Showing non-injectivity Let $X,Y$ be Banach spaces and let $A:X\rightarrow Y$ be a linear operator. Does it suffice to show that there exists a sequence $x_n\in X$ such that $\lim_{n\rightarrow\infty}Ax_n = 0$ with $||x_n||=1\quad\forall n\in\mathbb{N}$ to proof non-injectivity of $A$?
 A: There are examples of Banach spaces $X,Y$ along with bounded linear mappings $L:X\rightarrow Y$ and sequences $(x_{n})_{n}$ of elements in $X$ such that
$^{\lim}_{n\rightarrow\infty}L(x_{n})=0$ in the metric space induced by the norm on $Y$ but where $\|x_{n}\|=1$ for each $n$. For instance, if $X=Y$ and $X$ is a separable Hilbert space with orthonormal basis $(e_{n})_{n\geq 1}$, and
$A:X\rightarrow Y$ is the bounded linear operator defined by letting
$L(e_{n})=e_{n}/n$ for $n\geq 1$, then $\lim_{n\rightarrow\infty}L(e_{n})=0$ in the metric topology induced by the norm, but $\|e_{n}\|=1$ for each $n$.
The open mapping theorem for Banach spaces states that if $X,Y$ are Banach spaces and $L:X\rightarrow Y$ is a surjective bounded linear mapping, then the mapping $L$ is an open mapping. As a consequence, if $X,Y$ are Banach spaces and $L:X\rightarrow Y$ is a bijective linear continuous mapping, then $L$ is a homeomorphism (i.e. $L^{-1}$ is also continuous). As a consequence, if $L:X\rightarrow Y$ is a continuous linear surjection between Banach spaces with where $\lim_{n\rightarrow\infty}L(x_{n})=0$ with respect to metric generated by the norm but where $\|x_{n}\|=1$ for each $n$, then the mapping $L$ cannot be injective.
