# Closure of the space of step functions is Hilbert space

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?

• 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. Oct 4, 2018 at 17:42
• Lorenzo Quarisa thanks! Oct 4, 2018 at 17:49
• 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 $\left \langle \cdot,\cdot\right\rangle_H$. Oct 4, 2018 at 17:51
• Yes, closure by $\langle\cdot,\cdot\rangle_H$. So, in an inner product space closed sets are complete? Oct 4, 2018 at 17:54
• 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. Oct 4, 2018 at 18:47

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.
• And this is true because the function $R_H(t,s)$ is positive definite kernel. (This is known from the previous theory) Oct 5, 2018 at 7:48