Let $X$ be a normed linear space and $l:X \to \Bbb R$ be a linear functional. Show that if the Null space of linear functional $l$ is closed then $l$ is bounded.
Having trouble with this one. any help is much appreciated.
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Do you know that a functional is bounded if and only if it is continuous? – diracdeltafunk Apr 16 '23 at 19:12
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1Would this help? https://math.stackexchange.com/a/1255352/1104384 – Bruno B Apr 16 '23 at 19:31
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Yeah, I see it. It is just not true that the null space in my example is ${0}$. Somehow I assumed all $x_i$ to be non-negative. Gonna delete my comment now... – Lukas Apr 16 '23 at 19:42
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Let $Y=Ker(l)$, if $Y=X$, then we are done. Otherwise choose $x_0\notin Y$, we have $X=Y+\left[x_0\right]$ because $\forall x\in X$, we have $l(x-\frac{l(x)}{l(x_0)}x_0)=0$, so $x-\frac{l(x)}{l(x_0)}x_0\in Y$ and $x=(x-\frac{l(x)}{l(x_0)}x_0)+\frac{l(x)}{l(x_0)}x_0$. Therefore we may write $x=y+\lambda x_0$. Since $Y$ is closed, we have $d=\inf_{y\in Y}\lVert x_0-y\lVert>0$, so $\lVert x\lVert=\lVert y+\lambda x_0\lVert=|\lambda|\lVert \frac{y}{\lambda}+x_0\lVert\geq |\lambda d|$. Then $|l(x)|=|l(y+\lambda x_0)|=|\lambda l(x_0)|=|\frac{l(x_0)}{d}\lambda d|\leq |\frac{l(x_0)}{d}|\lVert x \lVert$, so $l$ is continuous with norm $\lVert l\lVert\leq |\frac{l(x_0)}{d}|$, therefore $l$ is bounded.
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