Open linear subspace of a Hilbert space.

Question

Does there exist any open linear (vector) subspace of a Hilbert space? I could not think of any example.

Actually, I was reading the book by Simmons, there almost in every theorem it assumed that "If M is a closed linear subspace".It seemed natural to me to think about subspaces which are not closed. I have an got an example which is not closed: Take the Hilbert space H = L^[0,1], with L^2 norm and the subspace set of all polynomials, it is not closed because it's closure is H and not open can be found here Set of all polynomials on [0, 1/2] is not open in C[0, 1/2]. Then I asked myself an example of to think of an open set. But I could lead myself nowhere, as I am not familiar with infinite dimensional vector space. Not closed does not necessarily mean open.

@metamorphy If you take a ball in R^n, and take an open ball around a point. then it contains at n linearly independent vectors, span of those vectors give you whole R^n. We know whole space is an open set but my point was in asking a proper subspace. Or what else you wanted to say? — epsilon_delta, Jan 02 '19 at 05:18
I'm telling you that there's no proper subspace. A nonempty open set contains some ball, the linear span of which is the whole space (this is true for any dimensions, including infinite). — metamorphy, Jan 02 '19 at 05:21
Sorry, please disregard my close vote; I misread the question. — angryavian, Jan 02 '19 at 05:27
@metamorphy I got the answer. I will write it and upload. thanks for your hint. — epsilon_delta, Jan 02 '19 at 07:17
Possible duplicate of An example of non-closed subspace of a Hilbert space? — Lord_Farin, Jan 02 '19 at 13:05
@Lord_Farin, sorry, I didn't understand, what you wanted to say. — epsilon_delta, Jan 02 '19 at 13:10
@BappaGhosh That this question is already answered before on Maths.SE, as you can see by following the link. Making this link formal (by closing as a duplicate) helps to avoid double efforts on the site. — Lord_Farin, Jan 02 '19 at 13:12
@Lord_Farin, I checked before I put the question. My question is, " Is there any open subspace of a Hilbert ." But that question reads non closed subspace of a Hilbert space. Now Non closed subspace doesn't mean open. — epsilon_delta, Jan 02 '19 at 13:21
@BappaGhosh You are right, I am sorry for the inconvenience! Should the question still get closed despite our comments, please ping me so that I can initiate re-opening. — Lord_Farin, Jan 02 '19 at 13:23
Sorry, I didn't get you. Why should the question get closed? — epsilon_delta, Jan 02 '19 at 13:34
@epsilon_delta: Once a close vote is cast on a Question, it goes into a Close Vote review queue. If a reviewer is not sufficiently wary, the distinction between open and non-closed may be overlooked. So it would be an honest mistake if the Question gets closed in this way, and we can vote to reopen it. — hardmath, Jan 02 '19 at 18:06

score 10 · Accepted Answer · edited Jan 02 '19 at 07:02

If $M \leq \mathcal{H}$ a subspace of a Hilbert space (or generally any normed space) is open, then it contains a ball around the origin $0 \in B_r(0) \subset M$, but for every (none-zero) vector $v \in \mathcal{H}$, we have $$ \frac{r}{2\Vert v \Vert} v \in B_r(0) \subset M $$ But M is a linear subspace so $ v \in M $. Thus the only open subspaces of $ \mathcal{H} $ are $ \mathcal{H} $ itself.

score 4 · Answer 2 · answered Jan 02 '19 at 06:15

No.

Let $N$ be a normed space and

$M \subsetneq N \tag 1$

a proper subspace. Then $M$ contains no nonempty open set. For if

$\emptyset \ne U \subset M \tag 2$

were open, with

$M \ni m \in U, \tag 3$

we could find $\rho > 0$ such that the open ball

$B(m, \rho) \subset U; \tag 4$

then picking any

$0 \ne v \in N \setminus M \tag 5$

the vector

$m + \alpha (v - m) \in B(m, \rho) \tag 6$

if $0 \ne \alpha \in \Bbb R$ is sufficiently small, since

$\Vert (m + \alpha (v - m)) - m \Vert = \Vert \alpha (v - m) \Vert = \vert \alpha \vert \Vert v - m \Vert < \rho \tag 7$

for

$\vert \alpha \vert < \dfrac{\rho}{\Vert v - m \Vert}; \tag 8$

but then

$m + \alpha(v - m) \in M, \tag 9$

whence

$\alpha(v - m) = m + \alpha(v - m) - m \in M, \tag{10}$

whence

$v - m \in M, \tag{11}$

whence

$v = v - m + m \in M, \tag{12}$

in contradiction to (5); therefore no $B(m, \rho)$ as in (5) can exist, and $M$ cannot be open, since it contains no open set.

Open linear subspace of a Hilbert space.

2 Answers2