Suppose we have a time series $(x_t\mid t\in \mathbb{Z})$ for which the partial sum process $X_T$ defined on the unit interval by $$ X_T(\xi)=\omega_T^{-1}\sum_{t=1}^{[T\xi]} (x_t-\mathbb{E}(x_t)),\quad \xi\in[0,1],$$ where $\omega_T^2=\mathrm{Var}(\sum_{t=1}^Tx_t)$, converges weakly to a standard Brownian motion $B$ as $T\rightarrow\infty$.
Here is my question: Why does the above setup imply that $\omega^2_T\sim T\omega^2$ for some $\omega^2\in(0,\infty)$?
I have read this statement in a paper, and the argument provided there is "otherwise the limit process cannot have the Brownian property $\mathrm{E}((B(s)-B(r))^2)=s-r$ for $0\leq r < s \leq 1$." However, I could not figure out a way to "formalize" this argument. Can anybody help me out there? I appreciate any help! Many thanks!!!
For every $\xi\leqslant1$, $\omega_{T\xi}X_{T\xi}(1)=\omega_TX_T(\xi)$. By hypothesis, $X_{T\xi}(1)\to B_1$ and $X_T(\xi)\to B_\xi$ in distribution. Since $B_\xi$ is distributed like $\sqrt{\xi}B_1$, this yields $\omega_{T\xi}\sim\omega_T\sqrt{\xi}$ when $T\to\infty$.
Let $\bar\omega_T=\omega_T/\sqrt{T}$. Thus, $\bar\omega_{T\xi}\sim\bar\omega_T$ when $T\to\infty$, for every $\xi\leqslant1$, and the assertion in the paper is that in this setting, $\bar\omega_T$ has a limit when $T\to\infty$. The case $\bar\omega_T=\log(1+T)$ seems to be a counterexample.