# Example of complete orthonormal set in an inner product space whose span is not dense

Let $X$ an inner product space and $A$ be an orthonormal set and $\overline{Span(A)}$ = $X$ then $A$ is Complete. But the converse is not true until we consider $X$ as a Hilbert space. I am searching of an example for that.

Here, I call a system/set $Z \subset X$ of vectors complete if $Z^\perp = \{0\}$.

My Attempt:

Lets take the polynomial vector space $P[x]$ over $\mathbb{R}$. Certainly it is not a Hilbert space. I choose the standard basis $\left \{1,x^2,x^3,... \right \}$ of $P[x]$. Using Gram–Schmidt process we can find an orthonormal basis. But i am stuck with the density part.

Please let me know how do i prove it.

Thank You.

• How exactly do you define completeness of an orthonormal set in an inner product space? – PhoemueX Oct 16 '15 at 20:55
• @PhoemueX: Basically i defined completeness of an orthonormal set by the density of the spanning set of that set i.e., if $A$ be an orthonormal set in an inner product space $X$ then $A$ is complete if $\overline{Span(A)}$ = $X$ – S.K Oct 16 '15 at 21:04
• @PhoemueX: Actually i am trying to find an example which shows that completeness of an orthonormal set $A$ does not implies $\overline{Span(A)}$ = $X$ in an inner product space $X$. This is true when space $X$ is Hilbert. – S.K Oct 16 '15 at 21:09
• But if you want to find an example for which $\overline{\rm{Span}(A)} \neq X$, then your definition of completeness must be obviously something different than $\overline{\rm{Span}(A)} =X$. (Otherwise, there can be no such example). – PhoemueX Oct 16 '15 at 21:14
• Yes you are right. I forgot to mention that completeness can also be defined like as: An orthonormal set $A$ is said to be complete if $A^{\perp}$ = $\left \{ 0 \right \}$. Sorry for that. – S.K Oct 16 '15 at 21:28

Here is a very general, abstract construction. Let $X$ be an incomplete seperable inner product space and let $Y$ be its completion. Since $X$ is incomplete, there is some $y \in Y \setminus X$. If we had $y \perp X$, this would imply (by density) that $y \perp Y$ and hence $y = 0\in X$, a contradiction. Hence, $y\not\perp X$. Thus, by renormalizing, there is $x_0 \in X$ with $\langle x_0, y\rangle =1$.

Now, the function $\varphi : Y \to \Bbb{K}, x \mapsto \langle x, y\rangle$ is a bounded functional on $Y$ and thus restricts to a bounded linear functional on $X$. Let $$M := \{x \in X \mid \varphi(x) = 0\} = X \cap \rm{ker}\,\varphi$$ and note that $M$ is closed in $X$.

Furthermore, $M \subset \rm{ker}\,\varphi$ is dense. Indeed, let $z \in \rm{ker}\,\varphi$. Then there is a sequence $(x_n)_n$ in $X$ with $x_n \to z$. Hence, $\varphi(x_n) \to \varphi(z) = 0$. Now let $x_n ' := x_n - \varphi(x_n) x_0$. Then $x_n ' \in M$ and it is easy to see $x_n \to z$.

Now, choose a countable dense set $(m_n)_n$ in $M$ and orthonormalize it using the Gram Schmidt procedure, producing an orthonormal set $(x_n)_n$ in $M$ with $$\overline{\rm{span}(x_n)_n} = \overline{\rm{span}(m_n)_n} = M.$$ Here, we take the closure in $X$. Note that the above is indeed true (we don't get all of $X$), since $M$ is closed in $X$. Thus, $(x_n)_n$ is not an orthonormal basis of $X$.

But $(x_n)_n$ is complete in your sense: Since if $x \in X$ satisfies $x \perp x_n$ for all $n$, then (by density) $x \perp M$. But as saw above, $M$ is dense in $\rm{ker}\,\varphi$, so that $x \perp \rm{ker}\,\varphi$ (where we consider $x$ as an element of the completion $Y$).

But it is easy to see $\rm{ker}\,\varphi = (\rm{span}(y))^\perp$, so that (recall that $Y$ is complete) we get $x \in ((\rm{span}(y))^\perp)^\perp = \rm{span}(y)$. Since $y \in Y \setminus X$ and $x \in X$, this implies $x=0$.

One can now certainly make this construction concrete by choosing e.g. $X = \Bbb{K}[X]$ and $Y = L^2([0,1])$ or $X = \ell_0$ (the finitely supported sequences) and $Y = \ell^2$, but I leave this to you as an exercise.