If $M^\perp$ consists only of the zero vector, then is $M$ total in $X$?

Question

Let $X$ be an inner product space, and let $M$ be a non-empty subset of $X$. Then $M$ is said to be total in $X$ if the span of $M$ is dense in $X$.

We have the following result:

If $M$ is total in $X$, then $M^\perp$ consists only of the zero vector in $X$.

Does the converse hold too?

I know that the anwer is in the affirmative if $X$ is also a Hilbert space.

What is the situation if $X$ is an inner product space but not a Hilbert space?

By definition, $$ M^\perp \colon= \{ \ x \in X \ \colon \ \langle x, v \rangle = 0 \ \mbox{ for all } \ v \in M \ \}.$$ And, $M^\perp$ is a (vector) subspace of $X$ and is a closed set in the metric space induced by the inner product.

Answer 1

Here is a counterexample. Define the subset $M$ of $l^2$ by $$ M = \{ e_n, n\ge 2\} . $$ Define $X$ to be the linear hull of $M\cup \{x\}$, where $x= (1,1/2,1/3,\dots,1/k,\dots)$. With the $l^2$-inner product, the space $X$ is a pre-Hilbert space.

Then $M^\perp =\{0\}$, $M^{\perp\perp}=X$, $x\not\in \bar M$.

Answer 2

$\textbf{Counterexample}$： Consider the pre-hilbert space $ \ell^2_{f}$ consisting of sequences of finite support in $\ell^2$,e.g.,$$ \ell^2_{f}=\{x=(x_1,x_2,\cdots,)|\text{there exists some } N\in \mathbb{N},\forall n>N,x_n=0\}\subset \ell^2 .$$ Let $$x_0 =(\frac{1}{1},\frac{1}{2},\frac{1}{3},\cdots)\in \ell^2$$ and $$M = \{x \in \ell^2_{f}\mid \langle x_0,x\rangle = 0\}.$$ Then one can check that $M\neq \ell^2_f$ but $ M^\perp=\{0\}$.