I have a mapping with is define as $T(u(t))x = E\Big(\sum_{i=1}^M \int_0^T u(t) \Delta_i(w,t)g_i(x) dt\Big) $ where $\Delta_i(w,t)$ is a stochastic process. Now I want to prove that the range of this mapping will have dimension $M$. First I integrated the mapping and $T(u(t))x = \sum_{i=1}^M Cg_i(x)$ which is a span of $g_i(x)$ which will have dimension at most $M$ if $g_i$'s are independent.
Now for the dimension of the range to be exactly $M$, I know I need to prove that each linearly independent $g_i$'s belong to the range of $T$. This is where I am having difficulties, I know that for the range of $T$ to be $g_i$, I need to use orthogonality so am thinking I can define $E(\Delta_i(w,t)) = h_i(t)$ and define say $w_i(t)$ to be an orthogonal function to $h_i$ (can this be done since $w_i(t)$ is arbitrary function on the domain) and do I just define their orthonrmals to get my results( am also confuse here since I have define $h_i$ to the expected function of $\Delta_i(w,t)$ so do I redefine it orthonormal and how do I incoporate the orthonormal in my mapping). That is can I do this \begin{equation} \begin{split} T(w_{i'})x & = \sum_i^M \langle{w_{i'},\Delta_i(w,t)}\rangle g_i(x)\\ & = E\Big(\int_0^T \sum_{i=1}^M w_{i'}(t)\Delta_i(w,t)g_i(x)dt\Big)\\ & = \Big(g_i(x)\int_0^T \sum_{i=1}^M w_{i'}(t)E(\Delta_i'(w,t))dt\Big)\\ & = \Big(g_i(x) \sum_{i=1}^M \int_0^T w_{i'}(t)h_i(t)dt\Big)\\ & = g_i(x) \end{split} \end{equation} where $w_{i'} $ is orthonormal of $w_i$ and $h_i$ is also orthonormal function such that if $i' = i$ the $(w_{i'} . h_i)$ = 1.
Please can someone help me? Also the domain of my function is define on positive real line and the range is just real line. Thank you