It is known that the limit of a sequence of non injective operators is not necessarily non-injective, for instance, the operator \begin{eqnarray\*}
T\_n &:&\ell ^{2}\rightarrow \ell ^{2} \\
x &\longmapsto &(x\_{1},\frac{x\_{2}}{2},...,\frac{x\_{n}}{n},0,0,...).
\end{eqnarray\*}
does the job.
My question is the following: what supplementary assumption can guarantee the non-injectivity of the limit operators of a sequence of non-injective operators acting from $E \rightarrow F$ ($E$ and $F$ are Hilbert spaces).


Thank you in advance.



Non-injectivity of the limit of non-injective sequence of operators
Let $(T\_n:E\to F)$ be a sequence of noninjective operators such that $\|T\_n-T\|\to 0$ as $n\to\infty$ for some bounded linear $T:E\to F$. Since $\ker(T\_n)\neq \{0\}$, for each $n\in\mathbb{N}$, we can pick $u\_n\in E$, $\|u\_n\|=1$ such that $T\_nu\_n=0$. Consequently,
$$\|Tu\_n\|\leq \|Tu\_n-T\_nu\_n\| \leq \|T-T\_n\|\|u\_n\| \to 0,$$
that is $0\in\sigma(T)$ and $(u\_n)$ is an approximate eigenvector for $T$.


Second, the unit ball of $E$ is weakly compact since $E$ is reflexive. An approximate eigenvector $(u\_n)$ of $T$ of the form above lies in the unit sphere of $E$, thus $(u\_n)$ has weak limit points. There are 2 disjoint cases:


1. Either $u\_n\to 0$ weakly for all approximate eigenvectors of the form above,
2. or there exists $(u\_n)$ that possess a nonzero weak limit point, say $u\in E$.


In the second case, there exists a subsequence $(u\_{n\_k})$ such that $u\_{n\_k}\to u\neq 0$ weakly in $E$ by Eberlein-Šmulian theorem. Next, for each $f\in F^\*$,
$$|f(Tu)| 
\leq |f(Tu - Tu\_{n\_k})| + |f(Tu\_{n\_k} - T\_{n\_k}u\_{n\_k})| 
\leq |foT(u-u\_{n\_k})| + \|f\|\|T-T\_{n\_k}\| 
\to 0$$
as $k\to\infty$. Thus, $f(Tu) = 0$ for all $f\in F^\*$, so $Tu=0$. Consequently, $T$ is not injective.


**We can conlude:** if there exists $u\_n\in\ker(T\_n)$, $\|u\_n\|=1$ such that $(u\_n)$ has nonzero weak limit points, then $T$ is not injective.


