The null subspace of $f, \Phi_0(f)$ is the set of all $|w\rangle$ such that $f(|w\rangle)=|0\rangle$.
Let $|f\rangle = \langle\tilde f; u\rangle^* \frac{|u\rangle}{\sqrt{\langle u\rangle}}$ then $|f\rangle$ is the uniqure vector corresponding to $\tilde f$
Proof:
$\langle f|u\rangle = \langle\tilde f; u\rangle$ so it works for $|u\rangle$f
For any other vector $|v\rangle$ consider $|w\rangle = |v\rangle - \langle\tilde f; v\rangle/\langle\tilde f;u\rangle |u\rangle$
so, $\langle\tilde f;w\rangle = \langle\tilde f;v\rangle - \langle\tilde f;v\rangle\langle\tilde f;u\rangle/\langle\tilde f;u\rangle = 0$ Hence, $|w\rangle \in \Phi_{0}(f)$ but since $|u\rangle \perp \Phi_0, \langle u|w\rangle = 0$ $\because \langle f|w\rangle =0$
So, any $|u\rangle\perp\Phi_0$ gives $|f\rangle$, but is it unique? i.e, if there are two $|u_i\rangle$ st $|f_1\rangle = \langle \tilde f;u_1\rangle |u_1\rangle/\langle u_i\rangle$ both obey $\langle\tilde f; v\rangle = \langle f_1|v\rangle = \langle f_2|v\rangle$ But then, $\langle f_1-f_2|v\rangle = 0$, and since $|v\rangle$ may not be zero, $f_1\ne f_2$