Why is this unbounded linear functional also has a closed kernel? 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?

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. 
