Functional Analysis – Non-Injectivity of the Limit of Non-Injective Sequence of Operators

fa.functional-analysisoperator-theoryreal-analysis

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.

Best Answer

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.

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