Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X_1$ and $X_2$ be real-valued square-integrable random variables defined on a probability space $(\Omega, {\cal F},P)$. For $i=1,2$, set $$ A_i := \{g(X_i)\in L^2 \mid g \text{ is some Borel measurable function with } \mathbb{E}g(X_i)=0 \} .$$ Note that $A_i$ forms a Hilbert subspace of $L^2(\Omega, {\cal F},P)$ for each $i$.

My question: does $$A_1 + A_2 := \{g_1(X_1) + g_2(X_2): g_i(X_i)\in A_i, \; i=1,2\}$$ equipped with the norm $||\cdot||_{L^2}$ also form a Hilbert subspace of $L^2(\Omega, {\cal F},P)$?

share|improve this question
    
Yes, the sum of two closed subspaces of a Hilbert space is again closed. –  Jim Belk Nov 26 '11 at 18:10
    
I think it can be done by considering orthonormal bases: If $H_1, H_2$ are closed subspaces of a Hilbert space $H$, then $H_1,H_2$ are Hilbert spaces in their own right. If the sum of them is indeed direct $H = H_1 \oplus H_2$ Then the norm of $(h_1, h_2) \in H_1 \oplus H_2$ is given by $\|(h_1,h_2)\|^2 = \|h_1\|^2 + \|h_2\|^2$. Completeness follows from the fact that $H_1,H_2$ are complete. –  user12014 Nov 27 '11 at 1:18
2  
@Jim No, I don't think this is true. –  Byron Schmuland Nov 27 '11 at 15:25
    
@PZZ Although the OP refers to a direct sum, the subspaces are not assumed to be orthogonal. I am interested in the problem as stated. –  Byron Schmuland Nov 27 '11 at 15:26
    
@Byron You are correct. I have posted an example of this below. –  Jim Belk Nov 29 '11 at 0:46
add comment

2 Answers 2

up vote 3 down vote accepted
+250

Building on my other answer, we can construct a counterexample to the original question.

Let $\{I_n \mid n\geq 3\}$, $\{I'_n \mid n\geq 3\}$, $\{J_n \mid n\geq 3\}$, and $\{J'_n\mid n\geq 3\}$ be mutually exclusive events satisfying $$ P(I_n) = P(I'_n) = \frac{1}{2n^2}\qquad\text{and}\qquad P(J_n) = P(J'_n)=\frac{1}{2n^4}. $$ Let $X_1$ and $X_2$ be the random variables $$ X_1(\omega) =\begin{cases}1/n & \text{if }\,\omega\in I_n \\ -1/n & \text{if }\,\omega\in I'_n, \\ 0 & \text{otherwise},\end{cases} \qquad\text{and}\qquad X_2(\omega) =\begin{cases}1/n & \text{if }\,\omega\in I_n\cup J_n, \\ -1/n & \text{if }\,\omega\in I'_n\cup J'_n, \\ 0 & \text{otherwise},\end{cases} $$ For each $n$, let $Y_n$ and $Z_n$ be the random variables $$ Y_n(\omega)=\begin{cases}n & \text{if }\,\omega\in I_n \\ -n & \text{if } \,\omega\in I'_n \\ 0 & \text{otherwise}\end{cases} \qquad\text{and}\qquad Z_n(\omega)=\begin{cases}n^2 & \text{if }\,\omega\in J_n \\ -n^2 & \text{if } \,\omega\in J'_n \\ 0 & \text{otherwise}\end{cases} $$ Then the functions $\{Y_n\mid n\geq 3\}$ and $\{Z_n \mid n\geq 3\}$ are orthonormal. Moreover, $Y_n \in A_1$ for each $n$, and $Y_n + \frac{1}{n} Z_n \in A_2$ for each $n$. It follows that $Z_n\in A_1+A_2$ for each $n$. However, the sum $$ \sum_{n=3}^\infty \frac{1}{n}Z_n $$ does not lie in $A_1+A_2$.

share|improve this answer
add comment

This is not intended as an answer. However, as Byron points out, the direct sum of two closed, non-orthogonal subspaces of a Hilbert space need not be closed. For others thinking about this question, here is an example of this phenomenon.

Let $\mathcal{H}$ be an infinite-dimensional Hilbert space, and let $\{\textbf{e}_1,\textbf{e}_2,\ldots\}$ and $\{\textbf{f}_1,\textbf{f}_2,\ldots\}$ be two mutually orthogonal sequences of orthonormal vectors in $\mathcal{H}$. Let $$ U \;=\; \text{closure}\bigl(\text{Span}\{\textbf{e}_n \mid n\in\mathbb{N}\}\bigr) $$ and let $$ V \;=\; \text{closure}\bigl(\text{Span}\bigl\{\textbf{e}_n+\tfrac{1}{n}\textbf{f}_n \mid n\in\mathbb{N}\bigr\}\bigr) $$ The subspaces $U$ and $V$ are closed by definition, and the intersection $U\cap V\;$ is trivial. However, the direct sum $U+V\;$ is not a closed subspace. In particular, observe that all of the vectors $\textbf{f}_n$ lie in $U+V$, but the sum $$ \sum_{n=1}^\infty \frac{1}{n}\textbf{f}_n $$ does not lie in $U+V$.  This gives a contradiction, since $\{1/n\}$ is an $\ell^2$ sequence.

share|improve this answer
    
Thanks for this! –  Byron Schmuland Nov 29 '11 at 0:52
    
Jim, why the sum is not in $U+V$? This is still not obvious to me. –  Kerry Nov 30 '11 at 1:54
    
@ChangweiZhou For every $v\in V$ and $n\geq 1$ we have $\langle v,\textbf{f}_n\rangle=\langle v,\textbf{e}_n\rangle/n$. If $u\in U$ and $v\in V$, then $n\langle u+v,\textbf{f}_n\rangle=n\langle v,\textbf{f}_n\rangle=\langle v,\textbf{e}_n\rangle$ is square summable. –  Byron Schmuland Nov 30 '11 at 2:56
    
@ByronSchmuland: This is clear. Thanks. –  Kerry Nov 30 '11 at 11:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.