Is the extension of injective bounded linear transformation a injective linear transformation

Let $X$ and $Y$ be two Banach spaces over $\mathbb{C}$

Let $V$ be a dense subspace of $X$

Let $T : V \to Y$ be a bounded linear operator

We know that $T$ can be uniquely extended to a bounded linear transformation $S : X \to Y$ such that $\left\| T \right\| = \left\| S \right\|$

I would like to know if is it true that

$$T \text{ is injective } \Longrightarrow S \text{ is injective }$$

thanks.

• It should be worth to note that your desired claim is true, if $T$ is an isometry. In fact, if $\mathcal E_0$ is a seminormed vector space, $\mathcal E_0$ is a dense subset of $\mathcal E_0$, $\mathcal F$ is a complete seminormed vector space, $F:=\mathcal F/\left\{f\in\mathcal F:\left\|f\right\|_{\mathcal F}=0\right\}$ and $\iota:\mathcal E_0\to\mathcal F$ is Lipschitz continuous, then there is a unique Lipschitz continuous $\overline\iota:\mathcal E\to F$. If $\iota$ is an isometry, then $\overline\iota$ is an isometry. If $\iota$ is linear, then $\overline\iota$ is linear. Jan 27, 2018 at 18:20
• @0xbadf00d thanks for your useful comment Jan 27, 2018 at 19:25

No. Using a Hamel basis of an infinite dimensional Banach space $Y$ one find a discontinuous linear functional $f$ on $Y$. Then $\|y\|' = \|y\|+|f(y)|$ is a strictly finer norm on $Y$ and the identity $(Y,\|\cdot\|') \to (Y\|\cdot\|)$ extends to the completion $X$ of $(Y,\|\cdot\|')$. This extension isn't injective because otherwise its inverse would be continuous by the closed graph theorem. But its restriction to the dense subspace $Y$ is injective.