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Let $(X, \langle\cdot, \cdot \rangle)$ be a Hilbert space with inner product $\langle\cdot, \cdot \rangle$ and induced norm $\rVert \cdot \lVert$.
We know that if $V\subseteq X$ is a closed subspace, then $(V^\perp )^\perp = V$, where $V^\perp$ denotes the orthogonal complement with respect to the inner product.

Now let $X:= L^2([0,1], \mathcal L^1, \lambda^1)$ be the set of Lebesgue square integrable functions on $[0,1]$.

Define $V:= \{\varphi \in X: \varphi(0) = 0 \}$. Clearly, $V$ is a subspace of $X$.

Now I showed that $V \neq(V^\perp )^\perp$, i.e. we have a strict inclusion, by explicitly constructing a function that is in $(V^\perp )^\perp$, but not in $V$.

To understand this better I tried to show that $V$ is not closed with respect to the $L^2$-norm. So I tried to construct a sequence $(\varphi_n)_{n\in \mathbb N}$ such that $\varphi_n(0) = 0$ for all $n\in \mathbb N$ but $\varphi_n \to \varphi$ with $\varphi(0) \neq 0$. However this was kind of tedious dealing with the square in the integrand. Is there an easy way to construct a sequence or some other, better/cleaner option to disprove closedness?

Staki42
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    Usually $L^2$ is defined as equivalence classes of functions that differ on a set of measure zero. Hence it doesn't make sense to speak of the value of an element of an element of $L^2$ at a point. – ziggurism Sep 19 '17 at 05:20
  • But if you consider some other space where it does make sense to talk about functions' values at points, i would expect the set of functions vanishing at a point to be a closed set. – ziggurism Sep 19 '17 at 05:21
  • For an example of a non-closed subspace, we could take the span of a basis, like the trig functions. – ziggurism Sep 19 '17 at 05:26
  • @ziggurism ${\varphi \in C[0,1] : \varphi(0) = 0}$ is not closed as a subset of $(C[0,1], |\cdot|_2)$ as an example of Ice sea shows. – mechanodroid Sep 19 '17 at 08:22
  • @ziggurism you are correct, I actually meant the subspace of continuous functions where it does make sense to talk about values at a point. Thanks for clarifying! – Staki42 Sep 19 '17 at 16:31
  • @lappen68 continuous functions are not complete. this is not a Hilbert space. – ziggurism Sep 19 '17 at 16:32
  • I see. My question was just intended to find a counterexample that ${\varphi \in C[0,1]: \varphi(0) = 0}$ is not closed in $(C[0,1], \lVert \cdot \rVert_2)$ which Ice sea constructed and therefore answered my question. Good thing that you pointed out my mistakes though. – Staki42 Sep 19 '17 at 16:37
  • @ziggurism So one more question: Say we have a inner product space $X$ which is not complete and a closed subspace $V \subseteq X$. Is the completeness condition necessary, i.e. can it happen that $(V^\perp)^\perp \neq V$? – Staki42 Sep 19 '17 at 16:42
  • @lappen68 I continue to believe that the set ${\phi\colon \phi(0)=0}$ actually is closed, in any reasonable space where this set can be defined. Ice sea's example sequence converges pointwise to a function with $f(0) = 0$. It doesn't show that the set is not closed. – ziggurism Sep 19 '17 at 16:46

4 Answers4

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For any subset $S$ of a inner product space, $S^\perp$ is closed(use continuity of the inner product). So $V^{\perp\perp}$ never equals to $V$ when $V$ is not closed.

C. Ding
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  • That was already a known fact for me, I just wanted to construct an explicit counter example to understand it better. – Staki42 Sep 19 '17 at 16:32
  • So, construct an explicit example of a non-closed set $S$. Say, the complement of the origin. Then compute $S^{\perp \perp}$. – GEdgar Sep 19 '17 at 18:14
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To get a complete inner product space from Lebesgue integrals, you must consider equivalence classes of functions defined up to sets of measure zero. Consequently predicates like $\phi(0)=0$ do not make sense for elements of $L^2([0,1])$.

We could instead ask the question about a space where evaluation does make sense, like the continuous functions $C([0,1])$ with uniform convergence or the space of pointwise convergence $\mathbb{R}^{[0,1]}$. Note that the space of continuous functions is not complete in the $L^2$ norm, so this is no longer a Hilbert space.

In both cases, the evaluation function (a function which takes a function and returns its value at a fixed input) $\text{ev}_0$ is continuous, hence the set $V=\{\phi\colon \phi(0)=0\}$ is the preimage of a point under a continuous function $V=\text{ev}_0^{-1}(0)$, hence it is closed. You cannot show this set is not closed, because it is in fact closed.

The example sequence given by Ice sea does not show that the set is not closed in $\mathbb{R}^{[0,1]}$, because it converges pointwise to a function with $\phi(0)=0$. It doesn't show that the set is not closed in $C([0,1])$, because it does not converge uniformly. And it does not show that the set is not closed in $L^2([0,1])$ because the set is not defined there.

