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.

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.  
