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On the space $l_0$ of sequences with finite non-zeros, define $f:l_0\to l_0$ to be: $$f(x)_k=kx_k,$$ then $f$ is famously unbounded. According to my textbook, that means $f$ must have a non-closed kernel. But in this particular case, isn't it that the kernel of $f$ is $\{0\}$, which is clearly closed?

xzhu
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    The kernel is necessarily non-closed only for a linear functional. A discontinuous linear operator can have a closed kernel. – Prahlad Vaidyanathan Dec 11 '13 at 17:50
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    A little bit more general: the implication "kernel closed $\Rightarrow$ continuity" holds true only if the target space is finite dimensional (see, e.g. http://math.stackexchange.com/questions/135791/the-kernel-of-a-continuous-linear-operator-is-a-closed-subspace). – Romeo Dec 11 '13 at 17:52
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    @Romeo: Yeah, your right, I forgot it has to be finite dimensional. Thanks. – xzhu Dec 11 '13 at 17:55
  • You are welcome. – Romeo Dec 11 '13 at 18:01
  • It should be "you're", sorry. – xzhu Dec 11 '13 at 18:04

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The answer was given in comments: because it's not a linear functional, but a linear operator with infinite-dimensional range. An important class of unbounded operators is closed unbounded operators; they have closed graph, and consequently, closed kernel. Thus, the given example is far from being exceptional.

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