Let $f_N(s,\omega)$ be some bounded jointly measurable function in the product space $L^1([0,T],\mathscr{L},\mu_L) \times (\Omega,\mathscr{F},P)$.
For such an $f_N$ we define a function $g_k$ for $k>0$ by $$g_k(t,\omega) = ke^{-kt} \int_0^t e^{ks} f_N(s,\omega)ds.$$ In this case, for any $\epsilon>0$, how do we find an $g_k$ for which
$\int_0^T E | f_N(t,\cdot) - g_k(t,\cdot)|^2 dt < \epsilon$?
This is part of a proof from Numerical Solution of Stochastic Differential Equations. I would greatly appreciate any help.
Now a function $f \in \mathcal{L}_{T}^{2}$ is generally not mean-square continuous, but we can approximate it arbitrarily closely in the norm (2.5) by one that is. To begin, we approximate $f$ by a bounded function $f_{N} \in \mathcal{L}_{T}^{2}$ defined by $$ f_{N}(t, \omega)=\max \{-N, \min \{f(t, \omega), N\}\} $$ for some $N>0 .$ Obviously $\left|f_{N}(t, \omega)\right| \leq N,$ with $f_{N}(t, \omega)=f(t, \omega)$ for those $(t, \omega)$ for which $|f(t, \omega)| \leq N .$ Moreover $$ \int_{0}^{T} E\left(\left|f_{N}(t, \cdot)-f(t, \cdot)\right|^{2}\right) d t \leq 4 \int_{0}^{T} E\left(|f(t, \cdot)|^{2}\right) d t<\infty $$ so by the Dominated Convergence Theorem 2.2 .3 applied to the functions $E\left(\left|f_{N}(t, \cdot)-f(t, \cdot)\right|^{2}\right)$ in $L^{1}\left([0, T], \mathcal{L}, \mu_{L}\right)$ it follows that $$ \int_{0}^{T} E\left(\left|f_{N}(t, \cdot)-f(t, \cdot)\right|^{2}\right) d t \rightarrow 0 \quad \text { as } \quad N \rightarrow \infty $$ Then for such an $f_{N}$ we define a function $g_{k}$ for $k>0$ by $$ g_{k}(t, \omega)=k e^{-k t} \int_{0}^{t} e^{k s} f_{N}(s, \omega) d s $$ From the properties of $f_{N}$ and the fact that the above integrand does not involve values of $f_{N}$ for times exceeding $t,$ it follows that $g_{k}$ is jointly $\mathcal{L} \times \mathcal{A}$ measurable and that $g_{k}(t, \cdot)$ is $\mathcal{A}_{t}$ -measurable for each $t \in[0, T] .$ Also from the bound on $\left|f_{N}\right|$ we have $$ \left|g_{k}(t, \omega)\right| \leq N\left(1-e^{-k t}\right) $$ so $E\left(g_{k}(t, \cdot)^{2}\right)$ is finite and integrable over $0 \leq t \leq T ;$ hence $g_{k} \in \mathcal{L}_{T}^{2} .$ Finally, the sample paths of $g_{k}$ satisfy $$ \left|g_{k}(t, \omega)-g_{k}(s, \omega)\right| \leq 2 k N|t-s| $$ and are thus continuous. In fact this bound also implies that $E\left(g_{k}(t, \cdot)^{2}\right)$ is continuous, that is $g_{k}$ is mean-square continuous. Consequently we can approximate it by a step function $f^{(n)} \in \mathcal{S}_{T}^{2}$ as in the first part of the proof. For any given $\epsilon>0$ we can choose $f_{N}, g_{k}$ and $f^{(n)}$ successively so that $$ \begin{array}{c} \left\|f-f_{N}\right\|_{2, T}<\frac{1}{3} \epsilon, \quad\left\|f_{N}-g_{k}\right\|_{2, T}<\frac{1}{3} \epsilon \\ \left\|g_{k}-f^{(n)}\right\|_{2, T}<\frac{1}{3} \epsilon \end{array} $$ Then by the triangle inequality $(1.4 .37)$ we have $$ \left\|f-f^{(n)}\right\|_{2, T}<\epsilon $$ which is what we were required to prove. $\square$
You can write $g_k$ as the convolution $g_k = f_N * \psi_k$, where $\psi_k(t) = k e^{-kt}\mathbf{1}_{t>0}$ is delta-like. Since $f_N(\cdot,\omega)\in L^2(0,T)$ for almost all $\omega$, then $$ ||g_k(\cdot,\omega) - f_N(\cdot,\omega)||_{L^2[0,T]} \to 0,k\to\infty, $$ for such $\omega$. Thanks to boundedness, $||g_k - f_N||^2_{2,T}\to 0$, $k\to\infty$ (I guess that $||f||^2_{2,T}$ is just the expectation of $||f||^2_{L^2[0,T]}$).