A counter-example to $KerT$ is closed iff $T$ is continuous when $T$ maps from $\mathbb{R}$ to $\mathbb{R^2}$ with respect to standard norms.

Question

I have proved:

Let $T:(X,||.||) \rightarrow (\mathbb{R},|.|_E)$ be a linear map.

$T$ is continuous iff $\operatorname{Ker}T$ is a closed subset of $X$.

I have noticed that proposition most probably goes wrong when we change $\mathbb{R}$ into any set. So I started to think to find a set for an counter-example which satisfies the following;

Let $T:(\mathbb{R},||.||_E) \rightarrow (\mathbb{R^2},||.||_E)$ be a linear map.

1. $\operatorname{Ker}T$ is closed but $T$ linear function discontinuous

or

2. $T$ linear function is continuous but $\operatorname{Ker}T$ is not closed in $X$.

I find couple of examples but didn't satisfy the followings. My best counter-example try is: $T^*:(\mathbb{R},||.||_E) \rightarrow (\mathbb{R^2},||.||_E)$

$x \mapsto ([x],0)$

Although $\operatorname{Ker}T^*=\{ x \in \mathbb{R}\quad |\quad T^*(x)=(0,0)\}= [0,1)$ is not closed set, the greater integer function is discontinuous at the integer points.

Thanks in advance for any help and guidance.

Answer 1

As mentioned in the comments, linear maps with finite-dimensional domain are always continuous.

You situation (ii) is impossible, since $\ker T=T^{-1}(\{0\})$ is closed if $T$ is continuous.

Situation (i) is definitely possible when $X$ (the domain) is infinite-dimensional, and the codomain is also infinite-dimensional. For instance, one can construct $T$ unbounded but injective, so $\ker T=\{0\}$.

Even for $T:X\to\mathbb R^2$, one can get $T$ unbounded but $\ker T$ closed. For instance, let $f,g:X\to\mathbb R$ be two unbounded functionals such that $\ker f\cap\ker g=\{0\}$. Put $T(x)=(f(x), g(x))$. Then $T$ is unbounded, and $\ker T=\ker f\cap\ker g=\{0\}$.