# Meaning of $\ker(x)$ when $x$ is an element of a Hilbert space

The follow question arose from the paper: The Hundal Example Revisited

We consider a separable Hilbert Space $X$ with countable ortho-normal basis $\{e_n\}_{n=1}^\infty$. The following is an excerpt from page 4 of [1]:

... the hyperplane $\ker(e_1)$ and...

My question is regarding the notation $\ker(e_1)$. Since it is described as a hyperplane I can only assume that it must be defined, for $a\in X$, as followed: $$\ker(a) := \{x\in X: \langle a,x \rangle = 0 \} .$$ I am familiar with using $\ker(T)$ as the kernel of an operator $T$, but have not seen it used for elements of the space. I guess we could view $\operatorname{ker}(a)$ as an abbreviation for $\ker(\langle a,\cdot\rangle)$, viewing $\langle a,\cdot\rangle$ as a mapping from $X$ to $\mathbb F$ ($X$'s scalar field).

Is my definition and motivation correct? If not, could someone provide a correct version?

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You are completely correct. It is $(\mathbb{F}e_1)^\perp$, or those vectors whose first coordinate (w.r.t. to given on-basis) vanishes. –  wildildildlife Jan 9 '12 at 23:53
You are correct. The inner product gives us a correspondence between elements of $X$ and linear functionals on $X$. It is the kernel of that linear functional that you want to use here. –  GEdgar Jan 10 '12 at 0:27