In an infinite dimensional real inner-product space , can any non-null orthogonal set of vectors in the space can be extended to an orthogonal basis?

Question

Let $V$ be an infinite dimensional real inner-product space , then is it true that any non-null orthogonal set of vectors in the space can be extended to an orthogonal basis ? Or at least is it true that every infinite dimensional real inner-product space has an orthonormal basis ?

Answer 1

It's curious that there were no real answers after you clarified that you were talking about orthonormal bases. (You should really refer to orthonormal bases instead of "orthogonal" bases, if only for clarity.)

One doesn't usually talk about orthonormal bases for an inner-product space except when the space is complete (that is, every Cauchy sequence converges, so the space is a Hilbert space).

If $H$ is a Hilbert space and $S$ is an orthonormal subset of $H$ then Zorn's lemma shows that $S$ can be extended to a maximal orthonormal set $B$. So $B$ is an orthonormal basis.

How you prove that a maximal orthonormal set in a Hilbert space is an orthonormal basis depends of which of various equivalent properties you're taking as the definition of "orthonormal basis". Let's say an onb is an orthonormal set with dense span. If $V$ is the span of a maximal orthonormal set $B$ and $V$ is not dense then the orthogonal complement $V^\perp$ is non-zero, contradicting the maximality of $B$.

That works in any finite dimensional inner product space, since such a space must be complete.

What about the question you actually asked, regarding an orthonormal subset of an infinte-dimensional inner product space? First we should note that without completeness the various conditions sometimes taken as the definition of onb are no longer equivalent; we assume the definition is as above, seems like the natural choice. In this case it's not true that every orthonomal set can be extended to an orthonomal basis.

The example is going to be a (non-closed) subspace of $\ell^2$. We say $x\in\ell^2$ if $x=(x_j)_{j=1}^\infty$ with $\sum|x_j|^2<\infty$; the standard inner product is $\langle x,y\rangle=\sum x_j\overline y_j$.

Let $v$ be the sequence $$v=(1/j)_{j=1}^\infty.$$For each $n$ let $e_n$ be the sequence consisting of all zeroes except for one one, in the $n$-th place. Let $V$ be the (finite-linear-combination) span of $$S=\{v\}\cup\{e_2,e_3,\dots\}$$(note that $e_1\notin S$). And let $$B=\{e_2,e_3,\dots\}.$$

Then $B$ is an orthonormal set that cannot be extended to an orthonormal basis.

Note first that if $p$ is a linear combination of the elements of $B$ then it's clear that $$||v-p||\ge1.$$So the span of $B$ is not dense; $B$ itself is not an orthonormal basis.

So it's enough to show that $B$ is a maximal orthonormal set. We need to show that if $x\in V$ and $x\perp B$ then $x=0$. This is clear: The fact that $x\perp B$ shows that $x_j=0$ for $j\ge 2$, so $x=(x_1,0,0,,\dots)$. Now since $x\in V$ we have $$x=\alpha v+\sum_{j=2}^Nc_je_j=x_1v+\sum_{j=2}^Nc_je_j;$$hence $$x_1/(N+1)=x_1v_{N+1}=x_{N+1}=0,$$so $x_1=0$, hence $x=0$.