# Is every subspace of an incomplete inner product space closed?

Let ($$V$$,$$\langle \cdot , \cdot \rangle$$) be an infinte-dimensional incomplete inner product space. Is it possible to find in it some subspace that is is not closed ? or is every subspace of this space closed ? consider finite and infinite dimensional subspaces.

• Suppose we take the span $E$ of an infinite orthonormal sequence. (Span = finite linear combinations). Is every subspace of $E$ closed in $E$? Aug 30, 2020 at 23:58
• I think your comment is the answer if I understood correctly. For example consider $C^0([0,2π])$ with $L^2([0,2π])$ inner product, which is an incomplete inner product space. Take $E$ = Span$\{a_0,\cos nx,\sin nx\ |\ n∈\mathbb{N}\}$. The generated space is all partial sums of the trigonometric series. Cauchy sequences of these may converge to a continuous function in $C^0([0,2π])$ but this limit is usually not in the span of $E$. So $E$ is not closed in $C^0([0,2π])$. Aug 31, 2020 at 7:52
• I was interested in the other part of the question. Is it true that every subspace of $E$ is closed in $E$? Aug 31, 2020 at 21:30
• Could you check this ? math.stackexchange.com/questions/3808778/… becuase I asked a similar question and tried to answer it. perhaps for infinite dimensional subspaces in complete or incomplete spaces it is not the case, that the subspace is always closed. I gave an example there Sep 1, 2020 at 9:27

A finite-dimensional subspace $$U$$ of $$V$$ is always closed: $$U$$ is complete with the induced norm (since all norms on $$U$$ are equivalent), so $$\|u_n - x\| \to 0$$ with $$u_n \in U, x \in V$$ implies that $$u_n$$ is a Cauchy sequence and hence converges to some $$u \in U$$. From this it is easy to see that $$x = u \in U$$ (because metric spaces are Hausdorff), i.e. $$U$$ is closed.
Infinite-dimensional subspaces may not be closed: Take for example $$c_{00}(\mathbb{C}) = \{(x_n)_{n \in \mathbb{N}} \in \mathbb{C}^{\mathbb{N}}: \exists\, m \in \mathbb{N} \text{ such that } x_n = 0 \, \forall {n \geq m}\}$$ and consider the inherited inner product of $$\ell^2(\mathbb{C})$$ on $$V := c_{00}(\mathbb{C}) + \mathbb{C} \cdot z$$ with $$z = (z_n)_{n \in \mathbb{N}} \in \ell^2(\mathbb{C}), \, z_n := \frac{1}{n}.$$
Then $$V$$ is not complete since it is not closed as a subset of $$\ell^2(\mathbb{C})$$. Furthermore $$c_{00}(\mathbb{C})$$ is dense in $$\ell^2(\mathbb{C})$$ and with this it is easy to see that $$c_{00}(\mathbb{C})$$ also is dense in $$V$$ with respect to the induced inner product. In particular, $$c_{00}(\mathbb{C})$$ is not closed in $$V$$ since $$z \notin c_{00}(\mathbb{C})$$.
• thanks for the answer. I just couldn't understand the argument about finite-dimensional subspaces, you say they are closed in the superset $V$, and another question if you don't mind, are these finite-dimensional subspaces in general complete if $V$ is not? Aug 31, 2020 at 12:17
• Yes, finite-dimensional normed spaces are always complete. From this it follows that they are closed in any space that they are included in, so in this example they are all closed in $V$. Aug 31, 2020 at 12:48