In Oksendal's book of Stochastic Differential Equations, the author does a reasoning similar to what's presented below.

The follwing three points are proved:

  1. I can find $\{\phi_n\}$ such that $E(\int_S^T(g-\phi_n)^2dt)\rightarrow 0$, as $n\rightarrow \infty$.
  2. I can find $\{g_n\}$ (each $g_n$ is a $g$ from point 1) such that $E(\int_S^T(h-g_n)^2dt)\rightarrow 0$, as $n\rightarrow \infty$.
  3. I can find $\{h_n\}$ (each $h_n$ is an $h$ from point 2) such that $E(\int_S^T(f-h_n)^2dt)\rightarrow 0$, as $n\rightarrow \infty$.

Then the author states that by the three points above, for any $f$ there is a sequence of $\{\phi_n\}$ such that $E(\int_S^T(f-\phi_n)^2dt)\rightarrow 0$, as $n\rightarrow \infty$.

How is he able to state that? I think he's probably using some simple relationship/inequality which allows him to conclude that, and I'm not getting it.

All of these functions have their own properties, stated in the book, which I didn't write, because in my opinion they would make the question bigger, without improving it.


He is essentially using the triangle inequality.

Fix some function $f$. Then by, 3., there is some $h_n$ with $\sqrt{\mathbb{E}\int_S^T (f-h_n)^2 \, dt} \leq \frac{1}{3n}$. Now, by 2., there is some $g_n$ with $\sqrt{\mathbb{E}\int_S^T (h_n-g_n)^2 \, dt} \leq \frac{1}{3n}$. Finally, by 1., there is $\phi_n$ with $\sqrt{\mathbb{E}\int_S^T (g_n-\phi_n)^2 \, dt} \leq \frac{1}{3n}$. Since $$\sqrt{\mathbb{E}\int_S^T f(t)^2 \, dt}$$ is a norm (the $L^2$-norm with respect to the product measure $\mathbb{P} \otimes \lambda$), it satisfies the triangle inequality. Writing $$f-\phi_n = (f-h_n)+(h_n-g_n)+(g_n-\phi_n)$$ we thus get

\begin{align*} \sqrt{\mathbb{E}\int_S^T (f-\phi_n)^2 \, dt} &\leq \sqrt{\mathbb{E}\int_S^T (f-h_n)^2 \, dt} + \sqrt{\mathbb{E}\int_S^T (h_n-g_n)^2 \, dt} \\ &\quad + \sqrt{\mathbb{E}\int_S^T (g_n-\phi_n)^2 \, dt} \\ &\leq 3 \frac{1}{3n}\end{align*}

and so $ \sqrt{\mathbb{E}\int_S^T (f-\phi_n)^2 \, dt} \to0$.

Remark: You can also read the statements a different (more analytic) way. Say, you have a normed space $X$ and two subsets $C$ and $D$. If $C$ is dense in $D$ and $D$ is dense in $X$, then $C$ is dense in $X$. Of course, you can iterate this (i.e. take another set which is dense in $C$ and so on). That's exactly Oksendal uses in his book. Statement 3. says that the functions $h$ (with certain properties) are dense (w.r.t. $L^2$-norm) in your original space of functions $f$. Statement 2. says that the functions $g$ (with certain properties) are dense (w.r.t. $L^2$-norm) in the set of functions $h$, and so on.

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  • $\begingroup$ Saz, thanks for the answer. I really appreciate it. I still have some questions though... 1) - The 2-norm as is defined in the book is $E(|X|^2)^{1/2}$. How does one go about to see that yours is also a norm? If we could use the Itô isometry, than I understand. However, at that point in the book, we only have isometry for simple functions. 2) - In the approximations that you use, shouldn't we use subsequences, i.e., for each $h_n$, we have $g_{n_i}$, and for each $g_{n_i}$ we can approximate it by $\phi_{n_{j_i}}$ 3) - I think there's a typo in the formula following "there's a $g_n$" $\endgroup$ – An old man in the sea. Jun 30 at 17:26
  • 2
    $\begingroup$ @Anoldmaninthesea. 1) This has nothing to do with Ito^'s isometry. For any measure $\mu$ the mapping $f \mapsto \sqrt{\int f^2 \, d\mu}$ defines a norm. Here, in our case, $\mu$ is the product measure of the probability measure $\mathbb{P}$ and Lebesgue measure $\lambda$. 2) No, we do not use subsequences here ... we do the whole argumentation for fixed $n$. E.g. after we have chosen $h_n$, we need to find some function $g$ "(with the properties of the "g-functions", which you did not state explitily) such that $\sqrt{\mathbb{E}\int |h_n-g|^2 \, dt} \leq \frac{1}{3n}$. This is possible because $\endgroup$ – saz Jun 30 at 18:30
  • $\begingroup$ of 2. (If you prefer think of $h$ instead of $h_n$... by 2. you can approximate this by some sequence of "g-functions"... in particular there is some $g$ (...or rather $g_n$) which is close to $h$ (i.e. $h_n$) in $L^2$ norm). Did you think about my remark? From my point of view, it's much easier to read the statements as denseness of certain subsets. 3)Yeah, right $\endgroup$ – saz Jun 30 at 18:32
  • $\begingroup$ Many thanks! Now I understand! ;) Btw, I did read your remark, and found it very intuitive. I just wasn't sure how it should be interpreted at the light of your answer. Now, I know. Thanks ;) +1 $\endgroup$ – An old man in the sea. Jun 30 at 19:14

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