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

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 $$0=t_0 and square integrable random variables $$\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

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

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 $$\lim_{n \to \infty} \int_0^{\infty} (f(t)-f_n(t))^2 \, dt =0.$$
• @user671288 Fix $t \in [0,N]$ for some $N \in \mathbb{N}$. Then $$|f(t)-f_n(t) = |f(t)-f(t_n)|, \qquad n \geq N$$ where $t_n$ is the smallest number of the form $k2^{-n}$ which is strictly larger than $t$; in particular $|t_n-t| \leq 2^{-n}$. Uniform continuity of $f$ now gives pointwise convergence. – saz May 5 at 16:07