For an example of a set that is not closed, just think of things which include only finite combinations, but not limits. For example, the space of polynomial functions $P([0,1])$ makes sense as a subset in any of these three function spaces. $e^x$ is the limit of a sequence of polynomials. Hence $e^x$ is in $(P([0,1])^\perp)^\perp$ but not $P([0,1])$.

Or consider the function space $L^2([-1,1])$. This space can be split into even functions $E=\{\phi\colon \phi(x) = \phi(-x)\}$ and odd functions $O=\{\phi\colon \phi(x) = -\phi(-x)\}$. The both subspaces are closed. A basis for the even functions is $\cos(n\pi x)$. The function $\left|1-x\right|$ is even. It's perpendicular to all odd functions. It's in $(\langle\cos(n\pi x)\rangle^\perp)^\perp,$ but not $\langle\cos(n\pi x)\rangle,$ since it's not a finite linear combination of cosine functions.

ziggurism
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  • I can make sense of what you are saying, however, here are two remarks/questions: 1) Why do we need uniform convergence? He shows that $f_n$ converges towards a function that is in $C[0,1]$ and $f(0) \neq 0$, how is that not sufficient to show that the set is not closed? 2) Here: https://math.stackexchange.com/questions/647817/orthogonal-complement-examples/647858#647858 Daniel Fischer gives an example of a non complete inner product space where $V$ closed but $V \neq (V^\perp)^\perp$ – Staki42 Sep 19 '17 at 17:30
  • @lappen68 You don't have to use uniform convergence. But you do have to choose some topology. I considered $L^2$, pointwise, and uniform. In Ice Sea's example, we have $f_n(0) = (1-0)^n-1$. The limit of this sequence is $0$, not $-1$. Ice sea's example is not showing what you want. – ziggurism Sep 19 '17 at 17:42
  • @lappen68 Looking at Daniel Fischer's example, I see that I must be mistaken about $(V^\perp)^\perp = \bar V$ in any inner product space. I apologize. Let me edit my answer. – ziggurism Sep 19 '17 at 17:51
  • You know the equivalent classes. But you didn't know what is dominated convergence theorem. – Ice sea Sep 19 '17 at 18:13
  • Then one of you has a contradiction in his answer. – Staki42 Sep 19 '17 at 18:21
  • @lappen68 what contradiction? – ziggurism Sep 19 '17 at 18:21
  • Ice sea still claims that the space is not closed (he edited his answer). – Staki42 Sep 19 '17 at 18:22
  • @lappen68 i guess icesea is talking about the space of functions with $L^2$ topology, but without identifying functions equal almost everywhere. This space is not Hausdorff and somewhat weird. It is not a Hilbert space. It is possible that the evaluation map is not continuous on this space, which would resolve the contradiction. – ziggurism Sep 19 '17 at 18:33
  • I see.I will think about it a little more. – Staki42 Sep 19 '17 at 18:36
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First, if the element in your space is considered to be an "equivalent classes", then $V=L^2[0,1]$. Obviously, it is closed. To make you problem well-defined, we consider $L^2[0,1]$ to be space of functions instead of equivalent classes.

It is easy to disprove the closendess in this case.
$f_n(x)=(1-x)^n-1$ for all $x\in[0,1]$. Then $f_n(x)\to f(x)=-1$ in $L^2[0,1]$ as $n\to\infty$. In addition, $f_n(0)=0$. To see the convergence, we only need the dominated convergence theorem. First $f_n(x)\to f(x)$ a.e. and they are are both bounded from above. So $f_n\to f$ in $L^2[0,1]$. i.e., $$ \int |f_n(x)-f(x)|^2 dx \to 0 $$

Ice sea
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Take the “prototype” of all Hilbert non finitely generated but separable space, namely $l^2$, the space of (complex) sequences $(a_n)$ such that $$ \sum_{n\ge0}|a_n|^2 $$ converges (for a real Hilbert space just use real sequences). The Hilbert form is $$ \langle (a_n),(b_n)\rangle=\sum_{n\ge0}\overline{a_n}b_n $$ (switch the conjugation if you use the opposite convention about linearity).

The subspace we want to consider is $V$ consisting of the “eventually null” sequences, that is all $(a_n)$ such that there exists $N$ with $a_n=0$ for every $n\ge N$.

It's easy to prove that $V$ is dense in $l^2$. Thus $V^{\perp\perp}=l^2$. Indeed, given $s=(a_n)\in l^2$, define $$ s^{(m)}(n)=\begin{cases} a_n & n\le m\\ 0 & n>m \end{cases} $$ Then $s^{(m)}\in V$ and $\lim_{m\to\infty}s^{(m)}=s$ (check it), so the statement about density follows. Since $V$ is dense, $V^{\perp}=\{0\}$.

egreg
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