I have an infinite dimensional Hilbert space $H$ and I can assume there is an orthonormal basis $S$ that is contained in a Hamel basis $F$, which is properly contained in the Hilbert space.
Given $u \in F$, I have the linear functional $\psi_u: H \rightarrow \mathbb{C}$ defined as follows. For any $v \in H$, write $v = au + \sum_{i=1}^n c_iw_i$ where $a, c_i \in \mathbb{C}$ and $w_i \in F \setminus \{u\}$. Then we define $\psi_u(v) = a$. The claim is that if $u \notin S$, then $\psi_u$ is not continuous.
I have already shown that $\psi_u$ is linear, but I'm struggling to show it is not continuous if $u \notin S$. I have been trying to show that $\psi_u$ is not bounded in the operator norm, but I haven't been able to find a sequence of norm $1$ vectors whose norms are unbounded.
I also cannot figure out $u$ not being in $S$ plays a role. Is it because if $u \notin S$, then it's possible to construct norm $1$ vectors such that $a$ is unbounded and we can choose $c_j$ and $w_j$ so that the vector has norm $1$? Expanding the norm of a vector using the inner product shows this could be true, but then I don't know to choose $c_j$ and $w_j$.
Any guiding hints would be appreciated.
remark that the kernel of the linear form $\psi_{u}$ is the span $F-\{u\}$ which contains $S$, So the kernel of $\psi_{u}$ is non-closed (since it is not identically zero) and dense,So $\psi_{u}$ cannot be continuous because the kernel of a continuous linear form must be closed as it is the inverse image of (the closed subset) $\{0\}$.In fact the closedness of the kernel characterize the continuity of linear maps with finite dimensional range.