Let $H$ be a complex inner product space and $\hat H$ its Hilbert space completion. Consider the bounded linear functional $\phi : H → \mathbb{C}, h ↦ 0$. Does Hahn-Banach tell us that its extension $\hat \phi$ to $\hat H$ must also be the zero map?
EDIT: for some more context, I'm studying the GNS representation $(H_\tau, \phi_\tau)$ of some $C^*$-algebra $A$, where $H_\tau$ is the Hilbert space completion of $A/N_\tau$ for $N_\tau := \{a ∈ A \mid \tau(a^*a) = 0\}$, and for each $a ∈ A$, $\phi_\tau(a)$ is the unique extension of the operator $\phi(a) = (b + N_\tau ↦ ab + N_\tau)$ to $H_\tau$. In my case, I have that $\phi(a)$ is the zero map, and I want to conclude $\phi_\tau(a)$ is as well.