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$X$ is a Banach space and $f$ be a linear functional from $X$ to $\mathbb{C}$. If $\operatorname{Ker} f$ is a $G_\delta$ set in $X$, then $f$ is continuous.

I know that if $\operatorname{Ker} f$ is closed, then $f$ is continuous. The argument is based on getting a ball of radius $r$ in $(\operatorname{Ker} f)^c$ and then showing that $|f(x)|\le \frac1r||x||$.

I was trying to extended this idea to the above problem, but failed. Any hint/ suggestions.

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    $\begingroup$ $\DeclareMathOperator{\Ker}{Ker}$In fact, if $f$ is discontinuous then $\Ker f$ is dense in $X$: Let $x_n \in X$ be such that $|f(x_n)|\geq n$, and take an arbitrary $x \in X$. The sequence $\left(x-\frac{f(x)}{f(x_n)}x_n\right)_{n=1}^\infty$ is in $\Ker f$ and converges to $x$. $\endgroup$ – mechanodroid Sep 11 '17 at 21:16
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Suppose $f$ is unbounded. Then its kernel is dense in $X$ (as noted by Mechanodroid). Let $U = X\setminus\operatorname{Ker} f$; by assumption this is a countable union of closed sets $F_n$, $n\in\mathbb N$. Each $F_n$ has empty interior, because $\operatorname{Ker}f $ is dense.

Pick any vector $x_0$ such that $f(x_0)\ne 0$. Note that every point $x\in X$ is either in $U$ or in $U+x_0$. Hence, $$ X = \left(\bigcup_{n=1}^\infty F_n \right) \cup \left(\bigcup_{n=1}^\infty (F_n +x_0)\right) $$ But a complete metric space cannot be written as a countable union of closed sets with empty interior, by Baire category theorem.

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  • $\begingroup$ If $x\in \operatorname{Ker} f$, then $x=x-x_0+x_0$. $f(x-x_0)=-f(x_0) \neq 0$. Hence $x-x_0 \in \cup F_n$ and hence $x \in \cup (F_n+x_0)$. Let me know if I have understood it correctly. $\endgroup$ – Sayan Sep 13 '17 at 15:49
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    $\begingroup$ Yes, that's what I meant. $\endgroup$ – user357151 Sep 13 '17 at 15:54

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