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Let $X$ be a linear normed space, and $L$ a nontrivial linear functional on $X$. Prove the following are equivalent:

$1) L$ is continuous

$2)$ The null space of $L$ is proper, closed linear subspace of $X$

$3)$ The null space of $L$ is not dense in $X$.

My Work:

I proved that $1)\Rightarrow 2)$ and $ 2)\Rightarrow 3)$, but stuck in proving $3)\Rightarrow 1)$. Since ker $L$ is not dense, there is $x\in X$ which is not a limit point of ker $L$. How does it proceed towards $L$ is continuous? Can anybody please give me a hint?

Extremal
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    What can you say about the set ${ x\in X : \lvert L(x)\rvert < 1}$ when the null space of $L$ is not dense? – Daniel Fischer Mar 26 '15 at 20:30
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    Not totally essential for this problem, and you may have realized this already, but let me point out the closure of a subspace $N$ is a subspace satisfying $N \subseteq \overline N \subseteq X$. If $N$ has codimension-$1$, then the only possibilities are $\overline N = N$ and $\overline N = X$. That is, a hyperplane is either closed or dense; there are no other possibilities. I think the equivalence of 1-3 is less suprising once you notice this. – Mike F Mar 26 '15 at 20:37

1 Answers1

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Since $L$ is linear, it suffices to show continuity at $0$. Assume $L$ is not continuous at $0$, and obtain from linearity that for any $a\in\mathbb{C}$ and any $\delta>0$, there is $x\in B(0,\delta)$ with $|L(x)|=|a|$, where $B(0,\delta)$ is the ball in $X$ with radius $\delta$, centered at $0$. Use linearity again to show that in fact, $x$ can be chosen to satisfy $L(x)=a$. It follows that the null space of $L$ is dense in $X$.

We elaborate the last statement. Let $y\in X$, and let $\delta>0$. Then there is $x\in B(0,\delta)$ with $L(x)=-L(y)$, and so $L(y+x)=0$. Since $y+x\in B(y,\delta)$, and $\delta$ can be as small as we wish, $y$ is a limit point of $\ker L$.

Amitai Yuval
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