Example of Normed Spaces $X$ and $Y$ and a linear operator $T$, so that $\ker (T)$ is not closed. So first, I thought about the zero transformation, however I saw it would not work. After that, I thought about the differentiation transformation, being $X$ and $Y$ the set of all infinitely differentiable functions, however $\ker(T)$ would be closed, since it would be the set of all constant functions, wouldn't it?
Can anyone help or just give a hint of something I am not thinking about? Thanks in advance.
 A: Let $V$ be a normed space. Then if $\tau: V \to \mathbb{C}$ is a linear map, the following two statements are equivalent:

*

*$\tau$ is continuous.

*$\ker(\tau)$ is closed in $V$.

This fact can be found in virtually every basic functional analysis book.

Hence, your example will be found once we construct a linear map $\tau: V \to \Bbb{C}$ such that $\tau$ is not continuous! To do this, we proceed as follows: let $V$ be your favorite  infinite-dimensional normed vector space with Hamel basis $\{e_n:n \in \mathbb{N}\}$. By rescaling the basis, we may assume that $\|e_n\| = 1$ for all $n \in \Bbb{N}$. Define a linear map $\tau:  V \to \mathbb{C}$ by
$\tau (e_n) = n$ for all $n \in \mathbb{N}$. Then
$$\|\tau\|= \sup_{\|v \| \leq 1} |\tau(v)| \geq n$$ for all $n \in \mathbb{N}$, so that $\|\tau\| = \infty.$ Hence, $\tau$ is not continuous and $\ker(\tau)$ is not closed in $V$.
A: Hint:
For a continuous operator $T$ this can never work,
because then the kernel is always closed.
Choose a linear operator $T$ which is not continuous
and which has a non-trivial kernel!
Then the chances are good that the kernel is not closed.
