In Nualart's book "The Malliavin Calculus and related topics" ,

denotes by $\mathcal{E}$ the set of step functions on $[0,T]$ and says that $\mathcal{H}=\overline{(\mathcal{E},\langle\cdot,\cdot\rangle_H)}$ is a Hilbert space, where $\langle\mathbb{1}_{[0,t]},\mathbb{1}_{[0,s]}\rangle_H=R_H(t,s)=\frac{1}{2}(t^{2H}+s^{2H}-|t-s|^{2H})$ and $H\in(0,1)$.

What I've done so far:

If $(\mathcal{E},\langle\cdot,\cdot\rangle_H)$ is dense in $(L^2([0,T]),\langle\cdot,\cdot\rangle_H)$ then $\mathcal{H}$ is Hilbert. This could happen if the two norms induced by the inner products are equaivalent . That means, $\|\mathbb{1}_{[0,t]}\|_{L^2([0,T])}=\sqrt t$ is equivalent to $\|\mathbb{1}_{[0,t]}\|_H=t^{H}$. This is true after some calculations.

My question is: Is my thought right? And if not why $\mathcal{H}$ is Hilbert?

  • $\begingroup$ If I understand correctly you are taking a vector space, equipping it with an inner product, and asking if the completion with respect to this inner product is a Hilbert space. This is true by definition of Hilbert space i.e. complete inner product space. $\endgroup$
    – Lorenzo Q
    Oct 4, 2018 at 17:42
  • $\begingroup$ Lorenzo Quarisa thanks! $\endgroup$ Oct 4, 2018 at 17:49
  • $\begingroup$ To be more precise it depends on what you mean by closure, if you took the closure with respect to the $L^2$ inner product then what you wrote makes sense. But I suspect that here the author simply means closure with respect to $\left \langle \cdot,\cdot\right\rangle_H$. $\endgroup$
    – Lorenzo Q
    Oct 4, 2018 at 17:51
  • $\begingroup$ Yes, closure by $\langle\cdot,\cdot\rangle_H$. So, in an inner product space closed sets are complete? $\endgroup$ Oct 4, 2018 at 17:54
  • 1
    $\begingroup$ Typically, you use the term 'closure' when you have a subset of a topological space. Here you are not considering $\mathcal{E}$ as a subset of some larger space. Thus the only reasonable meaning you can give to 'closure' is 'completion'. Notice that the closure of an inner product space as a subspace of its completion, is the completion itself, so the two notions agree. $\endgroup$
    – Lorenzo Q
    Oct 4, 2018 at 18:47

1 Answer 1


There is not much you have to do: you are giving a vector space with an inner product. It's completion is a Hilbert space. The only issue is to show that the inner product is such.

The definition is fine in the sense that the functions $\{1_{[0,t]}:\ t\in[0,T]\}$ are linearly independent, so you can define $\langle f,g\rangle_H$ for all $f,g\in\mathcal E$. What is not so clear to me is that you get an inner product: mainly, you need to show that $\langle f,f\rangle_H=0$ implies $f=0$. This means that if $f=\sum_{j=1}^n\alpha_j\,1_{[0,t_j]}$, $$ 0=\langle f,f\rangle_H=\frac12\,\sum_{k,j}\alpha_k\alpha_j(t_k^{2H}+t_j^{2H}-|t_k-t_j|^{2h}), $$ implies $\alpha_1=\cdots=\alpha_n=0$. I don't see an immediate way to show that, and that's the crucial issue to solve in your questions. If you show the above, the rest is straightforward.

There are a couple issues with your reasoning, too. It is not true that $t^{1/2}$ and $t^H$ are equivalent, unless $H=1/2$. If you had $H<1/2$ and $t^H\leq c\,t^{1/2}$, you have $t^{1/2-H} \geq1/c$ for all $t$ close to zero, which is false. Similarly with $H>1/2$.

Also, to test that two norms are equivalent you have to do it for all elements of the space, not just some. So you cannot do it for $1_{[0,t]}$ only, you have to do it for all linear combinations.

  • $\begingroup$ And this is true because the function $R_H(t,s)$ is positive definite kernel. (This is known from the previous theory) $\endgroup$ Oct 5, 2018 at 7:48
  • $\begingroup$ It is a way of establishing a theory of integration relative to fractional Brownian motion. Funny situation. One way to establish an abstract and general theory of integration is to go through the rough path theory. $\endgroup$
    – Zbigniew
    Jan 4, 2020 at 19:46

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