(Sketch): If (f=0), take (y=0). Otherwise, let (M = \ker f), a closed subspace. Since (f \neq 0), (M^\perp \neq 0). Pick (z \in M^\perp) with (f(z)=1). For any (x \in H), write (x = m + \alpha z) with (m \in M), (\alpha = f(x)). Then (f(x) = \alpha f(z) = \alpha). But (\langle x, z \rangle = \langle m, z \rangle + \alpha |z|^2 = \alpha |z|^2). So (\alpha = \frac\langle x, z \rangle^2). Hence (f(x) = \langle x, \fracz \rangle). Set (y = \fracz^2). Uniqueness: If (\langle x, y_1 \rangle = \langle x, y_2 \rangle) for all (x), then (\langle x, y_1-y_2 \rangle =0) for all (x), so (y_1-y_2=0). Also (|f| = |y|) by Schwarz.