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Let $(X_t)$ denote a process, where $X_t\in L^2(\Omega,F,P)$. Here, $L^2$ is a Hilbert space with inner product $\langle X,Y\rangle = E(XY)$.

Maybe a stupid question but is the closed span $$ \overline{\text{sp}}\left\{X_1,\ldots,X_n\right\} $$ a Hilbert space, too and are $X_1,X_2,\ldots,X_n$ lineary independent elements of this Hilbertraum?

I would say yes since it is a subvector space (zero is contained and closed under scalar multiplication and addition) and it is still complete (since it is topologically closed).

Am I right?

If yes, then $X_1,...,X_n$ form a base of this hilbert space and therefore are linearly independent.

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    $\begingroup$ Closed subsets of complete metric spaces are complete. $\endgroup$ – Shalop Oct 27 '15 at 18:04
  • $\begingroup$ So to sum it up: Yes, it is a Hilbert space since it is a sub-vector space and, moreover, topologically closed, hence complete. To the second question: $X_1,...,X_n$ are linear independent elements of this hilbert space $\overline{\text{sp}}\left\{X_1,...,X_n\right\}$ since they form a basis of it. - - Right? $\endgroup$ – M. Meyer Oct 27 '15 at 18:10
  • $\begingroup$ Of course not. For example let $X \in L^2$ and consider the constant process $X_n:=X$ for all $n$. Since the $X_n$ are all equal, they cannot be linearly independent in $L^2$. $\endgroup$ – Shalop Oct 27 '15 at 18:15
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Yes closed span of $X_1,\ldots,X_n$ is closed subset of complete space, so it is complete. And you can use inner product to from original space. Denote $$X_{n+1}=\sum_{i=1}^{n}X_i,$$ you can immediately see, that closed spans of $\{X_1,\ldots,X_n\}$ and $\{X_1,\ldots,X_{n+1}\}$ are the same, but the second set does not contain linearly independent elements. So $X_1,\ldots,X_n$ are linearly independent in the new space, only if they are linearly independent on the original space.

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  • $\begingroup$ Hm, but they form a base, don't they? I do not understand why they are not linear independent then, sorry. $\endgroup$ – M. Meyer Oct 27 '15 at 18:17
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    $\begingroup$ No, they do not form basis in general. $A$ is basis of space $X$ if elements of $A$ generates $X$ and elements $A$ are linearly independent. You did not assume, that $\{X_1,\ldots,X_n\}$ are linearly independent, so it general they could not form a basis. $\endgroup$ – iiivooo Oct 27 '15 at 18:25
  • $\begingroup$ So, for example, I cannot apply the Gram Schmidt process on $\left\{X_1,\ldots,X_n\right\}$, correct? $\endgroup$ – M. Meyer Oct 27 '15 at 18:26
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    $\begingroup$ You can, but if $\left\{X_1,\ldots,X_n\right\}$ are linearly dependent it stops after $k$ steps, $k < n$. Think about Shalops example, define $X_i=Y$ for all $i$. Then $span\left\{X_1,\ldots,X_n\right\}=span\{Y\}$ is clearly 1 dimensional space, so $\left\{X_1,\ldots,X_n\right\}$ could not be basis for $n>1$. $\endgroup$ – iiivooo Oct 27 '15 at 18:30

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