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 $A$?



Showing non-injectivity
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.


