Distance to the kernel of a functional in Hilbert space

Question

I need to prove that in Hilbert space $H$ the distance between point $z$ and the kernel $L$ of a linear functional $f$ is $d(z,L)=\frac{|f(z)|}{\|f\|}$. I know a rather sophisticated proof for Banach space but I was told that for Hilbert space the proof is much shorter. However, I cannot find it and will be thankful for any idea.

Answer 1

Let $\pi : H\rightarrow L$ be the orthogonal projection onto L. (this exists because H is a hilbert space), from a basic thm codim(L) = 1 and therefore for any $z\not\in L$: $$L^\perp=span<z-\pi(z)>$$

Then because $f$ achieves its maximum on $L^\perp$ (this should be easy, ask otherwise) it follows that $|f(z-\pi(z))|=||f||\cdot ||z-\pi (z)||$.

finally I think you know that $d(z,L)=||z-\pi (z)||$ if you do you should be able to calculate: $$|f(z)|=|f(z-\pi(z))|=||f||\cdot ||z-\pi (z)||=||f||\cdot d(z,L)$$