{\mathbb{R}^d} f(x+z) p_{t}(dx)$ is in $C_b(\mathbb{R}^d)$, i.e. continuous and bounded w.r.t. $z$, then the result follows from definition of weakly continuous. Boundedness of $g$ is easy to show: \begin{align} \lvert g(z)\rvert &\leq \int_{\mathbb{R}^d} \lvert f(x+z)\rvert p_{t}(dx) \\ &\leq M, \end{align} where $M\geq 0$ is the bound on $f$, and we use the fact that $p_t$ is a probability measure. For continuiuty of $g$, let $\varepsilon > 0$. Let $B \subset \mathbb{R}^d$ be a bounded set with measure $p_t(B) \geq 1 - \frac{\varepsilon}{4M}$. By continuity of $f$ and boundedness of $B$, we can find $\delta > 0$ such that $\lvert f(z) - f(y) \rvert < \frac{\varepsilon}{2}$ for any $z, y \in B$ with $0 \leq \lVert z - y \rVert < \delta$. Then, for $z, y \in \mathbb{R}^d$ with $\lVert z - y \rVert < \delta$, \begin{align} \lvert \int_{\mathbb{R}^d} f(x+z) p_{t}(dx) - \int_{\mathbb{R}^d} f(x+y) p_{t}(dx)\rvert &\leq \int_{\mathbb{R}^d} \lvert f(x+z) - f(x+y) \rvert p_{t}(dx) \\ &= \int_{B} \lvert f(x+z) - f(x+y) \rvert p_{t}(dx) + \int_{B^c} \lvert f(x+z) - f(x+y) \rvert p_{t}(dx) \\ &\leq \frac{\varepsilon}{2} + 2M \cdot \frac{\varepsilon}{4M} \\ &= \varepsilon \end{align}
Since we have established $g \in C_b(\mathbb{R}^d)$, then \begin{align} \lim_{s\searrow t} \int_{\mathbb{R}^d} f(y) p_s(dy) &=\lim_{u\searrow 0} \int_{\mathbb{R}^d} g(z) p_{u}(dz) \\ &= g(0) \\ &= \int_{\mathbb{R}^d} f(x) p_t(dx). \end{align} which is the desired result.
Is my proof correct?