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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?

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    $\begingroup$ 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 $f(0) = 0$.) $\endgroup$
    – David Gao
    Commented 2 days ago
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    $\begingroup$ In any case, the dimension is just $1$. The first isomorphism theorem implies$$(\text{ker} \, f)^\perp \simeq H/(\text{ker} \, f) \simeq f(H) = \mathbb{C}$$ $\endgroup$
    – David Gao
    Commented 2 days ago
  • $\begingroup$ @DavidGaoThanks!But there are still some obscure points. Why $f(H)=\mathbb{C}$, and why $(\ker f )^\perp \simeq H/\ker f?$ $\endgroup$
    – Jack J.
    Commented yesterday
  • $\begingroup$ 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? $\endgroup$
    – David Gao
    Commented yesterday
  • $\begingroup$ 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. $\endgroup$
    – David Gao
    Commented yesterday

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