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Suppose there exist a subset $M$ of an inner product space $X$, and the orthogonal complement of $M $ is the zero vector. If $X $ is a Hilbert Space then the span of $M $ will be dense in $X $, but this is not always true if $X $ is not a complete inner product space. Are there any examples of an orthonormal set in a non complete inner product space such that the orthorgonal complement of this set consists only of the zero vector, but this set is not a total orthonormal set? Thank you in advance.

Herr Warum
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  • Are you looking for an orthonormal set in an inner product space with orthonormal complement ${0}$ but which does not have a dense linear span? Because I doubt the existence of such a set. I'm not certain though. – SmileyCraft Dec 12 '18 at 18:18
  • Any orthonormal set $E $ in any inner product space $X $ thats non Hilbert, such that the orthogonal complement of $E $, denoted by $E^c $ is the zero vector and there exists no other $x\in X $ that is in $E^c $ and span $E $ is not dense in $X $. Just a thought when reading Functional Analysis by Kreyszig. – Herr Warum Dec 12 '18 at 18:31

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Let $E$ be the space of sequences of real numbers with finite support endowed with the $l^2$ inner product and define the linear form $\phi \colon E \to \mathbb{R}$ by $$\phi(x) = \sum_{n = 0}^\infty \frac{x_n}{n+1}, \quad x \in E.$$ Then $\phi$ is a continuous linear form, so that $F = \ker \phi$ is a closed subspace. Notice that $F \neq E$ (for instance take $x_0 = 1, x_1=0, x_2 = 0, \ldots$). In particular $F$ is not dense in $E$. Finally we can show that $F^\perp = \{0\}$ (see below for a more elegant solution). Indeed let $y \in F^\perp$ and fix $n \geq 0$. Define $x \in E$ by $$\begin{align*}x_n &= n+1, \\ x_{n+1} &= -(n+2),\\ x_k &= 0, \quad k \notin \{n,n+1\}.\end{align*}$$ Then $x \in F$ so that by definition of $y$ we have $0 = \langle x,y \rangle = (n+1)y_n - (n+2)y_{n+1}$, i.e. $$y_{n+1} = \frac{n+1}{n+2} \, y_n.$$ This is true for all $n\geq 0$. Since we assumed that $y$ has finite support, this is only possible if $y \equiv 0$.

As suggested in the comment by Hanno, another way to prove that $F^\perp = \{0\}$ is the following: notice that $\phi(x) = \langle x , a \rangle$ with $a = \left( \frac{1}{n+1} \right)_{n \geq 0} \in l^2- E$. Then in $l^2$, we have $\ker \phi = \{a\}^\perp$ and $(\ker \phi)^\perp = \text{span}\{a\}$, so $F^\perp = \text{span}\{a\} \cap E = \{0\}$.

Michh
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    Let me state two points (1) A bit misleading: You write "Notice that $F\neq E$", but the $x$ given inside the parentheses illustrates $\ker\phi\ne{0}$. (2) Conceptual: The linear form is $\phi(x) = \langle x,a \rangle$ with fixed $a = \left(\frac 1{n+1}\right)_{n=0,1,\dots}\in \ell^2(\mathbb N_0)\backslash E,$. Then in $\ell^2(\mathbb N_0)$ one has $(\ker\phi)^\perp = \langle a\rangle$, and $\langle a\rangle\cap E ={0}$. – Hanno Dec 29 '18 at 09:55
  • @Hanno (1) Yes you are correct about that and I will edit my post. (2) I chose not to talk about $l^2$ because in my eyes it is more confusing than anything else, but again you are correct and it is definitely a more elegant solution. Thanks for the remarks. – Michh Dec 29 '18 at 11:04
  • The title of the question would suggest that the OP is looking for an orthonormal subset with this property, not an entire subspace. Can you find an orthonormal basis for $F$? It exists by Gram-Schmidt but that isn't really explicit. – mechanodroid Dec 29 '18 at 11:42
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    @mechanodroid Reading off from the answer, the set ${e_n-\frac{n+2}{n+1}e_{n+1}\mid n\in\mathbb N_0}$ is a basis for F. Running Gram-Schmidt would yield an orthonormal one, but I do not feel any necessity to carry this through. – Hanno Dec 29 '18 at 12:03
  • @Hanno You're right, I agree. – mechanodroid Dec 29 '18 at 12:29