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** problem.** Let $X$ be an infinite-dimensional normed vector space and let $φ$ be a nonzero linear functional defined in $\,X.\,$ Then the following are equivalent. (i) $φ$ is bounded. (ii) The kernel of $φ$ is a closed linear subspace of X. (iii) The kernel of $φ$ is not dense in $X$.

My attempt.:(i)$\Rightarrow$(ii) because $φ$ is bounded,then it is continuous so The kernel of $φ$ is a closed linear subspace of $X$.

(ii)$\Rightarrow$(iii): because $φ$ is nonzero amd from (iii), (iii) is ture.

(iii)$\Rightarrow$(i): I don't know how to do it, i try to assmue $φ$ is not bounded,then we know $φ$ is not continuous at every point of $X$, i want use it to prove the kernel is dense in $X$,but I failed. I also try to use riesz lemma because from (iii) the closure of kernel is proper closed subspace of $X$,I still failed.

José Carlos Santos
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user158796
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1 Answers1

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(iii) im plies that the kernel is closed as shown here: $\ker f$ is either dense or closed when $f$ is a linear functional on a normed linear space

Now if $f$ is not bounded there exsits $(x_n)$ such that $|f(x_n)| >n\|x_n\|$ . Fix any $x_0$ with $f(x_0) \neq 0$ and consider $\frac {x_n}{f(x_n)} -\frac {x_0} {f(x_0)}$. This sequence lies in the kernel and its limit is $-\frac {x_0} {f(x_0)}$. The limit must be in the kernel but this is a contradiction to the choice of $x_0$.

Kavi Rama Murthy
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