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Let $f$ be a linear functional on a normed linear space $X$. Prove that $\ker f$ is either dense or closed in $X$.

Two possibilities can occur, i.e either $f$ is bounded or unbounded. If it is bounded then $f$ is continuous and hence $\ker f=f^{-1}(0)$ which is closed

How to show when $f$ is unbounded.

user642796
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2 Answers2

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Suppose $f$ is unbounded. Then there exists a sequence $\{x_n\}$ with $x_n\to0$ and $f(x_n)=1$ for all $n$.

Now choose any $y\in X$. Then $y-f(y)x_n$ is in $\ker f$, since $$f(y-f(y)x_n)=f(y)-f(y)=0.$$ Thus $y=\lim_n y-f(y)x_n$, where the elements in the sequence $\{y-f(y)x_n\}$ are in $\ker f$. So $\ker f$ is dense in $X$.

Martin Argerami
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Let $f$ be a non-trivial linear functional on $X$, and let $\mathcal{N}(f)$ be the null space of $f$. There exists $y \in X$ such that $f(y)=1$ because $f$ is non-trivial. Therefore, $$ x = f(x)y + (x-f(x)y) \in [\{y\}]+\mathcal{N}(f), \;\;\;(\dagger) $$ where $[\{y\}]$ is the one-dimensional linear space spanned by $y$. Hence $X/\mathcal{N}(f)$ is one-dimensional, and the quotient map is $$ x \mapsto f(x)y + \mathcal{N}(f). $$ If $\mathcal{N}(f)^{c}$ denotes the norm closure of $\mathcal{N}(f)$, then it follows that either $\mathcal{N}(f)^{c}=\mathcal{N}(f)$ or $\mathcal{N}(f)^{c}=X$. If $\mathcal{N}(f)^{c}=\mathcal{N}(f)$, then $X/\mathcal{N}(f)$ is a normed space, and the quotient map from $X$ to $X/\mathcal{N}(f)$ is continuous in norm, and this map agrees with $f$ because $x\mapsto f(x)y$ under this quotient map, because of $(\dagger)$. So $f$ is continuous if $\mathcal{N}(f)$ is closed. Conversely, if $f$ is continuous, then $\mathcal{N}(f)=f^{-1}\{0\}$ is closed. If $f$ is not continuous then $\mathcal{N}(f)^{c}=X$, which means that $\mathcal{N}(f)$ is a dense subspace of $f$ in $X$.

Disintegrating By Parts
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  • When you take the norm closure of the kernel $\mathcal{N}(f)$, how can you assert it is either closed or dense, but not "nowhere dense and not closed"? It seems, at that point, you only know the kernel is a proper subspace, and nothing more. – user760 Sep 14 '22 at 08:31
  • @user760 : $\mathcal{N}(f)$ has co-dimension $1$ in $X$. If you add any element to $\mathcal{N}(f)$ that was not already in $\mathcal{N}(f)$, then you get $X$. So, the closure of $\mathcal{N}(f)$ is either $\mathcal{N}(f)$ or it is $X$. There is no subspace strictly between $\mathcal{N}(f)$ and $X$. Is that what you were asking? – Disintegrating By Parts Sep 14 '22 at 17:49
  • Thanks! That makes sense! – user760 Sep 14 '22 at 18:15
  • @user760 : You are welcome! – Disintegrating By Parts Sep 14 '22 at 18:27
  • Then when the functional is discontinuous, the kernel is also codimension-1. And that extra dimension can be approximated by elements in the kernel. Also all the elements not in the kernel lie in this extra dimension. That means the kernel kind of "wraps up" around this extra dimension from all directions, but not intersecting with it. Cool! – user760 Sep 14 '22 at 18:37