Let $f$ be bounder linear functional on Hilbert space $H$ , then dimensional of orthogonal complement of null space is $1$

Question

Let $f$ be bounded linear functional (non trivial) on Hilbert Space $H$ . Let $null(f)$ be null space of $f$ in $H$. Then how to prove that dimension($null(f)^⊥$) is $1$.

Answer 1

Let $x$ be a fixed non-zero vector in $ker(f)^{\perp}$. Any other vector $y$ in $ker(f)^{\perp}$ can be written as $y=z+cx$ where $z=y-cx$ and $c$ is chosen such that $f(z)=f(y)-cf(x)=0$ (or $c=-\frac {f(y)}{f(x)}$). Now $z=y-cx\in ker(f)^{\perp}$ because $x, y \in ker(f)^{\perp}$. But $z \in ker (f)$. Hence $z=0$ and we get $y=cx$. Thus evey vector $y$ in $ker(f)^{\perp}$ is a scalar multiple of $x$.