\infty$, in general dependent on $t$, along which these vectors converge weakly in ${\cal H}$, $$\begin{aligned}
 \label{weakconv}
&& \hat{u}(t,\tau) \stackrel{\scriptsize{w}}{\longrightarrow}
\hat{u}(t)\,, \quad
  \tau^{-1/2}Q(\tau)\hat{u}(t,\tau)
   \stackrel{\scriptsize{w}}{\longrightarrow} g_0(t)\,,  \nonumber \\
&& t^{1/2}H(t\tau)^{1/2}P(\tau)\hat{u}(t,\tau)
   \stackrel{\scriptsize{w}}{\longrightarrow} h(t)\,,\end{aligned}$$ for some vectors $\hat{u}(t), \,  g_0(t)$ and $h(t)$ in ${\cal H}$. Note that the sequence $\{\tau_n(t)\}_{n=1}^{\infty}$ can be chosen the same for all three families.

&gt;From this result we see first that $Q(\tau)\hat{u}(t,\tau)
\stackrel{\scriptsize{s}}{\longrightarrow} 0$ uniformly in $t> 0$ as $\tau\rightarrow 0$, so that we have $Q\hat{u}(t)=0$ or $\hat{u}(t)=P\hat{u}(t) \in P{\cal H}$. For every $v \in
D[H^{1/2}]$ we have, with the limit taken along $\{\tau_n(t)\}$, $$\begin{aligned}
 \langle v, h(t)\rangle
 &\!=\!&  \lim\, \langle v,  t^{1/2}H(t\tau)^{1/2}P(\tau)\hat{u}
(t,\tau) \rangle\\
 &\!=\!& t^{1/2} \,\lim\, \langle H(t\tau)^{1/2}v,P(\tau)\hat{u}
(t,\tau) \rangle
 = t^{1/2} \langle H^{1/2}v, P\hat{u}(t)\rangle\,,\end{aligned}$$ because $H(t\tau)^{1/2}v
\stackrel{\scriptsize{s}}{\longrightarrow} H^{1/2}v$ as $\tau\rightarrow 0$. Hence $\hat{u}(t)=P\hat{u}(t)$ belongs to $D[H^{1/2}]$ and $h(t) =t^{1/2} H^{1/2}P\hat{u}(t)$ because $D[H^{1/2}]$ is dense by assumption. Furthermore, multiplying (\[fdecomp\]) by $\tau^{1/2}$ and taking the weak limit along the sequence $\{\tau_n(t)\}$ we get $g_0(t) = 0$. Similarly, multiplying (\[fdecomp\]) by $P(\tau)$ we have for every $v \in D[H^{1/2}P]$ $$\langle v, P(\tau)f \rangle
 = \langle v, P(\tau) \hat{u}(t,\tau) \rangle
  + \langle t^{1/2}H(t\tau)^{1/2}P(\tau)v,
     t^{1/2}H(t\tau)^{1/2}P(\tau)\hat{u}(t,\tau) \rangle.$$ Then taking the limit along the sequence $\{\tau_n(t)\}$ we get $$\langle v, Pf 