# banach spaces – Showing non-injectivity

There are examples of Banach spaces $$X,Y$$ along with bounded linear mappings $$L:Xrightarrow Y$$ and sequences $$(x_{n})_{n}$$ of elements in $$X$$ such that
$$^{lim}_{nrightarrowinfty}L(x_{n})=0$$ 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 $$(e_{n})_{ngeq 1}$$, and
$$A:Xrightarrow Y$$ is the bounded linear operator defined by letting
$$L(e_{n})=e_{n}/n$$ for $$ngeq 1$$, then $$lim_{nrightarrowinfty}L(e_{n})=0$$ 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 $$L:Xrightarrow Y$$ is a surjective bounded linear mapping, then the mapping $$L$$ is an open mapping. As a consequence, if $$X,Y$$ are Banach spaces and $$L:Xrightarrow Y$$ is a bijective linear continuous mapping, then $$L$$ is a homeomorphism (i.e. $$L^{-1}$$ is also continuous). As a consequence, if $$L:Xrightarrow Y$$ is a continuous linear surjection between Banach spaces with where $$lim_{nrightarrowinfty}L(x_{n})=0$$ with respect to metric generated by the norm but where $$|x_{n}|=1$$ for each $$n$$, then the mapping $$L$$ cannot be injective.