A terminal weak limit does not imply a path-space weak limit

Since $\sin (0)=\sin (\pi )=0$, both even and odd paths satisfy $w_n(1)=0$. Thus their terminal laws are all the point mass ${\delta }_0$ and in particular converge weakly to ${\delta }_0$ in $H$.

Define $\mathrm{\Phi }:\mathcal{C}\to \mathbb{R}$ by $\mathrm{\Phi }(w)=\min \{1,{\sup }_{0\le t\le 1}\mathrm{\parallel }w(t){\mathrm{\parallel }}_H\}$. The supremum norm inequality gives $\mathrm{\mid }{\sup }_t\mathrm{\parallel }w(t){\mathrm{\parallel }}_H-{\sup }_t\mathrm{\parallel }v(t){\mathrm{\parallel }}_H\mathrm{\mid }\le {\sup }_t\mathrm{\parallel }w(t)-v(t){\mathrm{\parallel }}_H=d(w,v),$ and truncation by $x\mapsto \min (1,x)$ is 1-Lipschitz on $[0,\mathrm{\infty })$. Hence $\mathrm{\Phi }$ is bounded and continuous. It equals $0$ on every even path and $1$ on every odd path, so its expectations under the path laws alternate between $0$ and $1$ and do not converge. By [F1], those path laws cannot converge weakly.

The example has convergence only after applying the terminal evaluation map; the bounded continuous path functional in step 1.2 detects the missing path-space convergence. Therefore endpoint convergence alone supplies no functional CLT conclusion, consistently with the scope in [F2].