Let $(X, \langle \cdot , \cdot \rangle)$ be an inner product space. For any orthonormal system $(e_i)_{i \in I}$ in $X$, consider the following two statements:
(i) $\text{span}\{e_i:i \in I\}$ is dense in $X$;
(ii) If $\langle x,e_i \rangle=0$ for some $x \in X$ and all $i \in I$ then $x=0$.
Turn (i) $\star$ (ii) into a true statement by replacing $\star$ with either $\Rightarrow$, $\Leftarrow$, or $\iff$. If your choice is $\Rightarrow$ (or $\Leftarrow$), give an example where $\Leftarrow$ (or $\Rightarrow$) fails.
Hint: Consider the span of $(1/j)$ and $e_n$, $n\geq 2$ in $l^2$.
(i) $\Rightarrow$ (ii): Let $x \in X$ be such that $\langle x,e_i \rangle=0$ for all $i \in I$. By the (conjugate) linearity of the inner product, $\langle x,a \rangle=0$ for all $a \in \text{span}\{e_i:i \in I\}$. By the continuity of the inner product, $\langle x,a \rangle=0$ for all $a \in \overline{\text{span}\{e_i:i \in I\}}=X$, which implies $x=0$.
I am having trouble with the reverse implication. I think it is false, but have not found a counterexample. I know these statements are equivalent in a Hilbert space, but I think completeness is essential in that case.
I don't think one can make a counterexample from the hint since $l^2$ is Hilbert space, so I don't understand the purpose of the hint.
Best Answer
Note that the hint said that you should consider the span of the standard basis vectors $e_n$ for $n \geq 2$ and of the vector $y = (1/n)_{n \in \Bbb{N}}$ in $\ell^2 (\Bbb{N})$.
Indeed, define $X := \mathrm{span} \big(\{e_n \colon n \in \Bbb{N} \} \cup \{y\}\big)$, and note that I really mean the linear algebra span, not the closed linear span.
I claim that if $x \in X$ satisfies $\langle x, e_n \rangle = 0$ for all $n \in \Bbb{N}_{\geq 2}$, then $x = 0$. Nevertheless, it is not true that the family $(e_n)_{n \geq 2}$ spans a dense subspace of $X$ (why?!).
To prove the claim, let $x \in X$, say $x = \alpha y + \sum_{i=2}^N c_i e_i$, where $\alpha \in \Bbb{K}$ and where some of the $c_i \in \Bbb{K}$ (either complex or real numbers, take your pick) might be zero. Assume that $\langle x, e_i\rangle = 0$ for $i \geq 2$. In particular, this means $$ 0 = \langle x, e_{N+1} \rangle = \alpha \langle y, e_{N+1} \rangle = \alpha / (N+1), $$ and hence $\alpha = 0$. Therefore, $x = \sum_{i=2}^N c_i e_i$ and $\langle x, e_i\rangle = 0$ for all $i \geq 2$, from which we easily get $x = 0$.