Let $V$ be a non-complete inner product space, and let $x_n$ be a weakly Cauchy sequence, i.e suppose $\langle x_n, y\rangle$ converges for every $y \in V$.
Is it true $x_n$ is bounded?
I know this is true when $V$ is complete (Hilbert), but this uses Bair theorem. So, I guess there should be a counter-example when completeness is not assumed.
Let $c_c$ be the space of finite sequences equipped with the $\ell^2$ inner product. Denote by $e_n$ the unit sequences. Then, $\langle n \, e_n, y\rangle = n \, y_n \to 0 = \langle 0, y\rangle$. However, $n \, e_n$ is unbounded.
However, note that $e_n$ does not converge weakly in the sense that $f(e_n) \to 0$ for all $f$ in the dual space.
In fact, $f(x_n) \to f(x)$ for all $f$ in the dual space implies that $x_n$ is bounded (by the uniform boundedness principle). This does not need completeness, since the dual space is automatically complete.