Finite sequence of random step processes such that $\lim_{n\to\infty}E(\int_{0}^{\infty}|f(t)-f_n(t)|^2dt)=0$ for $f(t)=e^{-t^2/4}$

Question

Let

$$f(t)=e^{-t^2/4}, \ \ \ t \ge 0$$

I want to show that $f$ is in $M^2$

where $M^2$ denotes the class of stochastic processes $f(t),t\ge0$ such that

$$E\left(\int_0^\infty|f(t)|^2dt\right)<\infty$$

and there is a sequence of $f_1,f_2,\ldots \in M^2_{\mathrm{step}}$ of random step processes such that

$$\lim_{n\to\infty}E\left(\int_{0}^{\infty}|f(t)-f_n(t)|^2dt\right)=0$$

$M^2_{step}$ is the set of random step processes. We call $f(t),t\ge0$ a random step process if there is a finite sequence of numbers $0=t_0<t_1<\ldots<t_n$ and square integrable random variables $\eta_0, \eta_1, \ldots, \eta_{n-1}$ such that

$$f(t)=\sum_{j=0}^{n-1}\eta_j1_{[t_j,t_{j+1})}(t)$$, where $\eta_j$ is $\mathcal F_{t_j}$-measurable for $j=0,1,\ldots,n-1$

The $E\left(\int_0^\infty|f(t)|^2dt\right)<\infty$ part is clear.

I need help for finding such a sequence $f_n$ of random step processes such that

$$\lim_{n\to\infty}E\left(\int_{0}^{\infty}|f(t)-f_n(t)|^2\right)=0$$

Answer 1

Hints: Set $$f_n(t) := \sum_{k=0}^{n 2^n} \exp \left[ - \frac{1}{4} \exp \left(- \frac{(k+1)}{2^n} \right)^2 \right] 1_{[k2^{-n},(k+1)2^{-n})}(t).$$

1. Check that $f_n(t) \leq f(t)$ for all $t \geq 0$. (Hint: $f$ is decreasing.)
2. Check that $f_n(t) \to f(t)$ as $n \to \infty$. (Hint: $f$ is uniformly continuous)
3. Conclude from the dominated convergence theorem that $\lim_{n \to \infty} \int_0^{\infty} (f(t)-f_n(t))^2 \, dt =0.$