Injectivity of a continuous extension of a densely defined function

Question

Suppose a linear function $T$ is defined on a dense subset of an algebra $\mathcal{A}$ such that $ker(T)=\{0\}$. Is it true that the continuous extension of $T$ to the whole algebra $\mathcal{A}$ is injective? I guess this should be true but i don't konw how to prove it.

Answer 1

I believe what you are asking is:

Let $\mathcal{A}$ be an algebra and $B\subseteq \mathcal{A}$ a dense subset of it. If $T: B\subseteq \mathcal{A} \rightarrow \mathcal{A}$ is a function such that $\ker T =\{0\}$ and which can be extended continuously and linearly on $\mathcal{A}$, is it true that this extension is also injective?

What I changed is that I moved the word "linear" from $T$ to its extension, the reason being that $T$ is defined on a subset $B$ and not a subspace. On the other hand, its extension acts on $\mathcal{A}$, so linearity makes sense here.

I also assumed that $T$ has a continuous extension to begin with, otherwise you may get counterexamples of operators which do not posses one. For example, take $B$ to be a dense Hamel basis of $\mathcal{A}$ and fix a $0\neq b_0\in \mathcal{A}$. Define $T: B\rightarrow \mathcal{A}$ by $T(b)=b_0$, for every $b\in B$. Then $T(b)\neq 0$, for every $b\in B$, but the unique linear extension of $T$ is not continuous: By the denseness of $B$, there exists a $(b_{i_n})_{n\in \mathbb{N}}$ with $b_{i_n}\rightarrow 2b_0$, which, if $T$ were continuous, would imply that $Tb_{i_n}=b_0\rightarrow 2Tb_0$. Similarly, there exists a $(b_{j_n})_{n\in \mathbb{N}}$ with $b_{j_n}\rightarrow b_0$, which would imply that $Tb_{j_n}=b_0\rightarrow Tb_0$, so $b_0=2b_0$, a contradiction.

But even under the assumption that $T$ has a continuous extension, the answer is still negative: Let $\mathcal{A}$ be a $C^*$ algebra with a Schauder basis $(e_n)_{n\in\mathbb{N}}$, for instance $\mathcal{A}=C[0,1]$, and consider the operator $T: \mathcal{A}\rightarrow \mathcal{A}$ for which $Te_1=0$ and $Te_n=e_n$, for $n\geq 2$.

Clearly $T$ is bounded, but it is not an injection since $\ker T=\langle e_1 \rangle$. Now let $B\subseteq \mathcal{A}$ be a dense Hamel basis which contains all the $e_n$'s. The set $B\cap \langle e_1\rangle$ is either empty, or the singleton $B\cap \langle e_1\rangle=\{\lambda_1 e_1\}$ (if it contained more than one element, then $B$ would be linearly dependent). The set $B'=B\setminus\{\lambda_1 e_1\}$ is disjoint from $\langle e_1 \rangle$, but it is still dense in $\mathcal{A}$ (otherwise $\{\lambda_1 e_1\}$ would be an isolated point of $\mathcal{A}$) and $T(b')\neq 0$ for every $b'\in B'$ as $\ker T = \langle e_1 \rangle$.