If $\dim (\operatorname{Ker}f)^{\perp} = 1$ then $f$ is continuous for $f$ linear functional on Hilbert space I need to prove the following result:


  If $\dim (\operatorname{Ker}f)^{\perp} = 1$ then $f$ is continuous for $f$ linear
  functional on Hilbert space.


I have the following ingredients in mind:


Use sequential characterization of continuity knowing that since $\operatorname{Ker}(f)$ is closed, projection theorem gives that $x_n = m_n + u_n \to x = m + u$ is equivalent to $m_n \to m \land u_n \to u$. 
Someone suggested me to further prove that $H = \operatorname{Ker}(f) \oplus \operatorname{Ker}(f)^{\perp} = \operatorname{Ker}(f) \oplus\operatorname{Img}(f)$ but I don't see how this is of any help. 


Any ideas?
Suppose $\dim(Ker(f)^{\perp})=1$, where $f$ is a linear functional. Then there exists a unit vector $x$ such that $x\perp Ker(f)$. In particular $f(x)\ne 0$. So,
$$
        f\left(y-\frac{f(y)}{f(x)}x\right) = 0,
$$
which forces the following for all $y$:
$$
             \left\langle y-\frac{f(y)}{f(x)}x,x\right\rangle = 0 \\
           \langle y,x\rangle = \frac{f(y)}{f(x)} \\
        f(y) = \langle y,\overline{f(x)}x\rangle,\;\; y\in H.
$$
So $f$ is a continuous linear functional because it has a Riesz representation.
Let $x$ non zero in the orhogonal of $Ker(f)$, the linear function $\langle x, .\rangle$ is continuous, this implies that its kernel $Ker(f)$ is closed and $f$ is continuous.

Closed kernel implies continuous linear functional : Zorn&#39;s Lemma