# On the dimension of the zero set of a bounded linear functional on a Hilbert space.

Let $$H$$ be a complex Hilbert space and let $$f\colon H \to \mathbb{C}$$ be a linear bounded functional. Since $$\ker f$$ is a closed vector subspace of $$H$$ we have that $$H=\ker f\oplus \ (\ker f)^\perp.$$ We suppose that $$f\ne 0$$, then $$\ker f \subset H$$ and then for the above relation we deduce that $$\{0\}\subset (\ker f )^\perp$$.

Can we say anything about $$(\ker f)^\perp$$ dimension?

I observe that $$(\ker f)^\perp=\{x\in H\;|\; f(x)\ne0\}=:(Z(f))^c$$

Can we what conlude anything on the dimension of the zero set of a bounded linear functional?

• That final line of equations is not what $(\text{ker} \, f)^\perp$ is - that’s the set-theoretic complement of $\text{ker} \, f$, not the orthogonal complement of $\text{ker} \, f$. (In particular, you have already observed that $0 \in (\text{ker} \, f)^\perp$, but clearly $f(0) = 0$.) Commented 2 days ago
• In any case, the dimension is just $1$. The first isomorphism theorem implies$$(\text{ker} \, f)^\perp \simeq H/(\text{ker} \, f) \simeq f(H) = \mathbb{C}$$ Commented 2 days ago
• @DavidGaoThanks!But there are still some obscure points. Why $f(H)=\mathbb{C}$, and why $(\ker f )^\perp \simeq H/\ker f?$ Commented yesterday
• For the first one, hint: $f$ is a linear map, so $f(H)$ is a subspace of $\mathbb{C}$. What are subspaces of $\mathbb{C}$? Since $f \neq 0$, which choice is the only possible one? Commented yesterday
• For the second one, just prove that, if $H = K \oplus L$, then the map $\pi: K \to H/L$, $\pi(k) = k + L$ is a linear isomorphism. Commented yesterday